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应用数理统计(英文)Answer to the 1st-11th

应用数理统计(英文)Answer to the 1st-11th
应用数理统计(英文)Answer to the 1st-11th

Answer of the first homework

1.2

(a) percent of patients in treatment group experienced a significant in symptoms is:

66/85×100%=77.64%;

Percent in control group is:

65/81×100%=80.24%.

(b) Symptomatic treatments appears to be more effective for sinusitis.

(c)The data do not provide convincing evidence, because the total of each group are

not equal and the percent of patients experienced a significant in symptoms are too similar.

I think It’s due to chance.

1.4(a)

(i) the cases: 202 black and 504 white adults.

(ii) the variables and their types:age,weight,height and body fat percentage are numerical types; sex and ethnicity are categorical types.

1.6

(a)each row represents a unique resident and his/her information;

(b)1691 participants were included;

(c) Numerical: age, amtWeekends and amtWeekdays. All of them are discrete;

Categorical: gender, marital, grossIncome and smoke. Only grossIncome is continuous, the others are variables.

1.9

(a) Study hours/week is explanatory variable; GPA is response variable.

(b) Short study hours/week leads to high proportion of lower GPA. With the study

hours/week becoming longer the proportion of lower GPA is fewer. And there are not definite relationship between the higher GPA and study hours/week.

(c) It’s an observation study;

(d) We cannot conclude the result that studying longer hours leads to higher GPAs.

1.12

(a) Higher percentage of internet users with higher Life expectancy;

(b) This type of study is observation;

(c) Internet using may leads to a optimistic attitude, so this attitude causes long life

expectancy.

1.15

It will be biased. It will overestimate the true value. Some of the students may come from the same family. The denominator is bigger than the true value, When we calculating the average.

1.21

(a) This type of study is experimental.

(b) Treatment group: exercise twice a week;

Control: not to exercise

(c) Yes it does, the blocking variable is age;

(d) No it doesn’t.

(e) The results of the study can be used to establish a causal relationship between

exercise and mental health. Because there are some other differences between these two groups having to control. Because it’s a random experiment.

No, it cannot be.

(f) Except for the exercise, some other differences must be controlled. For instance the

work and life environment should be controlled.

1.30

It’s a symmetric distribution.

1.34(a)

The mean of (1) is smaller than the (2), and the standard deviation of (1) is also smaller than the (2).Because in the two of these sequences only their last numbers are

different, and the (2) is bigger than the (1).

Answer of the 2nd homework

2.2

(a) The probability that the ball will land on a red slot on the next spin is

18/38=47.37%

(b) The probability that the ball will land on a red slot on the next spin is

18/38=47.37%

(c) I am equally confident of my answer to parts (a) and (b). Because each slot has an

equal chance of capturing the ball. 2.5

(a) The probability of getting all tails: (1/2)^10= 0.000977

(b) The probability of getting all heads:

(1/2)^10= 0.000977

(c) The probability of getting at least one tails: 1-(1/2)^10=0.999023 2.8

(a) They are not disjoint; (b)

(c) Percent of Americans live below the poverty line and only speak English at home :

14.6%-4.2%=10.4%

(d) Percent of Americans live below the poverty line or speak a language other than

English at home :

14.6%+20.7%-4.2%=31.1%

(e) Percent of Americans live above the poverty line and only speak English at home:

1- 31.1%=68.9%

(f) They are not independent. We can see from the Venn diagram that when one

happened, the other has the probability to happen or not happen. They have joint. 2.11

(a) The probability that a randomly chosen man has at least a Bachelor's degree is:

0.16+0.09=0.25

(b) The probability that a randomly chosen woman has at least a Bachelor's degree is:

0.17+0.09=0.26 (c) 0.25×0.26=0.065

Assumption A: Both are Bachelor's degree:

0.16×0.17=0.0272;

Assumption B: Both are Graduate or professional degree:

0.09×0.09=0.0081;

Assumption C: One is Bachelor's degree and the other is Graduate or professional

degree: 0.16×0.09+0.17×0.09=0.0297;

(d) I think it’s not reasonable. Two of the variables are independent. It exists several

Speak a language other than English at home and not live below the poverty line

Both Other

conditions above.

2.13

(a) is an invalid probability distribution. Because the total probability of the

distribution is not 1;

(b) is a valid probability distribution;

(c) is an invalid probability distribution. It’s simila r to (a);

(d) is an invalid probability distribution. Because the probability of the distribution

must be positive;

(e) is an valid probability distribution;

(f) is an invalid probability distribution. Because the probability of the distribution

must be positive;

Answer of the 3nd homework

2.33

(a) The expected number of smokers is:

13%×100=13

(b) I should not. Because the total number of the students is to small. It may produce a bias. 2.34

The expected winnings for a single game: E(X)=4

1()i i i x P X x ==∑

=0×0.5+5×0.25+10×0.2308+30×0.0192=4.134

4

22222

21

0 4.134)0.5(5 4.134)0.25(10 4.134)0.2308(30 4.134)0.0192()()

(29.52i i i x P X x σμ=-?+-?+-?+-?==-==∑ 2

5.

