Determine the number of significant figures in each

DMM Measurement Typesand Common TerminologyOverviewLearn how to correctly use and understand a digital multimeter (DMM). Article topics include display digits, AC and DC voltage measurements, AC and DC current measurements, resistance measurements, and continuity and diode testing. This tutorial is part of the Instrument Fundamentals series.Contentsw Display Digitsww Voltage Measurementswa. Input Resistanceb. Crest Factorc. Null Offsetd. Auto Zerow Current Measurementsww Resistance Measurementsww Additional Measurementswa. Continuity Testingb. Diode Testingw Noise Rejection Parametersww SummarywDisplay DigitsDigital multimeters (DMMs) can be useful for a variety of measurements. When choosinga DMM or understanding one you are using, the first things to be aware of are the displaydigits of the instrument.It is important that a DMM has enough digits to be precise enough for your application. Thenumber of display digits on a DMM is not related to the resolution, but can help determinethe number of significant values that can be displayed and read. DMMs are said to have acertain number of digits, such as 3 ½ digits or 3 ¾ digits. A full digit represents a digit thathas 10 states, 0 to 9. A fractional digit is the ratio of the maximum value the digit can attainover the number of possible states. For example, a ½ digit has a maximum value of one andhas two possible states (0 or 1). A ¾ digit has a maximum value of 3 with four possiblestates (0, 1, 2, or 3).Equation 1. DMMs often have fractional digits, which can display only a limited number of states.The fractional digit is the first digit displayed, with the full digits displayed after. For instance,on the 2 V range, the maximum display for a 3 ½ digit DMM is 1.999 V.Typically, ½ digit displays have full scale voltages of 200 mV, 2 V, 20 V, and 200 V while ¾ digitdisplays have full scale voltages of 400 mV, 4 V, 20 V, and 400 V.Voltage MeasurementsPractically every DMM has a DC and an AC measurement function. Voltage testing iscommonly used to test and verify the outputs of instruments, components, or circuits.Voltage is always measured between two points, so two probes are needed. Some DMMconnectors and probes are colored; red is intended for the positive point that you want toactually take a measurement of and black is intended for the negative point that is typically areference or ground. However, voltage is bidirectional, so if you were to switch the positiveand negative points, the measured voltage would simply be inverted.There are usually two different modes for measuring voltage: AC and DC. Typically, DC isdenoted with a V with one dashed line and one solid line while AC is denoted with a V with awave. Be sure to select the correct range and mode for your application.Figure 1. AC voltage (left) and DC voltage (right) measurements are commonly used to test and verify outputs ofinstruments, components, or circuits.There are several terms and concepts you should be familiar with when measuringAC or DC voltage.a. Input ResistanceAn ideal voltmeter has an infinite input resistance so that the instrument does not draw anycurrent from the test circuit. However, in reality, there is always some resistance that affectsmeasurement accuracy. T o minimize this problem, a DMM’s voltage measurement subsystems are often designed to have impedances in the 1s to 10s of MΩ. If you are measuring lowvoltages, even this resistance can be enough to add unacceptable inaccuracies to yourmeasurement. For this reason, lower voltage ranges often have a higher impedance optionsuch as 10 GΩ.With some DMMs, you can select the input resistance. For most applications, it can be saidthe higher the impedance, the more accurate the measurement. However, there are a fewcases where you might choose the lower impedance. For instance, a conduit that has manydifferent wires inside might have coupling across the wires. Even though the wires are openand floating, the DMM still reads a voltage. The higher impedance isn’t sufficient to eliminatethese ghost voltages, but a low impedance provides a path for this built-up charge and allows the DMM to correctly measure 0 V. An example of this at a lower voltage range is if you hadtraces close together on a circuit.b. Crest FactorWhen measuring AC signals (voltage or current), the crest factor can be an importantparameter when determining accuracy for a specific waveform. The crest factor is the ratio of the peak value to the rms value and is a way to describe waveform shapes. T ypically, the crest factor is used for voltages, but can be used for other measurements such as current. It istechnically defined as a positive real number, but most often it is specified as a ratio.Equation 2. The crest factor is a measure of how extreme the peaks are in a waveform.A constant waveform with no peaks has a crest factor of 1 because the peak value and therms value of the waveform are the same. For a triangle waveform, it has a crest factor of1.732. Higher crest factors indicate sharper peaks and make it more difficult to get anaccurate AC measurement.Figure 2. The crest factor of an AC signal can affect the accuracy.An AC multimeter that measures using true rms specifies the accuracy based on a sinewave. It indicates, through the crest factor, how much distortion a sine wave can have andstill be measured within the stated accuracy. It also includes any additional accuracy error for other waveforms, depending on their crest factor.For example, if a given DMM has an AC accuracy of 0.03 percent of the reading. Y ou aremeasuring a triangle waveform, so you need to look up any additional error with a crest factor of 1.732. The DMM specifies that for crest factors between 1 and 2, there is additional error of 0.05 percent of the reading. Y our measurement then has an accuracy of 0.03 percent +0.05 percent for a total of 0.08 percent of the reading. As you can see, the crest factor of awaveform can have a large affect on the accuracy of the measurement.c. Null OffsetMost DMMs offer the ability to do a null offset. This is useful for eliminating errors caused byconnections and wires when making a DC voltage or resistance measurement. First, youselect the correct measurement type and range. Then connect your probes together and wait for a measurement to read. Then select the null offset button. Subsequent readings subtractthe null measurement to provide a more accurate reading.d. Auto ZeroIn addition to performing a null offset, another way to improve voltage and resistancemeasurement accuracy is by enabling a feature called auto zero. Auto zero is used tocompensate for internal instrument offsets. When the feature is enabled, the DMM makes anadditional measurement for every measurement you take. This additional measurement istaken between the DMM input and its ground. This value is then subtracted from themeasurement taken, thus subtracting any offsets in the measurement path or ADC. Although itcan be very helpful in improving the accuracy of the measurement, auto zero can increase thetime it takes to perform a measurement.Current MeasurementsAnother common measurement function is DC and AC current measurements. Althoughvoltage is measured in parallel with the circuit, current is measured in series with the circuit.This means that you need to break the circuit—physically interrupt the flow of current—inorder to insert the DMM into the circuit loop to take an accurate measurement. Similar tovoltage, current is bidirectional. The notation is similar as well, but with an A symbol insteadof a V. The A stands for amperes, the unit of measure for current. Be sure to select thecorrect range and mode for your application.Figure 3. DC current (left) and AC current (right) measurements are helpful for troubleshootingcircuits or components.DMMs have a small resistance at the input terminals, and it measures the voltage. It thenuses Ohm’s law to calculate the current. The current is equal to the voltage divided by theresistance. To protect your multimeter, avoid switching out of the current measurementfunction when currents are flowing through the circuit. Y ou should also be careful not toaccidentally measure voltage while in the current measurement mode; this can cause thefuse to blow. If you do accidentally blow the fuse, you can often replace it. See yourinstrument’s instruction manual for detailed information.Resistance MeasurementsResistance measurements are commonly used to measure resistors or other componentssuch as sensors or speakers. Resistance measuring works by applying a known DC voltageover an unknown resistance in series with a small internal resistance. It measures the testvoltage, then calculates the unknown resistance. Because of this, test the device only whenit isn’t powered; otherwise, there is already voltage in the circuit and you can get incorrectreadings. Also keep in mind that a component should be measured before it is inserted intothe circuit; otherwise, you are measuring the resistance of everything connected to thecomponent instead of just the component by itself.One of the nice things about resistance is that it is nondirectional, meaning if you switch theprobes the reading is still the same. The symbol for a resistance measurement is an Ω, whichrepresents the resistance unit of measure. Be sure to select the correct range and mode foryour application. If the display reads OL, this means the reading is over the limit or greaterthan the meter can measure in that range. As discussed earlier, using the null offset canimprove your measurement readings.Figure 4. Resistance measurements are commonly used to measure resistors or other components. Additional MeasurementsMany DMMs offer two additional measurement functions: diode testing and continuity testing.a. Continuity TestingContinuity testing helps you identify when two points are electrically connected. This can bevery helpful when troubleshooting wire breaks, printed circuit board (PCB) traces, or solderjoints. When testing for continuity, it is essential to monitor exactly where the probes aretouching. As such, most DMMs emit a sound when they detect a closed circuit, so you don’thave to look up from your probes. As such, the symbol for continuity looks like a sound wave.Figure 5. Continuity testing helps you identify when two points are electrically connected.Continuity testing works just like a resistance measurement; as such, it is essential that your device not be powered when you are testing. It can also be helpful to make sure everything is connected first by brushing the test tips together to verify the beep. If you don’t hear asound, then check that the probes are firmly connected, your DMM has sufficient battery life, and that you are in the correct mode. Y ou should also look in your user manual to determine the level of