43σ= (b) I would be willing to pay nothing. Because I will pay nothing even if that I draw a red card and the probability is the biggest of all. 2.37

The expected return on this portfolio is :

(18%+9%-12%)/3=5%

2.41

Suppose that the Expects of Dodgers win and Padres win are equal. So:

5×0.46=E(X) Padres ×(1-0.46)

and

E(X) Padres=4.26

To make this a fair game I would need to bet on the Padres $4.26.

2.44

(a) Suppose: W=X+Y1+Y2+Y3

E(W)= E(X) +E(Y1) +E(Y2) +E(Y3)=48+3*2=54

Var(W)= Var(X+3Y)= Var(X)+9Var(Y)

and SD(W)= [Var(W)]1/2=1.25

(b) E(X-Y)= E(X) -E(Y)=48-2=46

SD(X-Y)= [Var(X-Y)]1/2= [Var(X)+ Var(Y)] 1/2=[1+0.0625] 1/2=1.03

(c) Because the variables X and Y are independent.

Var(X-Y)= Var(X)+(-1)2 Var(Y)= Var(X)+ Var(Y).

Additional questions for Chapter 2

PS:symbol of “·”present the answer.

Question 1

Which of the following events would you be most surprised by?

○exactly 3 heads in 10 coin flips

○exactly 3 heads in 100 coin flips

·exactly 3 heads in 1000 coin flips

Question 2

Which of the following states that the proportion of occurrences with a particular outcome converges to the probability of that outcome?

○ General addition rule

○Bayes’ theorem

·Law of large numbers

○ Law of averages

Question 3

Shown below are four Venn diagrams. In which of the diagrams does the shaded area represent A and B but not C?

○○

○·

Question 4

Each choice below shows a suggested probability distribution for letter grades in a class (Possible grades are A, B, C, or D or lower). Determine which is a proper probability distribution.

○ A: 20%, B: 0.40, C: 0.50, D or lower: -0.10

·A: 0.30, B: 0.30, C: 0.40, D or lower: 0

○A: 10%, B: 0.20, C: 0.50, D or lower: 0.10

○ A: 30%, B: 0.30, C: 0.30, D or lower: 0.30

Question 5

Which of the following statements is false?

○Two mutually exclusive outcomes (of the same event) cannot occur at the same time.

○ Two complementary outcomes (of the same event) cannot occur at the same time. ·Two independent events cannot occur at the same time.

○ Two disjoint outcomes (of the same event) cannot occur at the same time.

Answer of the 4th homework

3.2

N(μ=0,σ=1)

(a) Percentage is: 1-0.1292=0.8708;

(b) Percentage is: 0.5714;

(c) Percentage is almost to 1;

(d) Percentage is: 0.6915-0.3085=0.383.

3.4

(a) Men_N1: μ=4313,σ=583;

Women_N2: μ=5261,σ=807.

(b) Leo: Z1=(4948-4313)/583=1.0892;

Mary: Z2=(5513-5261)/807=0.3123.

It tells me that Mary has a better performance than Leo.

(c)Mary ranks better in his groups. Because that Z2 is more near to average score.

(d) Percent of the triathletes did Leo finish faster than his group:

1-0.8621= 0.1379

(e) Percent of the triathletes did Mary finish faster than his group:

1-0.6217=0.3783

(f) I think the answer of (c) would not change, and the others would change. Because that these answers are calculated by the model of Normal distribution.

3.6

(a) The fastest 5% is corresponded to the Z score about: -1.645.

So, cutoff T=4313-1.645×583=3354

(b) The slowest 10% is corresponded to the Z score about: 1.28.

So, cutoff T=5261+807×1.28=6294

3.7

N(μ=77,σ=5)

(a) Z=(83-77)/5=1.2 and so the probability is 1-0.8849=0.1151

(b) The coldest 10% is corresponded to the Z score about:-1.28

So, coldest T=77-5×1.28=70.6

3.12

N(μ=72.6,σ=4.78)

(a) Z=(80-72.6)/4.78=1.55 and so the probability is 0.9394

(b) Z=(60-72.6)/4.78=-2.64 and so the probability is 0.9394-0.0041=0.9353

(c) The fastest 5% is corresponded to the Z score about: 1.645

So, fast V=72.6+4.78×1.645=80.5

(d) Z=(70-72.6)/4.78=--0.54 and so the probability is 1-0.2946=0.7054

3.14

N(μ=100,σ)

(a) The top 2% is corresponded to the Z score about:2.06

Z=(132-100)/σ=2.06, and so theσ=15.53

Additional questions for Chapter 3

PS:symbol of “·”present the answer.

Question 1

[Quality control] At Heinz ketchup factory the amounts which go into bottles of ketchup are supposed to be normally distributed with mean 36 oz. and standard deviation 0.11 oz. Once every 30 minutes a bottle is selected from the production line, and its contents are noted precisely. If the amount of ketchup in the bottle is below 35.8 oz. or above 36.2 oz., then the bottle fails the quality control inspection. What percent of bottles pass the quality control inspection?