resistance required to trigger the sound as it varies from model to model.If you are testing a circuit that has a large capacitor, you may hear a quick beep and thensilence. This is because the voltage the DMM is applying to the circuit is charging up thecapacitor and, during that time, the DMM thinks it is a closed circuit when it isn’t really. b. Diode TestingDiode testing displays the forward voltage drop of the diode in volts. The symbol, notsurprisingly, is the diode symbol.Figure 6. Diode testing displays the forward voltage drop of the diode in volts.The DMM forces a small current through the diode and measures the voltage drop between the two test leads. When measuring a diode, you want the positive probe on the anode side and the negative on the cathode side. The voltage reading typically is about 0.7 V for silicon but can range from 0.5 to 0.9 V and still be a working diode. Germanium diodes are typically around 0.3 V.Figure 7. T ypically, test a diode with the positive probe on the anode side and the negative on thecathode side. However, switching them can also be illuminating.Next, switch probes so the negative is on the anode side and the positive is on the cathodeside. If the diode is working properly, the multimeter should show that there is an opencircuit indicated by OL.If a diode is defective, it can defect to be either a short or an open diode. If the diode hasfailed to open, the DMM shows OL in both the forward and reverse bias because the currentflowing through is zero and is an equivalent to an open circuit. If the diode is shorted, theDMM indicates 0 V as there is no voltage drop across the diode.Noise Rejection ParametersIt is always important to consider noise when taking a measurement. There are twoadditional parameters you should be familiar with to better understand your instrument andthe associated noise of the measurement.The normal-mode rejection ration (NMRR) describes the DMM’s ability to reject noise thatappears between the two input terminals or, in other words, the noise mixed in with themeasured signal. Most of this noise is a line frequency and its harmonics. NMRR, whichis often used to indicate the capability of the instrument to reject a power line noise of 50 or60 Hz, is valid only at the specified frequency and is useful when making DC measurements.Normal-mode noise can also be reduced through the use of shielding or filtering.The common-mode rejection ratio (CMRR) describes the DMM’s ability to reject noise that iscommon to both input terminals, such as from a noisy environment. Common-mode noise isusually less severe than normal-mode noise.NMRR and CMRR are typically specified at 50 Hz and 60 Hz, and CMRR is often specified ata DC value as well. Typical values are greater than 80 dB and 120 dB, respectively.SummaryThe number of display digits on a DMM is not related to the resolution, but can helpdetermine the number of significant values that can be displayed and read.For most applications, it can be said the higher the impedance, the more accurate thevoltage measurement.Higher crest factors indicate sharper peaks and make it more difficult to get an accurateAC measurement.Null offset can be used to eliminate errors caused by connections and wires whenmaking a DC voltage or resistance measurement.Auto zero is used to compensate for internal instrument offsets.Current measurements require you to break the circuit in order to insert the DMMinto the circuit loop.Accidentally measuring voltage while in the current mode can cause a fuse to blow.Resistance measurements and continuity testing should be taken when the circuitdoes not have power.The normal-mode rejection ratio (NMRR) describes the DMM’s ability to reject noisethat appears between the two input terminals.The common-mode rejection ratio (CMRR) describes the DMM’s ability to reject noisethat is common to both input terminals, such as from a noisy environment.。

a considerable number of the most
significant
“a considerable number of the most significant”的意思是“相当数量的最重要的”，这是一个比较简洁的表达方式，可以用在各种上下文中，例如：
1. 在学术研究中：“a considerable number of the most significant studies”（相当数量的最重要的研究）。

2. 在商业领域：“a considerable number of the most significant factors”（相当数量的最重要的因素）。

3. 在历史讨论中：“a considerable number of the most significant events”（相当数量的最重要的事件）。

4. 在艺术或文学评论中：“a considerable number of the most significant works”（相当数量的最重要的作品）。

请注意，具体的使用方式和上下文会影响这个短语的精确含义，但它通常用于强调某个领域或主题中一些非常重要或关键的元素、研究、事件、作品等。

Four Steps for Solving Quantitative ProblemsMaybe you have noticed that the sample problems in the first three chapters of this book are solved in a four-step process. The steps in the process are as follows.1.ANAL YZE2.PLANPUTE4.EVALUATENow it’s time to examine each step to learn how this process can help you solve more-complex problems, like those you will encounter in your chemistry course.1.ANAL YZEIn this first step, you should read the problem carefully and then reread it. You must determine what specific information and data you are given in the problem and what you need to find.Try to visualize the situation the problem describes. Look closely at the words in the problem statement for clues to understanding the problem. Are you work-ing with elements, compounds, or mixtures? Are they solids, liquids, or gases? What change is taking place? Is it a chemical change? What are the reactants? What are the products?It is always a good idea to collect and organize all of your information in a table, where you can see it at a glance. Include the things you want to find in the table, too. Be sure to include units with both the data and the quantities you must find. Scanning the quantities and their units will often provide clues about how to set up the problem. Remember, you may be given information not needed to solve the problem. You must analyze the data to determine what is useful and what is not.2.PLANIn the planning step, you develop a method to solve the problem. Always keep in mind what you want to find and its units. Chances are good that an approach that gives an answer with the correct units is the correct one to use. In any case, a setup that gives an answer with the wrong units is certain to be wrong. You may find it helpful to diagram your solution method. This process helps you organize your thoughts.As you work out your problem-solving method, write down a trial calculation without numbers but with units. When you complete your setup, see if the trial calculation will give you a quantity with the correct units. As stated above, if the setup gives the needed units, it is probably correct.During this planning process you may discover that you need more informa-tion, such as atomic masses from the periodic table, the boiling point of alcohol,or the density of tin. You will need to look up such information in the appropriate tables.PUTEIn this step, you follow your plan, set up a calculation using the data you have assembled, and compute the result.It is a good strategy to write out and check your calculation setup before you start working with your calculator. First reconfirm that the calculation will give a result with the correct units. Go through your calculation and lightly strikethrough the units that cancel. Be sure the remaining units are those that you want in your answer. Whenever possible, use your calculator in a way that lets you complete the entire problem without writing down numbers and then re-entering them.4.EVALUATEEveryone makes errors, but good problem solvers always develop strategies to check their work. Confidence in your problem-solving ability will come from knowing how to determine on your own whether your answers are correct.One evaluation strategy is to estimate the numerical value of the answer. In simple problems, you can probably do this in your head. With calculations having several terms, round off each numerical value to the nearest simple value, and then write and compute the estimation.Suppose you had to make the following calculation.ϭ?The numerical calculation can be estimated as follows.ᎏ6ϫ3020ᎏϭᎏ13200ᎏϭᎏ14ᎏϭ0.25Once you have done the actual computation, compare your result with theestimate. In this case, the calculation gives 0.242 g/cm 2, which is close to the esti-mated result. Therefore, it is likely that you made no mistakes in the calculation.Next, check that your answer is expressed to the correct number of significant figures. Look at the data values you used in the calculation. Usually, significant figures will be limited by the measurement that has the fewest significant figures. Finally, ask yourself the simple question, does this answer make sense based on what you know? If you are calculating the circumference of Earth, an answer of 50 km is obviously much too small. If you are calculating the density of air, a value of 340 g/cm 3is much too large because air density is usually less than 1g/cm 3. You will detect many errors by asking if the answer makes sense.28.8 g ᎏᎏᎏ6.30 cm ϫ18.9 cmSample Problem 1A 10.0% sodium hydroxide solution has a density of 1.11 g/mL. What volume in liters will 2280 g of the solution have?SolutionANAL YZEWhat is given in the problem?the density of the sodium hydroxide solution ing/mL, and the mass of the solution whose vol-ume is to be determinedWhat are you asked to find?the volume of the specified mass of solution Next, bring together in a table everything you might need in the problem. Include what you want to find in the table. The fact that the solution is 10.0% sodiumhydroxide is unimportant in the solution of this problem. This piece of data need not appear in your table. Notice that the problem asks you for a volume in liters and that density is given in grams per milliliter. You will need a factor to convert between these units. Therefore, include in the table any relationships between units that will be helpful.PLANWhat steps are needed to calculate the volume of 2280 g of the solution?Apply the relationship D ؍ᎏm V ᎏ. Rearrange to solve for V,substitute data, and convert to liters.Solve the density equation for V .V ϭᎏm Dᎏϭvolume in mL mass of solution in g ᎏᎏᎏdensity of solution in g/mlItemsData Density of solution1.11 g/mL Mass of solution2280 g Volume of solution? L Relationship between mL and L 1000 mL ϭ1 LTo change the result to liters, multiply by the conversion factor.ϫϭvolume in L COMPUTEEVALUATEAre the units correct?Yes; units canceled to give liters.Is the number of significant figures correct?Yes; the number of significant figures is correct because data were given to three significant figures.Is the answer reasonable?Yes; the calculation can be approximated as 2000/1000 ؍2. Also, considering that 2000 g of water occupies 2 L, you would expect 2 L of a slightly more dense material to have a mass slightly greater than 2000 g.Practice1.Gasoline has a density of 0.73 g/cm 3. How many liters of gasoline would be required to increase the mass of an automobile from 1271 kg to 1305 kg? ans: 47 L2.A swimming pool measures 9.0 m long by3.5 m wide by 1.75 m deep. What mass of water in metric tons (1 metric ton ϭ1000 kg) does the pool contain when filled? The density of the water in the pool is 0.997 g/cm 3.ans: 55 metric tons3.A tightly packed box of crackers contains 250 g of crackers and measures 7.0cm ϫ17.0 cm ϫ19.0 cm. What is the average density in kilograms per liter of the crackers in the package? Assume that the unused volume is negligible.ans: 0.11 kg/L ϫ2280 g 1.11 g/mL 1 L 1000 mLϭ 2.05 L 1 L ᎏ1000 mLmass of solution in g ᎏᎏᎏdensity of solution g/mlAdditional ProblemsSolve these problems by using the Four Steps for Solving Quantitative Problems. 1.The aluminum foil on a certain roll has a total area of 18.5 m2and a mass of1275 g. Using a density of 2.7 g per cubic centimeter for aluminum, determine the thickness in millimeters of the aluminum foil.2.If a liquid has a density of 1.17 g/cm3, how many liters of the liquid have amass of 3.75 kg?3.A stack of 500 sheets of paper measuring 28 cm ϫ21 cm is 44.5 mm high andhas a mass of 2090 g. What is the density of the paper in grams per cubiccentimeter?4.A triangular-shaped piece of a metal has a mass of 6.58 g. The triangle is 0.560mm thick and measures 36.4 mm on the base and 30.1 mm in height. What is the density of the metal in grams per cubic centimeter?5.A packing crate measures 0.40 m ϫ0.40 m ϫ0.25 m. You must fill the cratewith boxes of cookies that each measure 22.0 cm ϫ12.0 cm ϫ5.0 cm. How many boxes of cookies can fit into the crate?6.Calculate the unknown quantities in the following table. Use the followingrelationships for volumes of the various shapes.Volume of a cube ϭlϫlϫlVolume of a rectangle ϭlϫwϫhVolume of a sphere ϭ4/3␲r3Volume of a cylinder ϭ␲r2ϫhD m V Shape Dimensionsa. 2.27 g/cm3 3.93 kg? L cube? m ϫ? m ϫ? mb. 1.85 g/cm3? g? cm3rectangle33 mm ϫ21 mm ϫ7.2 mmc. 3.21 g/L? kg? dm3sphere 3.30 m diameterd.? g/cm3497 g? m3cylinder7.5 cm diameter ϫ12 cme.0.92 g/cm3? kg? cm3rectangle 3.5 m ϫ1.2 m ϫ0.65 m7.When a sample of a metal alloy that has a mass of 9.65 g is placed into agraduated cylinder containing water, the volume reading in the cylinderincreases from 16.0 mL to 19.5 mL. What is the density of the alloy sample in grams per cubic centimeter?8.Pure gold can be made into extremely thin sheets called gold leaf. Supposethat 50.0 kg of gold is made into gold leaf having an area of 3620 m2. Thedensity of gold is 19.3 g/cm3.a.How thick in micrometers is the gold leaf?b.A gold atom has a radius of 1.44 ϫ10Ϫ10m. How many atoms thick is thegold leaf?9.A chemical plant process requires that a cylindrical reaction tank be filledwith a certain liquid in 238 s. The tank is 1.2 m in diameter and 4.6 m high.What flow rate in liters per minute is required to fill the reaction tank in the specified time?10.The radioactive decay of 2.8 g of plutonium-238 generates 1.0 joule of energyas heat every second. Plutonium has a density of 19.86 g/cm3. How manycalories (1 cal ϭ4.184 J) of energy as heat will a rectangular piece of pluto-nium that is 4.5 cm ϫ3.05 cm ϫ15 cm generate per hour?11.The mass of Earth is 5.974 ϫ1024kg. Assume that Earth is a sphere of diame-ter 1.28 ϫ104km and calculate the average density of Earth in grams per cubic centimeter.12.What volume of magnesium in cubic centimeters would have the same massas 1.82 dm3of platinum? The density of magnesium is 1.74 g/cm3, and the density of platinum is 21.45 g/cm3.13.A roll of transparent tape has 66 m of tape on it. If an average of 5.0 cm oftape is needed each time the tape is used, how many uses can you get from a case of tape containing 24 rolls?14.An automobile can travel 38 km on 4.0 L of gasoline. If the automobile isdriven 75% of the days in a year and the average distance traveled each day is86 km, how many liters of gasoline will be consumed in one year (assume theyear has 365 days)?15.A hose delivers water to a swimming pool that measures 9.0 m long by 3.5 mwide by 1.75 m deep. It requires 97 h to fill the pool. At what rate in liters per minute will the hose fill the pool?16.Automobile batteries are filled with a solution of sulfuric acid, which has adensity of 1.285 g/cm3. The solution used to fill the battery is 38% (by mass) sulfuric acid. How many grams of sulfuric acid are present in 500 mL ofbattery acid?。

《英语测试》题库及答案《英语测试》题库及答案1．dentify the scales represented by the following kinds of data.2．iven the following raw score distribution on a spelling test of 25 items, calculate (a) ordinal ratings, (b) percentage scores, (c) percentile ranks: 6, 8, 9, 11, 12, 12, 14, 15, 19, 19, 21, 25. 3．If you administered a multiple-choice test of phrasal verb usage to a class of ten students and got the following distribution of scores: 11, 13, 16, 16, 18, 20, 20, 25, 27, 29, what would be the corresponding z-scores, T-scores, and normal distribution area proportions if the standard deviation is 5.95?4．In a normal distribution, what proportion of the scores lie between plus and minus one standard deviation? Between plus and minus two standard deviations?5．Compute item difficulty and the proportion incorrect for each of the items in the following scoring matrix. Identify items for rejection.Items1 2 3 4 51 1 0 1 1 12 1 1 0 0 03 1 1 0 0 1Examinees 4 1 0 1 0 15 0 1 0 0 06 1 1 1 1 07 1 1 0 0 18 0 1 0 1 09 1 1 0 0 110 0 1 1 0 16．Look at the following paired scores on listening comprehension (X) and general proficiency (Y) for ten students.X Y25 1813 2115 2219 2717 3116 2221 2924 3322 2621 29Compute the correlation coefficient for these scores. Please do a one-tailed Pearson test to determine whether it is significant on the level of significance of <.05?7．Nine students have taken two different aptitude tests and got the following test result. Compute the Pearson product moment correlation coefficient. And determine whether it is significant on the level of significance of <.05?Student No. 01 02 03 04 05 06 07 08 09Test A 56 60 61 62 65 67 71 71 74Test B 34 26 62 30 30 28 34 36 408．Compute the point biserial correlation coefficient for each of the five items with total score. 9．What do we mean by correlation in statistics?10．Please state the difference between the classroom tests and external tests.11．Make a simple clarification on the Classical True Score Model and enumerate its demerits. 12．Please write five multiple-choice items on some grammatical elements (tense, agreement, etc.).TABLE B z-Scores Corresponding to Cumulative Area Proportions of the Normal Distribution z Area z Area z Area z Area -4.0 .0000 -1.9 .0287 0.1 .5398 2.1 .9821 -3.9 .0001 - 1.8 .0359 0.2 .5793 2.2 .9861 -3.8 .0001 - 1.7 .0446 0.3 .6179 2.3 .9893 -3.7 .0001 - 1.6 .0548 0.4 .6554 2.4 .9918 - 3.6 .0002 - 1.5 .0668 0.5 .6915 2.5 .9938 -3.5 .0002 - 1.4 .0808 0.6 .7257 2.6 .9953 -3.4 .0003 - 1.3 .0968 0.7 .7580 2.7 .9965 -3.3 .0005 -1.2 .1151 0.8 .7881 2.8 .9974 -3.2 .0007 - 1.1 .1357 0.9 .8159 2.9 .9981 - 3.1 .0010 - 1.0 .1587 1.0 .8413 3.0 .9987 - 3.0 .0013 - 0.9 .1841 1.1 .8643 3.1 .9990 -2.9 .0019 - 0.8 .2119 1.2 .8849 3.2 .9993 -2.8 .0026 -0.7 .2420 1.3 .9032 3.3 .9995 - 2.7 .0035 - 0.6 .2743 1.4 .9192 3.4 .9997 -2.6 .0047 - 0.5 .3085 1.5 .9332 3.5 .9998 -2.5 .0062 - 0.4 .3446 1.6 .9452 3.6 .9998 -2.4 .0082 -0.3 .3821 1.7 .9554 3.7 .9999 - 2.3 .0107 - 0.2 .4207 1.8 .9641 3.8 .9999 - 2.2 .0139 - 0.1 .4602 1.9 .9713 3.9 .9999 -2.1 .0179 0.0 .50002.0.97724.01.0000-2.0.0228For values between the listed points, you may interpolate according to the following example: Problem: Find the area proportion for a z-score of 1.473. Solution: z interval area interval observed z interval 1.5 .9332 1.473 - 1.4 -.9192 - 1.4.1.0146.073.073 is to .1 as X is to .01400140.1.073.X=,0102.1.0140.073.=?=X .0102 + .9192 = .9294, corresponds to z-score of 1.473TABLE E Critical Values of the Pearson Product-Moment Correlation Coefficient*Level of significance for one-tailed test.05 .025 .01 .005 .0005Level of significance for two-tailed testDf=N - 2 .10 .05 .02 .01 .0011 .9877 .9969 .9995 .9999 1.00002 .9000 .9500 .9800 .9900 .99903 .8054 .8783 .9343 .9587 .99124 .7293 .8114 .8822 .9172 .97415.6694 .7545 .8329 .8745 .95076 .6215 .7067 .7887 .8343 .92497 .5822 .6664 .7498 .7977 .89828 .5494 .6319 .7155 .7646 .87219 .5214 .6021 .6851 .7348 .847110 .4973 .5760 .6581 .7079 .823311 .4762 .5529 .6339 .6835 .801012 .4575 .5324 .6120 .6614 .780013 .4409 .5139 .5923 .6411 .760314 .4259 .4973 .5742 .6226 .742015.4124 .4821 .5577 .6055.724616 .4000 .4683 .5425 .5897 .708417 .3887 .4555.5285 .5751 .693218 .3783 .4438 .5155 .5614 .678719 .3687 .4329 .5034 .5487 .665220 .3598 .4227 .4921 .5368 .652425 .3233 .3809 .4451 .4869 .597430 .2960 .3494 .4093 .4487 .554135 .2746 .3246 .3810 .4182 .518940 .2573 .3044 .3578 .3932 .489645 .2428 .2875 .3384 .3721 .464850.2306 .2732 .3218 .3541 .443360 .2108 .2500 .2948 .3248 .407870 .1954 .2319 .2737 .3017 .379980 .1829 .2172 .2565 .2830 .356890 .1726 .2050 .2422 .2673 .3375100 .1638 .1946 .2301 .2540 .321113．What is the difference between criterion referenced test and norm referenced test?14．What is computerized adaptive testing?15．What is item difficulty? How to compute item difficulty?16．How are item difficulties, item discriminability represented in item response theory? 17．What general relationship exists between test reliability and the number of items on the test? 18．What does it mean if we say a correlation coefficient is significant at the p <.05 level? 19．What are the sources of measurement errors in language test?20．What is the meaning of measures of dispersion?21．List five common problems at the item writing stage. Provide one example of each problem type and correct the item if necessary.22．What is the essential difference between scatterplots of low-correlated variables and those of high-correlated variables? 23．Please compute the correlation coefficient of the following item and determine whether it is24．In a general proficiency test, mean=95.120, standard deviation=10.858, please compute the z-score, T-score, normal distribution area proportions of the following scores: 120, 108, 100, 95, 83, 71, 61.《英语测试》作业参考答案1．Identify the scales represented by the following kinds of data:Solution:Raw scores from a listening comprehension test Ordinal scaleAdjectives on a word-association test Nominal scalePercentile scores from a spelling test Ordinal scaleSpeed of note-taking in words per minute Ratio scaleI.Q.-equivalent scores on a vocabulary test Interval scalez-scores on the TOEFL Interval scaleThe number of instrumentally motivated students in class Nominal scale2．Given the following raw score distribution on a spelling test of 25 items, calculate (a) ordinal ratings, (b) percentage scores, (c) percentile ranks: 6, 8, 9, 11, 12, 12, 14, 15, 19, 19, 21, 25.Solution:Raw scores Ordinal ratings Percentage Scores Percentile Ranks25 1st 100 95.821 2nd 84 87.519 3rd 76 75.019 4th 76 75.015 5th 60 62.514 6th 56 54.212 7th 48 41.712 8th 48 41.711 9th 44 29.29 10th 36 20.88 11th 32 12.56 12th 24 4.23．