(a) 1.82%

(b) 3.44%

(c) 6.88%

·93.12%

(e) 96.56%

Question 2

Which of the following is false? Hint: It might be useful to sketch the distributions. ·The Z score for the median of a symmetric distribution is approximately 0. (b) Regardless of the shape and skew of a distribution, Z scores are still defined and can be calculated for observations from that distribution.

(c) If you calculated the Z score for the median of a righ t skewed distribution, you’d most likely get a positive number.

(d) If observations don’t come from a distribution that’s nearly normal, we can’t calculate percentiles based on the Z table.

Question 3

Suppose weights of checked baggage of airline passengers follow a nearly normal distribution with mean 45 pounds and standard deviation 3.2 pounds. Most airlines charge a fee for baggage that weigh in excess of 50 pounds. What percent of airline passengers are expected to incur this fee?

(a) 0.9406

·0.0594

(c) 0.1032

(d) 0.0756

Question 4

Scores on a standardized test are nearly normally distributed with a mean of 100 and a standard deviation of 20. If these scores are converted to standard normal Z scores, which of the following statements will be correct?

·The mean will be 0, and the median should be roughly 0 as well.

(b) The mean will equal 0, but the median cannot be determined.

(c) The mean of the standardized Z scores will equal 100.

(d) The mean of the standardized Z scores will equal 5.

Question 5

ACT scores are distributed nearly normally with mean 21 and standard deviation 5. Jim, who scored a 24 on his ACT. Which of the following is true?

(a) Jim's Z score is -0.6

·Jim scored better than approximately 72.57% of ACT takers.

(c) 72.57% of ACT takers scored better than Jim.

(d) Jim's percentile score is 60%.

Answer to the 5th homework

3.17

Φ(μ±σ) =68%:69.26~86.14, in this area, the number of students is 14 and the percentage is 14/20=70% ;

Φ(μ±2σ) =95%:60.82~94.58, in this area, the number of students is 19 and the percentage is 19/20=0.95

Φ(μ±3σ) =99.7%:52.38~103.02, in this area, the number of students is 20 and the percentage is 20/20=100%

These scores is approximately follow the 68-95-99.7% Rule.

3.19

Yes,these data appear to follow a normal distribution.

The histogram looks normal, and the normal probability plot close to straight and stable.

Additional questions for Chapter 3

PS:symbol of “·”present the answer.

Question 1

Which of the following is false?

·Majority of Z scores in a right skewed distribution are negative.

(b) In skewed distributions the Z score of the mean might be different than 0.

(c) For a normal distribution, IQR is less than 2SD.(50<68)

(d) Z scores are helpful for determining how unusual a data point is compared to the rest of the data in the distribution.

Question 2

Which of the following, on its own, is the least useful method for assessing if the data follow a normal distribution?

(a) Check if the points are on a straight line on a normal probability plot.

(b) Check if 68% of the data are within 1 SD of the mean, 95% of data are within 2 SDs of the mean, and 99.7% of data are within 3 SDs of the mean.

(c) Check if the distribution is unimodal and symmetric.

·Check if the mean and median are equal.

Question 3

Distribution of number of hours of sleep college students get has a mean of 7 hours and a standard deviation of 1 hour. 72% of students sleep between 6 and 8 hours,

92% of students sleep between 5 and 9 hours, and 99% sleep between 4 and 10 hours. Which of the following is true?

(a) The distribution is nearly normal.

·The distribution is more variable than a normal distribution with mean 7 and standard deviation 1.

(c) The distribution is less variable than a normal distribution with mean 7 and standard deviation 1.

Question 4

SAT scores are distributed nearly normally with mean 1500 and standard deviation 300. According to the 68-95-99.7% rule, which of the following is false?

(a) Roughly 68% of students score between 1200 and 1800 on the SAT.

(b) Roughly 95% of students score between 900 and 2100 on the SAT.

(c) Roughly 99.7% of students score between 600 and 2400 on the SAT.

·No students can score below 600 on the SAT.

Question 5

A doctor collects a large set of heart rate measurements that approximately follow a normal distribution. He only reports 3 statistics, the mean = 110 beats per minute, the minimum = 65 beats per minute, and the maximum = 155 beats per minute. Which of the following is most likely to be the standard deviation of the distribution?

(a) 5

·15

(c) 35

(d) 90

Question 6

(a) Make a histogram of the following random samples x1,…,x40. Do you think data comes from a symmetrical distribution? Does it skew to the left or right?

x

I don’t think they come from a symmetrical distribution. It skews to the left .

(b) Make a QQ-plot of these random samples. Do you believe the data is approximate to a normal distribution?

Standard Normal Quantiles

Q u a n t i l e s o f I n p u t S a m p l e

I think the data is approximate to a normal distribution.

(c)You may also try (a) and (b) by log transformation of sample data, that is, log(x1),…,log(x40).