If you administered a multiple-choice test of phrasal verb usage to a class of ten students and got the following distribution of scores: 11, 13, 16, 16, 18, 20, 20, 25, 27, 29, what would be the corresponding z-scores, T-scores, and normal distribution area proportions if the standard deviation is 5.95?Solution:Rawscores z-scores T-scores Normal distribution area proportions29 1.597 65.97 94.4927 1.261 62.61 89.6325 0.925 59.25 82.2420 0.084 50.84 53.3520 0.084 50.84 53.3518 -0.252 47.48 40.0516 -0.588 44.12 27.8116 -0.588 44.12 27.8113 -1.093 39.07 13.7311 -1.429 35.71 7.654In a normal distribution, what proportion of the scores lie between plus and minus one standard deviation? Between plus and minus two standard deviations?Solution:In a normal distribution, 68.3% of scores lie between plus and minus one standard deviation;95.5% of scores lie between plus and minus two standard deviation.5．Compute item difficulty and the proportion incorrect for each of the items in the following scoring matrix. Identify items for rejection.Items1 2 3 4 51 1 0 1 1 12 1 1 0 0 03 1 1 0 0 1Examinees4 1 0 1 0 15 0 1 0 0 06 1 1 1 1 07 1 1 0 0 18 0 1 0 1 09 1 1 0 0 110 0 1 1 0 1Solution:Item 1 2 3 4 5Difficulty 0.7 0.8 0.4 0.3 0.6Proportion incorrect 0.3 0.2 0.6 0.7 0.4Items 2 and 4 can be rejected, because it seems that item 2 is too easy and item 4 is too difficult.6．Look at the following paired scores on listening comprehension (X) and general proficiency (Y) for ten students:X Y25 1813 2115 2219 2717 3116 2221 2924 3322 2621 29Compute the correlation coefficient for these scores. Please do a one-tailed Pearson test to determine whether it is significant on the level of significance of <.05?Solution:r=0.273.The critical value for Pearson’ r of the s ignificance of <.05, one tailed test when the degree of freedom equals to 8 is0.5494. Since r=0.273<0.5494, it is not significant.7．Nine students have taken two different aptitude tests and got the following test result. Compute the Pearson product moment correlation coefficient. And determine whether it is significant on the level of significance of <.05?Student No. 01 02 03 04 05 06 07 08 09Test A 56 60 61 62 65 67 71 71 74Test B 34 26 62 30 30 28 34 36 40Solution:r=-0.0176.Critical value is 0.6664. And r=-0.0176<0.6664, so it is not statistically significant.8．Look at the following scoring matrix for twenty students:Compute the point biserial correlation coefficient for each of the five items with total score.Solution:Item One: 0.639Item Two: 0.112Item Three: 0.422Item Four: 0.512Item Five: 0.1299．What do we mean by correlation in statistics?Solution:Correlation is a statistical relation between two or more variables such that systematic changes in the value of one variable are accompanied by systematic changes in the other. Generally, correlation does not suggest cause-effect relationship. 10．Please state the difference between the classroom tests and external tests.Solution:The aim of the classroom test is different from that of the external examination. External examinations are generally concerned with evaluation for the purpose of selection, the classroom test is concerned with evaluation for thepurpose of enabling teachers to increase their own effectiveness by making adjustments in their teaching to enable certain groups of students or individuals in the class to benefit more.11．Make a simple clarification on the Classical True Score Model and enumerate its demerits.Solution:The classical theory is the earliest theory of measurement. The classical theory is also referred to as the classical reliability theory because its major task is to estimate the reliability of the observed scores of a test. That is, it attempts to estimate the strength of the relationship between the observed score and the true score. It is also sometimes referred to as the true score theory because its theoretical derivations are based on a mathematical model known as the true score model. In a testing situation, we first devise a scaling rule to change a set of responses into a numerical observed score. Next we infer that the observed score faithfully reflects a true score. Finally, we infer that the true score truthfully reflects the quantity of a construct, which may or may not exit.Classical True Score Model has the following demerits: 1) the values of commonly used item statistics in test development such as item difficulty and item discrimination depend on the particular examinee samples in which they are obtained; 2) comparisons of examinees on an ability measured by a set of test items comprising a test are limited to situations in which examinees are administered the same (or parallel) test items; 3) one of the fundamental concepts, test reliability, is defined in terms of parallel forms; 4) it provides no basis for determining how an examinee might perform when confronted with a test item; 5) it presumes that the variance of errors of measurement is the same for all examinees.12．Please write five multiple-choice items on some grammatical elements (tense, agreement, etc.).Omitted.13．What is the difference between criterion referenced test and norm referenced test?Solution:Characteristically criterion-referenced tests are devised before the instruction itself is designed. The test must match teaching objectives perfectly, so that any tendency of the teacher to “teach to the test” would be permissible in that attaining objectives would thereby be assured. A criterion or cut-off score is set in advance (usually 80 to 90 percent of the total possible score), and those who do not meet the criterion are required to repeat the course.Students are not evaluated by comparison with the achievement of other students, but instead their achievement is measured with respect to the degree of their learning or mastery of the prespecified content domain.Norm-referenced or standardized tests are quite different from criterion-referenced tests in a number of respects; although, once again, some of the identical items may be used under certain conditions. By definition, a norm-referenced test must have been previously administered to a large sample of people from the target population(e.g., 1,000 or more). Acceptable standards of achievement can only be determined after the test has beendeveloped and administered. Such standards are found by reference to the mean or average score of other students from the same population.14．What is computerized adaptive testing?Solution:Adaptive testing, or computer adaptive testing, is a process of test administration in which test items are selected on the basis of the examinee’s response to previously administered items. Most commonly, such an approach would, for an examinee who experiences success with a given item, result in the purposeful presentation of an item of greaterWith each response, the computer makes a revised estimate of the examinee's ability; each revived estimate becomes more reliable. The test is terminated when the estimate reaches a specified level of reliability. We can say that computerized adaptive testing takes as its theoretical basis the Item Response Theory (IRT) or the latent trait theory, which makes the general assumption that examinee performance on a pool of test items is assumed to depend on a limited number of psychological traits called abilities. 15．What is item difficulty? How to compute item difficulty? Solution:Item difficulty is determined as the proportion of correct responses, signified by the letter “p”. The formula for item difficulty then isNCp r∑=where, p = difficulty, proportion correct,∑C r = the sum of correct responses, N = the number of examinees.16．How are item difficulty, item discriminability represented in item response theory? Solution:The idea of item difficulty as a location index will be examined first. In Figure 1, three item characteristic curves are presented on the same graph. All have the same level of discrimination but differ with respect to difficulty. The lefthand curve represents an easy item because the probability of correct response is high for low-ability examinees and approaches 1 for high-ability examinees. The center curve represents an item of medium difficulty because the probability of correct response is low at the lowest ability levels, around .5 in the middle of the ability scale and near 1 at the highest ability levels. The righthand curve represents a hard item. The probability of correct response is low for most of the ability scale and increases only when the higher ability levels are reached. Even at the highest ability level shown (+3), the probability of correct response is only .8 for the most difficult item.The concept of discrimination is illustrated in Figure 2. This figure contains three item characteristic curves having the same difficulty level but differing with respect to discrimination. The upper curve has a high level of discrimination since the curve is quite steep in the middle where the probability of correct response changes veryrapidly as ability increases. Just a short distance to the left of the middle of the curve, the probability of correct response is much less than .5, and a short distance to the right the probability is much greater than .5. The middle curve represents an item with a moderate level of discrimination. The slope of this curve is much less than the previous curve and the probability of correct response changes less dramatically than the previous curve as the ability level increases. However, the probability of correct response is near zero for the lowest-ability examinees and near 1 for the highest-ability examinees. The third curve represents an item with low discrimination. The curve has a very small slope and the probability of correct response changes slowly over the full range of abilities shown. Even at low ability levels, the probability of correct response is reasonably large, and it increases only slightly when high ability levels are reached.17．What general relationship exists between test reliability and the number of items on the test?Solution:Reliability is affected by the number of items in the test. We can readily understand how it happens that with more items in the test a greater range of scores is possible, and thus examinees are more widely dispersed along the scoring continuum. In this way it can be said that we have greater person separability and less likelihood that examinees would change rank order on repeated administrations of the test.18．What does it mean if we say a correlation coefficient is significant at the p <.05 level?Solution:“P < 0.05” means that based on the test, there is less than a 5% chance that we are wrong in rejecting the null hypothesis. And we are generally safe to say that the two variables under test are correlated.19．What are the sources of measurement errors in language test?Solution:Usually when we administer and score a test, some error of measurement is present. A variety of kinds of measurement error can introduce fluctuations in observed scores and thus reduce reliability. These different kinds of measurement error can be caused from the test-takers’ side, the scorers’ side and the administration of the test. A variety of changes may take place within the test-taker that either will introduce error on repeated administrations or will change the test-taker’s true s core from time one to time two. If true scores change, correlations between repeated sets of observed scores will go down, causing us to underestimate the reliability of the test. Temporary changes in the examinee may introduce measurement error. Influences such as fatigue, sickness, emotional disturbance, and practice effect may cause the examinee's observed score to temporarily deviate from his or her true score, or that score which reflects his or her actual ability. Subjectivity in scoring or mechanical errors in the scoring process may introduce inconsistencies in scores and produce unreliable measurement. These kinds of inconsistencies usually occur within or between the raters themselves. Inconsistencies in the administrative process may introducemeasurement error and thus reduce test reliability. This is most observable in situations where the test is administered to different groups in different locations or on different days, but may also appear within one administration in one location. 20．What is the meaning of measures of dispersion?Solution:Measures of dispersion describe how the data varies or is dispersed (spread out). The two most commonly used measures of dispersion are the range and the standard deviation. Rather than showing how data are similar, they show how data differs (its variation, spread, or dispersion).21．List five common problems at the item writing stage. Provide one example of each problem type and correct the item if necessary.Solution:1. Mixed ResponseItems are sometimes prepared to test a specific skill or competence, but close examination of the response options reveals that the options actually measure ability in skill areas other than those intended. For example: John flowers to the party last night.a) carries C) liftsb) carried d) liftedIt should be better if we have,a) carries C) is carryingb) carried d) has carried2. Length CuesFrequently the longest and most explicit option available is the obvious answer. Consider the following example:In the story, the merchant was unhappy because ita) rained. c) was windy.b) was dark. d) was windy and rainy and he had forgotten his raincoat.3. Nonsense DistractorsFor most purposes, nonsense distractors are to be avoided. Nonsense options have two basic problems. First, they tend to be weak distractors. Second, they frequently have negative “wash-back” on instruction; i.e., the students may learn errors from the examination itself. Consider the following example:They said theya. had gone. c. have went.b. had go. d. had went.Obviously distractors (b), (c), and (d) contain structures that do not occur in grammatically acceptable English for classroom use.4. Trick QuestionsFor some teachers it is always a temptation to include trick quesions on an examination. It is difficult to say whether the teacher's motive is to display cleverness, to embarrass the students for revenge, or merely to ensure test difficulty.The point is that such questions make for both inaccurate measurement and poor pedagogy. Consider the followingWhen is it not appropriate not to be absent from class?a) When you are sick.b) When you are young.c) While class is in session.d) Whenever the teacher is angry.5. Connnon Knowledge ResponsesParticularly when testing the skill of reading comprehension, it may happen that an item may test common knowledge. In this way the correct response may be chosen without comprehension of the reading passage. An example would be the following: We learn from this passage that Napoleon wasa) British c) Polishb) French d) German22．What is the essential difference between scatterplots of low-correlated variables and those of high-correlated variables? Solution:In scatterplots, if the variables are highly correlated, the dots fall very close together in a comparatively narrower or more elongated ellipse than those low-correlated variables.23．Please compute the correlation coefficient of the following item and determine whether it is significant on the level of significance of <.05? And then determine its discriminability.Solution:We compute the Pear son’s Product Moment Correlation coefficient and arrive at r=0.3007. And we do a one-tailed test with p<.05, degree of freedom being 16-1=15. And the critical value is 0.4124. But because r=0.3007<0.4124, the correlation is not statistically significant. So this item lacks a good discriminability.24．In a general proficiency test, mean=95.120, standard deviation=10.858, please compute the z-score, T-score, normal distribution area proportions of the following scores: 120, 108, 100, 95, 83, 71, 61.Solution:Raw score z-score T-score normal distribution area proportion120 2.291 72.91 98.9 108 1.186 61.86 88.22 100 0.449 54.49 67.33 95 -0.011 49.89 49.56 83 -1.116 38.84 13.22 71 -2.221 27.79 1.32 61 -3.142 18.58 0.08。

第五章环境和生物间的相互作用Interactions: Environments and Organisms The science of ecology is the study of the ways organisms interact with each other, and with their nonliving surroundings.Ecology deals with the ways in which or ganisms are adapted to their surroundings,how they make use of these surroundings, and how an area is altered by the presence and activities of organisms.All organisms are dependent on other organisms in some way.Everything that affects an organism during its lifetime is collectively known as its environment.Abiotic factors can be organized into several broad categories: energy, nonliving matter, and processes that involve the interactions of nonliving matter and energy.All organisms require a source of energy to maintain themselves. The ultimate source of energy for almost all organisms is the sun.All forms of life require atoms of elements such as carbon, nitrogen, and phosphorus, and molecules such as water to construct and maintain themselves. Organisms constantly obtain these materials from their environment. The atoms become part of an organism's body structure for a short time period, and eventually all of them are returned to the environment through respiration, excretion, or death and decay.The structure and location of the space organisms inhabit is also an important abiotic aspect of their environment.Important ecological processes involve interactions of matter and energy.The climate (average weather patterns over a number of years) of an area involves energy in the form of solar radiation interacting with the matter that makes up the Earth.The biotic factors of an organism's environment include all forms of life with which it interacts.Although organisms interact with their surroundings in many ways, certain factors may be critical to a particular species' success. A shortage or absence of this factor restricts the success of the species; thus, it is known as a limiting factor.The limiting factor for many species of fishes is the amount of dissolved oxygen in the water.The environment influences the organism, and organisms affect the environment.The habitat of an organism is the space that the organism inhabits, the place where it lives (its address).The niche of an organism is the functional role it has in its surroundings(its profession).Genes are distinct pieces of DNA that determine the characteristics an individual displays.A population is considered to be all the organisms of the same kind found within a specific geographic region.A species is a population of all the organtisms potentially capable of reproducingnaturally among themselves and having offspring that also reproduce.The process that leads to this close fit between the characteristics organisms display and the demands of their environment is known as natural selection.natural selection is the mechanism that causes evolution to occur.When we look at the effects of natural selection over time, we can see considerable change in the characteristics of a species and kinds of species present. Some changes take thousands or millions of years to occur. Others, such as resistance to pesticides, can occur in a few years.Scientists have continuously shown that this theory of natural selection can explain the development of most aspects of the structure, function, and behavior of organisms. It is the central idea that helps explain how species adapt to their surroundings. When we discuss environmental problems, it is helpful to understand that species change and that as the environment is changed, either naturally or by human action, some species will adapt to the new conditions while others will not.The environment in which organisms exist does not remain constant over long time periods. Those species that lack the genetic resources to cope with a changing environment go extinct. Extinction is the loss of an entire species and is a common feature of the evolution of organisms.Natural selection is constantly at work shaping organisms to fit a changing environment.It is clear that humans have had a significant impact on the extinction of many kinds of species.Wherever humans have modified the environment for their purposes (farming, forestry, cities, hunting, and introducing exotic organisms), species are typically displaced from the area.If large areas are modified, entire species may be displaced. Ultimately, humans are also subject to evolution and the possibility of extinction as well.Coevolution is the concept that two or more species of organisms can reciprocally influence the evolutionary direction of the other. In other words, organisms affect the evolution of other organisms.Since all organisms are influenced by other organisms, this is a common pattern.Ecologists look at organisms and how they interact with their surroundings.One common kind of interaction called predation occurs when one organism, known as a predator, kills and eats another, known as the prey.A second type of interaction between species is competition, in which two organisms strive to obtain the same limited resource.Symbiosis is a close, long-lasting, physical relationship between two different species. There are three different categories of symbiotic relationships: parasitism, commensalism, and mutualism.If we examine our activities, we can see that we have complicated interactions with other organisms.Predator----Humans throughout the world use animals as