Standard Normal Quantiles

Q u a n t i l e s o f I n p u t S a m p l e

QQ Plot of Sample Data versus Standard Normal

x

频数直方图与正态分布密度函数(拟合)

Yes, I think they come from a symmetrical distribution. And the data is approximate to a normal distribution.

Answer to the 6th homework

4.1

(a) Mean. Each student reports a numerical value: a number of hours.

(b) Mean. Each student reports a number, which is a percentage, and we can average over these percentages.

(c) Proportion. Each student reports Yes or No, so this is a categorical variable and we use a proportion.

(d) Mean. Each student reports a number, which is a percentage like in part (b). (e) Proportion. Each student reports whether or not he got a job, so this is a categorical variable and we use a proportion. 4.3

(a) The point estimate is the sample mean:13.65.The median is 14 (b) 1.91SD =,3115132IQR Q Q =-=-=

(c) A load of 16 credits is not unusually high. A load of 18 credits is unusually high .Because 1613.652 1.9117.47<+?=,and 1813.652 1.9117.47>+?= .

(d) It would not be surprising. The second sample is about 0.4 standard error from 13.65 credits .In other words, 14.02 credits does not seem to be implausible given that original sample was relatively close to it. (e) We can use SE. And 0.191

SE =

= 4.5

(a) The variability he would he expect to see in the mean prices of repeated sample is 2.89

SE =

= (b) Yes, it’s true. The price $80 is about $5 standard error from $75 .In other words $80 does not seem to be implausible given that John’s sample was relatively close to it.

Z = 1.73, which indicates that the two values are not unusually distant from each other when accounting for the uncertainty in John's point estimate. 4.6 (a) 0.93

SE =

= (b) 2.15Z =-, which indicates that the two values are unusually distant from each other when accounting for the uncertainty in John's point estimate. I think this number is unreasonable.

Additional questions for Chapter 4

PS: symbol of “·” present the answer. Question 1

Researchers studying anthropometry collected body girth measurements and skeletal diameter measurements, as well as age, weight, height and gender, for 507 physically active individuals. The histogram below shows the sample distribution of heights in centimeters, and the table shows sample statistics calculated based on this sample. Which of the following is not necessarily true ?

(a) The sample median is 170.3 cm.

(b) The sample mean is 171.1 cm.

·The population mean is 171.1 cm.

(d) The point estimate for the population mean is 171.1 cm.

Question 2

Researchers studying anthropometry collected various body and skeletal measurements for 507 physically active individuals. The histogram below shows the sample distribution of heights in centimeters. If the 507 individuals are a simple random sample - and let’s assume they are - then the sample mean is a point estimate for the mean height of all active individuals. What measure do we use to quantify the variability of such an estimate? Compute this quantity using the data from this sample and choose the best answer below.

(a) standard error = 0.019

(b) standard deviation = 0.019

(c) standard deviation = 0.417

(d) mean squared error = 0.105

·standard error = 0.417

Question 3

For the standard deviation σ or s and the standard error SE, which of the following is the correct set of descriptions?

(a) σ: variability in sample data

SE: variability in point estimates from different samples of the same size and from same population

s: variability in population data

(b) σ: variability in sample data

s: variability in point estimates from different samples of the same size and from same population

SE: variability in population data

·s: variability in sample data

SE: variability in point estimates from different samples of the same size and from same population

σ: variability in population data

(d) SE: variability in sample data

σ: variability in population data

s: variability in point estimates from different samples of the same size and from same population

Question 4

Four plots are presented below. The plot at the top is a distribution for a population. The mean is 60 and the standard deviation is 18. Also shown below is a distribution of

(1) a single random sample of 500 values from this population,

(2) a distribution of 500 sample means from random samples of each size 18,

(3) a distribution of 500 sample means from random samples of each size 81. Determine which plot (A, B, or C) is which.

(a) (1) one sample, n = 500 - Plot C

(2) 500 samples, n = 18 - Plot B

(3) 500 samples, n = 81 - Plot A ·(1) one sample, n = 500 - Plot B

(2) 500 samples, n = 18 - Plot C

(3) 500 samples, n = 81 - Plot A

(c) (1) one sample, n = 500 - Plot A

(2) 500 samples, n = 18 - Plot B

(3) 500 samples, n = 81 - Plot C

(d) (1) one sample, n = 500 - Plot C

(2) 500 samples, n = 18 - Plot A

(3) 500 samples, n = 81 - Plot B

(e) (1) one sample, n = 500 - Plot A

(2) 500 samples, n = 18 - Plot C

(3) 500 samples, n = 81 - Plot B

Answer to the 7th homework 4.11

(a) False.

(b) False (c) True.

(d) False.

(e) False.

(f) True.

(g) False.

4.12

(a) False.

(b) False. (c) False.

(d) True.

(e) True.

(f) False.

(g) True.

4.13

We assumes that is a normal model distribution,and first find z*=1.65 such that

90% of the distribution falls between -z*and z* in the standard normal model, N(μ=0,

σ=1). 0.14y SE ===.

The 90% confidence interval can then be computed as :

1.65(

2.97,

3.431)y y SE ±?→

That means we are 90% confident that the average number of exclusive relationships these 203 undergraduates have been in betwe en 2.97 and 3.431。 4.14

We assumes that is a normal model distribution ,and first find z*=1.96 such that 95% of the distribution falls between -z*and z* in the standard normal model, N(μ=0,

σ=1). 0.063y SE ===。

The 95% confidence interval can then be computed as :

1.96(23.316,23.564)y y SE ±?→

That means we are 95% confident that the average number of exclusive relationships these 203 undergraduates have been in between 23.316 and23.564。 4.15(a)

(a) H0 : μ = 8 (On average, New Yorkers sleep 8 hours a night.)

HA : μ < 8 (On average, New Yorkers sleep less than 8 hours a night.) 4.16(b)

H 0: 462μ= the average GRE Verbal Reasoning score has not changed since 2004. H A : 462μ≠ the average GRE Verbal Reasoning score has changed since 2004.

Additional questions for Chapter 4

PS: symbol of “·” present the answer. Question 1

The General Social Survey (GSS) is a sociological survey used to collect data on demographic characteristics and attitudes of residents of the United States. In 2010, the survey collected responses from over a thousand US residents. The survey is conducted face-to-face with an in-person interview of a randomly-selected sample of adults. One of the questions on the survey is “For how many days during the past 30 days was your mental health, which includes stress, depression, and problems with emotions, not good?”

Based on responses from 1,151 US residents, the survey reported a 95% confidence interval of 3.40 to 4.24 days in 2010. Given this information, which of the following statements would be most appropriate to make regarding the true average number of days of “not good” mental health in 2010 for US residents?

(a) For all US residents in 2010, based on this 95% confidence interval, we would not

A company offering online speed reading courses claims that students who take their courses show a 5 times (500%) increase in the number of words they can read in a minute without losing comprehension. A random sample of 100 students yielded an average increase of 415% with a standard deviation of 220%.

Calculate a 95% confidence interval for the average increase in number of words students can read in a minute without losing comprehension. Choose the closest answer.

· (371.88, 458.12)

(b) (412.09, 417.91) (c) (411.37, 418.63)

(d) (378.7, 451.3)

Answer to the 8th homework

4.17 H0: 2x hours = HA: 2x hours >

4.18 H0:23.44x yearsold = HA:23.44x yearsold ≠

4.21

We can set up

the null hypothesis for this test as

a skeptical perspective: the lifespan last 7hours. The alternative hypothesis: the lifespan last less than 7hours. We can write these hypotheses as

H0: 7μ=

HA: 7μ<

The standard error can be estimated from the

sample standard deviation and the sample size:

0.144x SE ===

The hypothesis test will be evaluated using a significance level of α= 0.05

N(μ=7,σ=1.25)

Z=(6.85-7)/0.144=-1.042

Using the normal probability table, the left area is found to be 0.2984. Because p-value = 0.2984>0.05 =α,we receive the null hypothesis. 4.23 (a)

We assume that sample mean is normal.

Test conclusion Do not reject H0 Reject H0 in favor of HA

Truth

H0 True Okay Type 1Error

HA True Type 2 Error Okay

H0: 127x = HA: 127x ≠ (b)

The standard error can be estimated from the sample standard deviation and the

sample size:

4.875x SE ==

The hypothesis test will be evaluated using a significance level of α= 0.05.

N(μ=127,σ=4.875)

Z=(137.5-127)/4.875=2.15

Using the normal probability table, the right area is found to be 0.0158.

Because p-value =left tail + right tail= 2×0.0158<0.05 =α,we reject the null hypothesis. And in favor of HA.

The change in wait times statistically is not significant. (c)

The conclusion of the hypothesis test would change if the significance level was changed to α= 0.01. Because p-value = 2×0.0158>0.01 =α 4.26

Using the normal probability table, Z score corresponding to the p-value=0.05 are found to be 1.65.

(x 34)/(10 1.65Z =-=

So the samples mean x 36.05= 4.31(b) H0: 18x = HA: 18x >

And we assume that sample mean is normal.

The standard error can be estimated from the sample standard deviation and the

sample size:

0.361x SE ===

The hypothesis test will be evaluated using a significance level ofα= 0.05.

Z=-=

(19.2518)/0.361 3.46

Using the normal probability table, the right area is found to be 0.0003.

Because p-value =0.0003<0.05 =α,we reject the null hypothesis. And in favor of HA.

4.32(b)

x=

H0: 500%

x<

HA: 500%

And we assume that sample mean is normal.

The standard error can be estimated from the sample standard deviation and the sample size:

22%

SE==

x

The hypothesis test will be evaluated using a significance level ofα= 0.05.

Z=-=-

(4.155)/0.22 3.86

Using the normal probability table, the left area is found to be less than 0.0002. Because p-value =0.0004<0.025 =α,we reject the null hypothesis. And in favor of HA.

The true average improvement is less than 500%.

Additional questions for Chapter 4

PS:symbol of “·”present the answer.