food.Herbivore----Humans rely on many kinds of plants as their primary source of food.Scavenger----Scavenging involves finding and consuming animals that are already dead.Commensalism-----Humans find themselves on both sides of commensal relationships.Parasitism----Although humans do not live in or on other living things, we do engage in relationships that are parasitic in nature.Mutualism----Humans have many mutualistic relationships with plants and animals.Competition----Humans are in competition with all other organisms on Earth. As we convert land and aquatic resources to our uses, we deprive other organisms of what they need to survive.A community is an assemblage of all the interacting populations of different species of organisms in an area.An ecosystem is a defined space in which interactions take place between a community, with all its complex interrelationships, and the physical environment.While it is easy to see that the physical environment places limitations on the kinds of organisms that can live in an area, it is also important to recognize that organisms impact their physical surroundings.Every system has parts that are related to one another in specific ways.Producers are organisms that are able to use sources of energy to make complex, organic molecules from the simple inorganic substances in their environment.Primary consumers, also known as herbivores, are animals that eat producers (plants or phytoplankton) as a source of food.Secondary consumers or carnivores are animals that eat other animals.Decomposers are organisms that use nonliving organic matter as a source of energy and raw materials to build their bodies.Whenever an organism sheds a part of itself, excretes waste products, or dies, it provides a source of food for decomposers.Since decomposers carry on respiration, they are extremely important in recycling matter by converting organic matter to inorganic material.Many small animals, fungi, and bacteria fill this niche.A keystone species is one that has a critical role to play in the maintenance of specific ecosystems.Some species have pivotal roles, and their elimination or severe reduction can significantly alter ecosystems.The energy stored in the molecules of producers is transferred to other organisms when the producers are eaten.Each step in the flow of energy through an ecosystem is known as a trophic level. Producers (plants, algae, phytoplankton) constitute the first trophic level, and herbivores constitute the second trophic level. Carnivores that eat herbivores are the thir,trophic level, and carnivores that eat other ". carnivores are the fourth trophic level.As energy flows through an ecosystem, it passes through several levels known as trophic levels. Each trophic level contains a certain amount of energy. Each time energy flows to another trophic level, approximately 90 percent of the useful energy islost, usually as heat to the surroundings. Therefore, in most ecosystems, higher trophic levels contain less energy and fewer organisms.The passage of energy from one trophic level to the next as a result of one organism consuming another is known as a food chain.When several food chains overlap and intersect, they make up a food web.All matter is made up of atoms. These atoms are cycled between the living and nonliving portions of an ecosystem. The activities involved in the cycling of atoms include biological, geological, and chemical processes. Therefore, these nutrient cycles are often called biogeochemical cycles.All living things are composed of organic molecules that contain atoms of the element carbon. The carbon cycle includes the processes and pathways involved in capturing inorganic carbon-containing molecules, converting them into organic molecules that are used by organisms, and the ultimate release of inorganic carbon molecules back to the abiotic environment.Fossil fuels (coal, oil, and natural gas) are part of the carbon cycle as well.At one time, these materials were organic molecules in the bodies of living organisms.The organisms were buried and the organic compounds in their bodies were modified by geologic forces. Thus, the carbon atoms present in fossil fuels were removed temporarily from the active,short-term carbon cycle. When we burn fossil fuels, the carbon reenters the active carbon cycle. Another very important nutrient cycle, the nitrogen cycle,involves the cycling of nitrogen atoms between the abiotic and biotic components and among the organisms in an ecosystem.Because atmospheric nitrogen is not usable by plants, nitrogen-containing compounds are often in short supply and the availability of nitrogen is often a factor that limits the growth of plants. The primary way in which plants obtain nitrogen compounds they can use is with the help of bacteria that live in the soil. Bacteria, called nitrogen-fixing bacteria, are able to convert the nitrogen gas (N2) that enters the soil into ammonia that plants can use.Bacteria and other types of decay organisms are involved in the nitrogen cycle also. Dead organisms and their waste products contain molecules, such as proteins, urea, and uric acid, that contain nitrogen. Decomposers break down these nitrogen-containing organic molecules, releasing ammonia, which can be used directly by many kinds of plants. Still other kinds of soil bacteria called nitrifying bacteria are able to convert ammonia to nitrite, which can be converted to nitrate. Plants can use nitrate as a source of nitrogen for synthesis of nitrogen-containing organic molecules.Finally, bacteria known as denitrifying bacteria are, under conditions where oxygen is absent, able to convert nitrite to nitrogen gas (N2), which is ultimately released into the atmosphere. These nitrogen atoms can reenter the cycle with the aid of nitrogen-fixing bacteria.The phosphorus cycle differs from the carbon and nitrogen cycles in one important respect. Phosphorus is not present in the atmosphere as a gas. The ultimate source of phosphorus atoms is rock. In nature, new phosphorus compoundsare released by the erosion of rock and become dissolved in water. Plants use the dissolved phosphorus compounds to construct the molecules they need. Animals obtain the phosphorus they need when they consume plants or other animals. When an organism dies or excretes waste products, decomposer organisms recycle the phosphorus compounds back into the soil.Phosphorus compounds that are dissolved in water are ultimately precipitated as deposits. Geologic processes elevate these deposits and expose them to erosion, thus making these deposits available to organisms.Two activities have caused significant changes in the carbon cycle: burning fossil fuels and converting forests to agricultural land.One consequence of these actions is that the amount of carbon dioxide in the atmosphere has been increasing steadily since humans began to use fossil fuels extensively. It has become clear that increasing carbon dioxide is causing changes in the climate of the world, and many nations are seeking to reduce energy use and prevent deforestation.The burning of fossil fuels has also altered the nitrogen cycle.When fossil fuels are burned, the oxygen and nitrogen in the air are heated to high temperatures and a variety of nitrogen-containing compounds are produced.If too much nitrogen or phosphorus is applied as fertilizer or if they are applied at the wrong time, much of this fertilizer is carried into aquatic ecosystems.The presence of large amounts of these nutrients in either freshwater or saltwater results in increased rates of growth of bacteria, algae,and aquatic plnts. Increases in the number of these organisms can have many different effects. Many algae are toxic, and when their numbers increase significantly,fish are killed and incidents of human poisoning occur. An increase in the number of plants and algae in aquatic ecosystems also can lead to low oxygen concentrations in the water. When these organisms die, decomposers use oxygen from the water as they break down the dead organic matter. This lowers the oxygen concentrations and many organisms die.Everything that affects an organism during its lifetime is collectively known as its environment. The environment of an organism can be divided into biotic (living) and abiotic (non-living) components.The space an organism occupies is known as its habitat, and the role it plays in its environment is known as its niche.I The niche of a species is the result of natural selection directing the adaptation of the species to a specific set of environmental conditions.Organisms interact with one another in a variety of ways.A community is the biotic portion of an ecosystem that is a set of interacting populations of organisms. Those organisms and their abiotic environment constitute an ecosystem.About 90 percent of the energy is lost as it passes from one trophic level to the next. This means that the amount of biomass at higher trophic levels is usually much less than that at lower trophic levels.The flow of atoms through an ecosystem involves all the organisms in thecommunity. The carbon, nitrogen, and phosphorus cycles are examples of how these materials are cycled in ecosystems.Key Terms :abiotic factors非生物因子biogeochemical cycles 生物地球化学循环biomass 生物量biotic factors 生物因子carbon cycle 碳循环community 群落competition 竞争consumer 消费者decomposer 分解者ecology 生态学ecosystem 生态系统environment 环境evolution 进化extinction 灭绝food chain 食物链food web 食物网genes 基因habitat 生境limiting factor 限制性因素natural selection 自然选择niche 小生境nitrogen cycle 氮循环population 种群primary consumer 初级消费者range of tolerance 耐受范围secondary consumer 次级消费者species物种trophic level 营养级Review Questions:1. Define environment.3. How is natural selection related to the concept of niche?5. How is an ecosystem different from a community?8. What are some different trophic levels in an ecosystem?9. Describe the carbon cycle, the nitrogen cycle, and the phosphorus cycle.第六章生态系统及群落的种类Kinds of Ecosystems and Communities Ecosystems are dynamic, changing units.The concept that communities proceed through a series of recognizable, predictable changes in structure over time is called succession. The relatively stable, long-lasting community that is the result of succession is called a climax community.Primary succession is a successional progression that begins with a total lack of organisms and bare mineral surfaces or water.Secondary succession is much more commonly observed and generally proceeds more rapidly, because it begins with the destruction or disturbance of an existing ecosystem. Fire, flood, windstorm, or human activity can destroy or disturb a community of organisms.The general trend in succession is toward increasing complexity and more efficient use of matter and energy compared to the successional communities that preceded them.The principal concepts of land succession can be applied to aquatic ecosystems. Except for the oceans, most aquatic ecosystems are considered temporary. Certainly, some are going to be around for thousands of years, but eventually they will disappear and be replaced by terrestrial ecosystems as a result of normal successional processes. All aquatic ecosystems receive a continuous input of soil particles and organic matter from surrounding land, which results in the gradual filling in of shallow bodies of water such as ponds and lakes.The same processes and activities drive both primary and secondary succession. The major difference is that secondary succession occurs when an existing community is destroyed but much of the soil and some of the organisms remain. A forest fire, a flood, or the conversion of a natural ecosystem to agriculture may be the cause.As settlers removed the original forests or grasslands and converted the land to farming, the original "climax" community was destroyed. Eventually, as poor farming practices destroyed the soil, many farms were abandoned and the land was allowed to return to its "original" condition.Biomes are terrestrial climax communities with wide geographic distribution.The distribution of terrestrial ecosystems is primarily related to precipitation and temperature.A lack of water is the primary factor that determines that an area will be a desert. Deserts are areas that generally average less than 25 centimeters(10 inches) of precipitation per year.Since the rate of evaporation is high, plant growth and flowering usually coincide with the periods when moisture is available. Deserts are also likely to be windy. We often think of deserts as hot, dry wastelands devoid of life. However, many deserts are quite cool during a major part of the year.Another misconception about deserts is that few species of organisms live in the desert. There are many species, but they typically have low numbers of individuals.The desert has many kinds of animals. However, they are often overlooked because their populations are low, numerous species are of small size, and many are inactive during the hot part of the day. They also aren't seen in large, conspicuous groups.Grasslands, also known as prairies or steppes, are widely distributed over temperate parts of the world. As with deserts, the major factor that contributes to the establishment of a grassland is the amount of available moisture. Grasslands generally receive between 25 and 75 centimeters (10 to 30 inches) of precipitation per year.Most of the moist grasslands of the world have been converted to agriculture , since the rich, deep soil that developed as a result of the activities of centuries of soil building is useful for growing cultivated grasses such as corn (maize) and wheat. The drier grasslands have been converted to the raising of domesticated grazers such as cattle, sheep, and goats. Therefore, little undisturbed grassland is left, and those fragments that remain need to be preserved as refuges for the grassland species that once occupied huge portions of the globe.Tropical parts of Africa, South America, and Australia have extensive grasslands spotted with occasional trees or patches of trees. This kind of a biome is often called a savanna. Although savannas receive 50 to 150 centimeters (20 to 60 inches) of rain per year, the rain is not distributed evenly throughout the year. Typically, a period of heavy rainfall is followed by a prolonged drought. This results in a very seasonally structured ecosystem. The plants and animals time to their reproductive activities to coincide with the rainy period, when limiting factors are least confining. The predominant plants are grasses, but many drought-resistant, flat-topped, thorny trees are common.The Mediterranean shrublands are located near an ocean and have wet, cool winters and hot, dry summers. Rainfall is 40 to 100 centimeters (15 to 40 inches) per year. As the name implies, this biome is typical of the Mediterranean coast and isalso found in coastal southern California, the southern tip of Africa, a portion of the west coast of Chile, and southern Australia. The vegetation is dominated by woody shrubs that are adapted to withstand the ot, dry summer.Another biome that is heavily influenced by seasonal rainfall is known as the tropical dry forest.Tropical rainforests are located near the equator in Central and South America, Africa, Southeast Asia, and some islands in the Caribbean Sea and Pacific Ocean. The temperature is normally warm and relatively constant. There is no frost, and it rains nearly every day. Most areas receive in excess of 200 centimeters (80 inches) of rain per year. Some receive 500 cennmeters (200 inches) or more. Because of the warm temperatures and abundant rainfall, most plants grow very rapidly; however, soils are usually poor in nutrients because water tends to carry away any nutrients not immediately taken up by plants.Tropical rainforests are under intense pressure from logging and agriculture.Many of the countries where tropical rainforests are present are poor and seek to obtain jobs and money by exploiting this resource.Forests in temperate areas of the world that have a winter-summer change of seasons typically have trees that lose their leaves during the winter and replace them the following spring. This kind of forest is called a temperate deciduous forest and is typical of the eastern half of the United States, parts of south central and south- eastern Canada, southern Africa, and many areas of Europe and Asia. These areas generally receive 75 to 100 centimeters (30 to 60 inches) of relatively evenly distributed precipitation per year.In contrast to tropical rainforests, where individuals of a tree species are scattered throughout the forest, temperate deciduous forests generally have many fewer species, and many forests may consist of two or three dominant tree species.These forests are home to a great variety of insects, many of which use the leaves and wood of trees as food.Throughout the southern half of Canada, parts of northern Europe, and much of Russia, there is an evergreen coniferous forest known as the taiga, northern coniferous forest, or boreal forest. The climate is one of short, cool summers and long winters with abundant snowfall. The winters are extremely harsh and can last as long as six months. Typically, the soil freezes during the winter. Precipitation ranges between 25 and 100 centimeters (10 to 40 inches) per year. However, the climate is typically humid because there is a great deal of snowmelt in the spring and generally low temperatures reduce evaporation. The landscape is typically dotted with lakes, ponds, and bogs.North of the taiga is the tundra, a biome that lacks trees and has a permanently frozen subsurface soil. This frozen soil layer is known as permafrost. Because of the permanently frozen soil and extremely cold, windy climate (up to 10 months of winter), no trees can live in the area.Because of the very short growing season, damage to this kind of ecosystem is slow to heal, so the land must be handled with great care.Terrestrial biomes are determined by the amount and kind of precipitation and by temperatures. Other factors, such as soil type and wind, also play a part. Aquatic ecosystems also are shaped by key environmental factors. Several important factors are the ability of the sun's rays to penetrate the water, the depth ofthe water, the nature of the bottom substrate, the water temperature, and the amount of dissolved salts.An important determiner of the nature of aquatic ecosystems is the amount of salt dissolved in the water.Those that have little dissolved salt are called freshwater ecosystems, and those that have a high salt content are called marine ecosystems.Coral reef ecosystems are produced by coral animals that build up-shaped external skeletons around themselves.Because they require warm water, coral ecosystems are found only near the equator Coral ecosystems also require shallow, clear water since the algae must have ample sunlight to carry on photosynthesis.Coral reefs are considered one of the most productive ecosystems on Earth.Mangrove swamp ecosystems occupy a region near the shore. The dominant organisms are special kinds of trees that are able to tolerate the high salt content of the ocean.The trapping of sediment and the continual extension of mangroves into shallow areas result in the development of a terrestrial ecosystem in what was once shallow ocean.An estuary is a special category of aquatic ecosystem that consists of shallow, partially enclosed areas where freshwater enters the ocean.Estuaries are particularly productive ecosystems because of the large amounts of nutrients introduced into the basin from the rivers that run into them. This is further enhanced by the fact that the shallow water allows light to penetrate to most of the water in the basin.Estuaries are especially important as nursery sites for fish and crustaceans such as flounder and shrimp.Freshwater ecosystems differ from marine ecosystems in several ways. The amount of salt present is much less, the temperature of the water can change greatly, the water is in the process of moving to the ocean, oxygen can often be in short supply, and the organisms that inhabit freshwater systems are different.Freshwater ecosystems can be divided into two categories: those in which the water is relatively stationary, such as lakes, ponds, and reservoirs, and those in which the water is running downhill, such as streams and rivers.Large lakes have many of the same characteristics as the ocean.Farming and construction expose soil and release nutrients, as do other human activities such as depositing sewage into streams and lakes. Deep, clear, cold, nutrient-poor lakes are low in productivity and are called oligotrophic lakes. Shallow, murky, warm, nutrient-rich lakes are called eutrophic lakes.The dissolved oxygen content of the water is important since the quantity of oxygen determines the kinds of organisms that can inhabit the lake.When organic molecules enter water, they are broken down by bacteria andfungi. These decomposer organisms use oxygen from the water as they perform respiration. The amount of oxygen used by decomposers to break down a specific amount of organic matter is called the biochemical oxygen demand (BOD).Just as estuaries are a bridge between freshwater and marine ecosystems, swamps and marshes are a transition between aquatic and terrestrial ecosystems.Many swamps and marshes are successional states that eventually become totally terrestrial communities.summary:Ecosystems change as one kind of organism replaces another in a process called succession.Ultimately, a relatively stable stage is reached, called the climax community.Major regional terrestrial climax communities are called biomes.The primary determiners of the kinds of biomes that develop are the amount and yearly distribution of rainfall and the yearly temperature cycle.Aquatic ecosystems can be divided into marine (saltwater) and freshwater ecosystems.Coral reefs are tropical marine ecosystems dominated by coral animals. Mangrove swamps are tropical marine shoreline ecosystems dominated by trees. Estuaries occur where freshwater streams and rivers enter the ocean. They are usually shallow, very productive areas. Many marine organisms use estuaries for reproduction.KeyTerms:biochemical oxygen demand (BOD) 生化需氧量biome 生物群系climax community 顶级群落coral reef ecosystem 珊瑚礁生态系统desert 沙漠estuary 河口eutrophic lake 富营养湖freshwater ecosystem 淡水生态系统grassland 草地mangrove swamp ecosystem 红树林沼泽生态系统marine ecosystem 海洋生态系统marsh 草本沼泽plankton 浮游生物phytoplankton 浮游植物zooplankton 浮游动物primary succession 原生演替secondary succession 次生演替swamp 木本沼泽tropical rainforest 热带雨林Review Questions:1.Describe the process of succession. How does primary succession differ from secondary succession?4. What two primary factors determine the kind of terrestrial biome that will develop in an area?6. What areas of the ocean are the most productive?9. List three differences between freshwater and marine ecosystems.10. What is an estuary?Why are estuaries important?。