Question 1

Two-sided alternative hypotheses are phrased in terms of:

(a) < or >

(b)≤ or ≥

·≠

(d)≈ or =

Question 2

A Type 1 error occurs when the null hypothesis is

(a) rejected when it is false

·rejected when it is true

(c) not rejected when it is false

(d) not rejected when it is true

Question 3

If it’s relatively riskier to reject the null hypothesis when it might be true, should a

smaller or a larger significance level be used?

(a) larger

·smaller

Question 4

The nutrition label on a bag of potato chips says that a one ounce (28 gram) serving of potato chips has 130 calories and contains ten grams of fat, with three grams of saturated fat. A random sample of 35 bags yielded a sample mean of 134 calories with a standard deviation of 17 calories. We are evaluating whether these data provide convincing evidence that the nutrition label does not provide an accurate measure of calories in the bags of potato chips at the 10% significance level. Which of the following is correct?

(a) The p-value is approximately 8%, which means we should fail to reject the null hypothesis and determine that these data do not provide convincing evidence the nutrition label does not provide an accurate measure of calories in the bags of potato chips.

·The p-value is approximately 8%, which means we should reject the null hypothesis and determine that these data provide convincing evidence the nutrition label does not provide an accurate measure of calories in the bags of potato chips.

(c) The p-value is approximately 16%, which means we should reject the null hypothesis and determine that these data provide convincing evidence the nutrition label does not provide an accurate measure of calories in the bags of potato chips.

(d) The p-value is approximately 16%, which means we should fail to reject the null hypothesis and determine that these data do not provide convincing evidence the nutrition label does not provide an accurate measure of calories in the bags of potato chips.

Answer to the 9th homework

4.33

(a) The distribution is unimodal and strongly right skewed with a median between

5 and 10 years old. Ages range from 0 to slightly over 50 years old, and the middle 50% of the distribution is roughly between 5 and 15 years

old. There are potential outliers on the higher end.

(b) When the sample size is small, the sampling distribution is right skewed, just like the population distribution. As the sample size increases, the sampling distribution gets more unimodal, symmetric, and approaches normality. The variability also decreases. This is consistent with the Central Limit Theorem.

4.36

With the sample size increase the standard error will becoming small. So

(1) Plot B; (2) Plot C; (3) Plot A

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清华大学-杨虎-应用数理统计课后习题参考答案2

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应用数理统计课后习题参考答案

习题五 1 某钢厂检查一月上旬内的五天中生产的钢锭重量,结果如下:(单位:k g) 日期重旦量 1 5500 5800 5740 5710 2 5440 5680 5240 5600 4 5400 5410 5430 5400 9 5640 5700 5660 5700 10 5610 5700 5610 5400 试检验不同日期生产的钢锭的平均重量有无显著差异? ( =0.05) 解根据问题,因素A表示日期,试验指标为钢锭重量,水平为 5. 2 假设样本观测值y j(j 123,4)来源于正态总体Y~N(i, ),i 1,2,...,5 检验的问题:H。:i 2 L 5, H i : i不全相等. 计算结果: 注释当=0.001表示非常显著,标记为*** '类似地,=0.01,0.05,分别标记为 查表F0.95(4,15) 3.06,因为F 3.9496 F0.95(4,15),或p = 0.02199<0.05 ,所 以拒绝H。,认为不同日期生产的钢锭的平均重量有显著差异 2 考察四种不同催化剂对某一化工产品的得率的影响,在四种不同催化剂下分别做试验 解 根据问题,设因素A表示催化剂,试验指标为化工产品的得率,水平为 4 . 2 假设样本观测值y j(j 1,2,..., nJ来源于正态总体Y~N(i, ), i 1,2,...,5 .其中样本容量不等,n分别取值为6,5,3,4 .

日产量 操作工 查表 F O .95(3,14) 3.34,因为 F 2.4264 F °.95(3,14),或 p = 0.1089 > 0.05, 所以接受H 。,认为在四种不同催化剂下平均得率无显著差异 3 试验某种钢的冲击值(kg Xm/cm2 ),影响该指标的因素有两个,一是含铜量 A ,另 一个是温度 试检验含铜量和试验温度是否会对钢的冲击值产生显著差异? ( =0.05 ) 解 根据问题,这是一个双因素无重复试验的问题,不考虑交互作用 设因素A,B 分别表示为含铜量和温度,试验指标为钢的冲击力,水平为 12. 2 假设样本观测值y j (i 1,2,3, j 1,2,3,4)来源于正态总体 Y j ~N (j , ),i 1,2,3, j 1,2,3,4 .记i 为对应于A 的主效应;记 j 为对应于B j 的主效应; 检验的问题:(1) H i 。: i 全部等于零,H i — i 不全等于零; (2) H 20 : j 全部等于零,H 21: j 不全等于零; 计算结果: 查表F 0.95(2,6) 5.143 ,局.95(3,6) 4.757 ,显然计算值F A , F B 分别大于查表值, 或p = 0.0005 , 0.0009均显著小于0.05,所以拒绝H i°,H 20,认为含铜量和试验温度 都会对钢的冲击值产生显著影响作用 . 4 下面记录了三位操作工分别在四台不同的机器上操作三天的日产量: 检验的问题:H 0: 1 计算结果: H i : i 不全相等