第1周部分作业答案4. (1) 已知近似值*x 有5位有效数字，试求其相对误差限.(2) 已知近似值*x 的相对误差限是0.03%，问*x 至少有几位有效数字？解 设*x 是x 近似值，且*120.10mn x a a a =±⨯ ，其中n a a a ,,,21 是0到9之间的自然数，11a ≥，m 为整数。

(1) 因为*x 有5位有效数字，所以由定理1.2.1 (1) 知：*5144*1111111010100.005%222x x a a x-+---≤⨯=⨯⨯≤⨯=,即*x 的相对误差限为0.005%.(2) 因为*121210.100.100.(1)10m m m n n x a a a a a a a =±⨯=⨯<+⨯ ，且*x 的相对误差限为0.03%，所以***3311*0.03%0.(1)100.30.(1)100.510mm m x x x xx a a x----=⨯<⨯+⨯=⨯+⨯<⨯,故由定义1.2.3知：即*x 至少有3位有效数字。

3. (1) Suppose the approximation *x has 5 significant digits, try to find therelative error bound of *x .(2) Determine the minimal number of significant digits of the approximation *xsuch that the relative error bound of *x is 0.03%.Solution. Suppose*1210.10,19,09,2,3,,mn i x a a a a a i n=±⨯≤≤≤≤= ,is the approximation to x .(1) Since *x has 5 significant digits, part (1) of Theorem 1.2.1 gives*5144*1111111010100.005%222x x a a x-+---≤⨯=⨯⨯≤⨯=,that is, the relative error bound of *x is 0.005% from Definition 1.2.3.(2) Since*121210.100.100.(1)10mmmn n x a a a a a a a =±⨯=⨯<+⨯ ,and the relative error bound of *x is 0.03%, we have***3311*0.03%0.(1)100.30.(1)100.510mm m x x x xx a a x----=⨯<⨯+⨯=⨯+⨯<⨯,that is, *x retains at least 3 significant digits from Definition 1.2.5.。

SIGNIFICANT FIGURES, CALCULATIONS, AND READING MEASUREMENTS LABPurpose: At the core of all science is the ability to take accurate and precise measurements and to do calculations. Central to these concepts are significant figures, because they help tell us how good measurements are. In this lab you will gain experience with all of these concepts.Background: You are going to need some background to do this experiment on the different topics:Significant figures:∙You have probably covered significant figures in lecture, but because it is such a difficult concept for most students, it is being summarized again here. Significant figures are nothing more than the amount of digits we get from a measurement. The better a measurement is, the more significant figures you will have. This also means that for numbers that you don’t measure, such as conversion factors, there are no significant figures. For example, if you measure a sample ona balance and the balance reads 2.543 g, then there are 4 significant figures. If you measure thesame sample on a more sensitive balance, it may read 2.5438 g. The second measurement has 5 significant figures, and reflects a better measurement.∙An important note to make is that the very last digit in a measurement is always the least certain. Two people reading the same length on a ruler may see 1.24 cm vs. 1.25 cm. All three numbers are significant figures, but the last number has a bit of uncertainty. Both values are correct because it is expected that the last number is the uncertain one.∙To measure with significant figures, you count ALL digits. Each one is considered significant. For example, a ruler that reads 1.36 cm has 3 significant figures. A beaker with 12.52 mL of liquid has 4 significant figures. Scientific equipment has markings that let you read its measurement toa certain number of significant figures. The last value in your reading will ALWAYS be anestimate between the markings. This is why no two people will necessarily have the same exact number.o Ruler example:Here we see a ruler with markings every number, but none in between. Themeasurement between 6 and 7 looks to be a little before the half‐way point.Acceptable answers for this measure of distance would be anywhere from 6.2‐6.4.o Volume example: When you read volumes in glassware, youread a point called the meniscus. When you add liquid toglassware, the interface between the air and the liquid beingmeasured bends to form a shape like a “U”. This “U” shape isthe meniscus. Always read the bottom of the meniscus so that it is at your eye level.The meniscus here is right around the 11.5 marking. Possible answers would be 11.49‐11.52. You read a thermometer in the same way.To do calculations with significant figures, there are a number of rules to learn. Although painful, it is best to commit these to memory as soon as possible. There is a different set of rules for addition/subtraction and multiplication/division. In most cases you can use standard rounding rules. If the final number is a 5, we round that to the closest even number.o In addition and subtraction, pick the number with the fewest decimal places. Use that same number of decimal places in your final answer.Examples: 12.0 + 12.13 = 24.1 0.02‐0.0032=0.02o In multiplication and division, you need to determine which numbers are indeed significant. Remember that if a number starts with zeros, those aren’t significant. Anyother zeros are. Count all digits in the numbers. In your final answer, use as manysignificant figures as the number with the fewest digits. If you end up with an answerwith zeros at the end, it is best to write that in scientific notation if the answer isambiguous.Examples: 0.00352 X 125=0.440 143/0.02010=7110 = 7.11 X 103Equipment: Burets, Erlenmeyer flasks, graduated cylinders, beakers, rulers, balances, thermometers, metal rod samples.Procedure:I.There are stations set up around the lab that you will visit. Read the equipment at eachstation and record your values on the data sheet. You can start with any station. When youare finished, do the calculations on the post‐lab sheet.a.Station 1: Reading buretsBurets are long pieces of glass with calibrated markings on the side. They are markedbackwards, where the 0‐mL mark is at the top, and the 50‐mL mark is at the bottom.These are used to measure the amount of liquid that is delivered by subtractingnumbers. Read and record the volumes on the burets.b.Station 2: Reading Erlenmeyer flasksErlenmeyer flasks are pieces of glassware that are tapered up at the top and graduatedalong the sides. These are so large that a meniscus can’t usually be read. These are notused to measure exact volumes, but more estimates of large volumes. Read and recordthe volumes on the Erlenmeyer flasks.c.Station 3: Graduated cylindersGraduated cylinders are used to measure more exact volumes of liquid. They are thinglass tubes with markings up the side, similar to a buret, except they start at 0‐mL andgo up. These will show a meniscus, so be careful when you are reading them. Read andrecord the volumes on the graduated cylinders.d.Station 4: BeakersBeakers are cylindrical glass containers which are commonly used for crudemeasurements and mixing chemicals. Beakers come in many sizes, from less than 10‐mL to over several liters! Read and record the volumes on the beakers.e.Station 5: ThermometersIn science we tend to use mercury or alcohol thermometers. We measure all of ourchemicals based off the Celsius (o C) scale. Read and record the temperatures of thebeakers of water set out.f.Station 6: BalancesBalances are great for measuring the mass of objects. Use the balance to measure andrecord the mass (weigh) the metal rods available.g.Station 7: RulersRulers are not used as often in chemistry as the other equipment you have seen in thisexperiment, however, depending on the discipline, rulers prove a valuable device formeasuring distance. Measure the object using the available rulers.SIGNIFICANT FIGURES, CALCULATIONS, AND READING MEASUREMENTS PRE‐LABName:_________________________________After reading the introduction to this lab, answer the following questions to prepare you for the experiment. You may want to refer to lecture materials.1.Significant figures. For the numbers below, state how many significant figures there are.a. 4.0258 _________b. 1.024 _________c.0.2600 _________d.1345.1 _________e.1000.1 _________2.Calculations. For the following problems, calculate the answer using the correct number ofsignificant figures.a. 1.2 0.13 _________b. 1.2 0.13 _________c.485.369 0.124 _________d.0.0012 0.0008 _________e. 1.265 1.02 82 _________3.Go online and look up the following glassware. Draw an example of the glassware in the spacebelow.BuretGraduated CylinderBeakerErlenmeyer FlaskSIGNIFICANT FIGURES, CALCULATIONS, AND READING MEASUREMENTS POST‐LABName:__________________________________________Partner(s):_______________________________________Data Sheet1.Station 1: Reading buretsa.Meniscus reading for buret “A” _________________mLb.Meniscus reading for buret “B” _________________mL2.Station 2: Erlenmeyer flasksa.Volume reading for flask “A” _________________mLb.Volume reading for flask “B” _________________mL3.Station 3: Graduated cylindersa.Meniscus reading for graduated cylinder “A” _________________mLb.Meniscus reading for graduated cylinder “B” _________________mL4.Station 4: Beakersa.Volume reading for beaker “A” _________________mLb.Volume reading for beaker “B” _________________mL5.Station 5: Thermometersa.Temperature reading for thermometer “A” _________________o Cb.Temperature reading for thermometer “B” _________________o C6.Station 6: Balancesa.Mass of object “A” _________________gb.Mass of object “B” _________________g7.Station 7: Rulersa.Length of object “A” _________________cmb.Length of object “B” _________________cmPost‐lab questions. Show your work:1.For station 1, add together the buret volumes.2.Multiply your buret volumes and subtract them from 100.1 mL.3.For station 2, subtract volume “B” from volume “A”.4.For station 3, multiply the volumes in the graduated cylinders.5.Subtract “A” from “B” in the graduated cylinders, then add 0.223 mL.6.For station 4, subtract volume “A” from volume “B”.7.For station 5, divide temperature “A” by “B”.8.For station 6, add the masses.9.For station 7, subtract “B” from “A”.10.Do the following calculation:19.31 39.1111.9311.32 0.94。