应用数理统计习题答案 西安交大 施雨

应用数理统计答案 学号: 姓名: 班级:

目录 第一章数理统计的基本概念 (2) 第二章参数估计 (14) 第三章假设检验 (24) 第四章方差分析与正交试验设计 (29) 第五章回归分析 (32) 第六章统计决策与贝叶斯推断 (35) 对应书目:《应用数理统计》施雨著西安交通大学出版社

第一章 数理统计的基本概念 1.1 解:∵ 2 (,)X N μσ ∴ 2 (,)n X N σμ ∴ (0,1)N 分布 ∴(1)0.95P X P μ-<=<= 又∵ 查表可得0.025 1.96u = ∴ 2 2 1.96n σ= 1.2 解:(1) ∵ (0.0015)X Exp ∴ 每个元件至800个小时没有失效的概率为: 800 0.00150 1.2 (800)1(800) 10.0015x P X P X e dx e -->==-<=-=? ∴ 6个元件都没失效的概率为: 1.267.2 ()P e e --== (2) ∵ (0.0015)X Exp ∴ 每个元件至3000个小时失效的概率为: 3000 0.00150 4.5 (3000)0.00151x P X e dx e --<===-? ∴ 6个元件没失效的概率为: 4.56 (1)P e -=- 1.4 解:

i n i n x n x e x x x P n i i 1 2 2 )(ln 2121)2(),.....,(1 22 =-- ∏∑ = =πσμσ 1.5证: 2 1 1 2 2)(na a x n x a x n i n i i i +-=-∑∑== ∑∑∑===-+-=+-+-=n i i n i i n i i a x n x x na a x n x x x x 1 2 2 2 2 11) ()(222 a) 证: ) (1111 1+=+++=∑n n i i n x x n x ) (1 1 )(1 1 11n n n n n x x n x x x n n -++=++=++

英语作文讲义

互动问答: 听见音,敲1;看见老师,敲2;两者都OK,敲3. 1. 在屏幕前的你是家长还是学生? 2. 你喜欢写作文吗? 3. 你在写英语作文中遇到的障碍是什么?

趣味问答: 1.Q: Which letter is a part of your face? 2. Q: Which letter is a question? 3. Q :Who is closer to you, your mum or your dad ? 4. Q: How many sides does a house have ?Ii ——eye Yy ——why Mum, because dad is father (同音词farther 更远的). Two , inside and outside .

讲在前面的: 1. 我们致力于打造服务济南市六年级学生的系列课程,以提升学习兴趣、培养学习习惯、提高学习成绩为重,帮助学生做好六升初的过渡衔接。 2. 本次集训营也为系列课程,一共8讲,希望学生和家长准时观看收听,认真记好笔记,集齐8次课程笔记,有资格参加笔记大赛,赢取奖品。

1. 不要刷屏,好好听讲 2. 手边笔记本,随听随写

Welcome to Amanda’s class!

更多模板烦请登陆:https://www.doczj.com/doc/8a2361803.html,/ppt/ 版权声明感谢您下载觅知网平台上提供的PPT 作品,为了您和觅知网以及原创作者的利益,请勿复 制、传播、销售,否则将承担法律责任!觅知网将对作品进行维权,按照传播下载次数进 行十倍的索取赔偿!1.在觅知网出售的PPT 模板是免版税类(RF :Royalty-Free )正版受《中国人民共和国著 作法》和《世界版权公约》的保护,作品的所有权、版权和著作权归觅知网所有,您下载的是PPT 模板素材的使用权。 2.不得将觅知网的PPT 模板、PPT 素材,本身用于再出售,或者出租、出借、转让、分销、发布或者作为礼物供他人使用,不得转授权、出卖、转让本协议或者本协议中的权利。 自我介绍 姓 名:张蕾(Amanda)简介:10年朴新教学经验累计学员6000+人次,累计课时8000+。做过教学管理工作,教材编写工作。获奖经历:2016年集团优秀教师奖2018年教师赛课一等奖2019年组内阶段赛课一等奖

2.考研英语二作文2020年讲义

考研英语二写作讲义 2020年考研英语(二)考试大纲中作文部分的描述如下: 该部分由A、B两节组成,主要考查考生的书面表达能力。共25分。 A节:题型有两种,每次考试选择其中的一种形式。 备选题包括: 1)考生根据所给情景写出一篇约100词(标点符号不计算在内)的应用性短文,包括私人和公务信函、备忘录、报告等。 2)考生根据所提供的汉语文章,用英语写出一篇80~100词的该文摘要。 考生在答题卡2上作答,共10分。 B节: 要求考生根据所规定的情景或给出的提纲,写出一篇200词以上的英语说明文或议论文。提示情景的形式为图画、图表或文字等。考生在答题卡2上作答。共20分。 英语二2020年起才有小作文,总共考过2次,都是书信,均为两种类型的混合信:2020年是感谢+邀请,2020年是祝贺+建议。2020年复习重点依然是书信,注意类型的混合搭配。 ◆书信类: 基本模式 书信分为三个部分来写:称呼、正文和签名。 书信格式如下:

常用万能句:I am Zhang Wei, I am writing to you for the purpose of +V-ing. 或者:I am _______, I am writing this letter to +V. 表达建议:I am Li Ming. I am writing to you for the purpose of expressing my deepest concern about____________(具体内容题目给出). I sincerely wish you consider my suggestion. 表达感谢;I am ____(具体名字题目会给出). I am writing to you for the purpose of expressing my sincere thanks. 表达道歉;My name is ______. I am writing to you for the purpose of making an apology due to my carelessness. 表达邀请;I am______(具体名字题目会给出), I am writing to you for the purpose of honorably inviting you to _______ 表达不满;I am_____(具体名字题目会给出). I am writing to you for the purpose of making a complaint about your_____ (具体事情题目会给出) 表达祝贺;I am____(具体名字题目给出). I am writing to you to express sincere congratulations on your success and happiness! 表达求职:I am Li Ming, who graduates from Shandong University. I am writing you for the purpose of obtaining the position of ____(具体职位题目会给出) 表达辞职;My name is _______(具体名字题目会给出). I am writing to you for the purpose of quitting my present job. 万能建议段:It must be pointed out that the situation of ____________ is going from bad to worse. Therefore, as far as I am concerned,first of all, Governments departments are required to take effective measures to solve this serious problem. Secondly, different classes in the society need to cooperate closely to pay attention to this problem. At last, as individuals, we should care about it and set up examples for others. 万能祝贺段:As a matter of fact, I know that you have devoted a great deal of precious time as well as energy to it. During the past valuable time and experience, you have obtained not only rich work experience but also wide-spreading interpersonal relationship. Therefore, the past experience has laid solid foundation for the present success. I wish you to accept my heartfelt(发自内心的) congratulations. In fact, I have also learned a lot from your achievement this time. You have set up a brilliant example for us to make success and realize our dreams in the near future. 万能求职段:As a matter of fact, I am quite sure that I am extremely qualified for this position. To begin with, my major learned in the college matches with this position very much. In the second place, I have got three years’ rich experience concerning this position. Thirdly, my character is not only careful but also patient. In addition, I am very open-minded person. Therefore, I consider I am the right person for this position.

应用数理统计课后习题 清华大学出版社 杨虎 钟波第三章作业参考答案

第 三 章 作 业 参 考 答 案 2、解:计算矩估计:2 1)1(1 ++= +?= ? αααα dx x x EX , 令 X EX =++= 2 1αα ,解得 1 2-1?1-=X X α ; 计算极大似然估计:α α αα α)()1()1()()(1 1 1 ∏∏∏ ===+=+= = n i i n n i i n i i x x x f L )ln()1ln()(ln 1 ∏=++=?n i i x n L ααα0 )ln(1 )(ln 1 =++= ??? ∏=n i i x n L αα α 解得 ) ) ln(1(?1 2∏=+-=n i i x n α ; 将样本观测值代入,得到估计值分别为0.3077?1=α ,0.2112?2=α。 6、 解:(1)由例3.2.3可知,μ的极大似然估计分别为 X =μ ?, 05.0)(1)(=-Φ-=>μA A X P )645.1(95.0)(Φ==-Φ?μA 645 .1+=?μA ,由46页上极大似然估计的不变性可知645.1??+=μA ; (2)由例3.2.3可知,2 σμ,的极大似然估计分别为 ∑=-= =n i i X X n X 1 2 2 ) (1 ??σ μ,, 05.0)( 1)(=-Φ-=>σ μ A A X P )645.1(95.0)( Φ==-Φ?σ μ A σ μ645.1+=?A ,由46页上极大似然估计的不变性可知σμ?645.1??+=A 。 8、解:计算2 2 2 2222)()()(σσ μC n S CE X E CS X E -+ =-=-,由题意则有 2 2 2 2 μσ σ μ=-+ C n ,解得n C 1= 。

研究生《应用数理统计基础》庄楚强 四五章部分课后答案

4-45. 自动车床加工中轴,从成品中抽取11根,并测得它们的直径(mm )如下: 10.52,10.41,10.32,10.18,10.64,10.77,10.82,10.67,10.59,10.38,10.49 试用W 检验法检验这批零件的直径是否服从正态分布?(显著性水平05.0=α) (参考数据:) 4-45. 解:数据的顺序统计量为: 10.18,10.32,10.38,10.41,10.49,10.52,10.59,10.64,10.67,10.77,10.82 所以 6131 .0][)()1(5 1 ) (=-= -+=∑k k n k k x x a L , 又 5264.10=x , 得 38197 .0)(11 1 2 =-∑=i i x x 故 984.0) (11 1 2 2 =-= ∑=i i x x L W , 又 当n = 11 时,85.005.0=W 即有 105.0<

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