math117finalreview-fall2014

8/16/2019 Math117FinalReview-Fall2014

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Nation Women in

Parliament (%)

Female Economic

Activity

Iceland 33.3 87

Australia 28.3 79 Canada 24.3 83

Japan 10.7 65

United States 15.0 81

New Zealand 32.2 81

MATH 117 - Elements of Statistics

Fall 2014 Final Exam Review for Chapters 1 – 11, 15

From the textbook: Statistics: The Art and Science of Learning from Data (3rd ed.), by Alan Agresti and Christine Franklin

1. UW Student Survey In a University of Wisconsin (UW)

study about alcohol abuse among students, 100 of the

40,858 members of the student body in Madison were

sampled and asked to complete a questionnaire. One

question asked was, “On how many days in the past

week did you consume at least one alcoholic drink?”a. Identify the population and the sample.

b. For the 40,858 students at UW, one characteristic of

interest was the percentage who would respond

“zero” to this question. For the 100 students

sampled, suppose 29% gave this response. Does this

mean that 29% of the entire population of UW

students would make this response? Explain.

c. Is the numerical summary of 29% a sample statistic,

or a population parameter?

2. Median versus mean sales price of new homes In

December 2010, the US Census Bureau reported that themedian US sales price of new homes was $241,500.

Would you expect the mean sales price to have been

higher or lower? Explain.

3. Female Heights According to a recent report from the

US National Center for Health Statistics, females

between 25 and 34 years of age have a bell -shaped

distribution for height, with a mean of 65 inches and

standard deviation of 3.5 inches.

a. Give an interval within which about 95% of the

heights fall.

b. What is the height for a female who is 3 standard

deviations below the mean? Would this be a ratherunusual height? Why?

4. High School Graduation Rates The distribution of high

school graduation rates in the United States in 2004 had

a minimum value of 78.3 (Texas), first quartile of 83.6,

median of 87.2, third quartile of 88.8, and maximum

value of 92.3 (Minnesota) (Statistical Abstract of the

United States, 2006).

a. Report the range and the interquartile range.

b. Would a box plot show any potential outliers?

Explain.

5. Blood Pressure A World Health Organization study (the

MONICA project) of health in various countries reported

that in Canada, systolic blood pressure readings have a

mean 121 and a standard deviation of 16. A reading

above 140 is considered to be high blood pressure.

a. What is the z-score for a blood pressure reading of

140? How is this z-score interpreted?

b. The systolic blood pressure values have a bell-

shaped distribution. Report an interval within which

about 95% of the systolic blood pressure values fall.

6. Life After Death for Males and Females In a recent General

Social Survey, respondents answered the question, “Do you

believe in a life after death?” The table shows the responses

cross-tabulated with gender.

Opinion About Life After Death by Gender

Gender Opinion About Life After Death

Yes No

Male 621 187

Female 834 145

a. Construct a table of conditional proportions.

b. Summarize results. Is there much difference between

responses of males and females?

7. Women in Government and Economic Life The OECD

(Organization for Economic Cooperation and Development)

consists of advanced, industrialized countries that accept theprinciples of representative democracy and a free market

economy. For the nations outside of Europe that are in the

OECD, the table shows UN data from 2007 on the percentage

of seats in parliament held by women and female economic

activity as a percentage of the male rate.

a. Treating women in parliament as the response variable,

prepare a scatterplot and find the correlation. Explain

how the correlation relates to the trend shown in the

scatterplot.

b. Use software or a calculator to find the regression

equation. Explain why the y-intercept is not meaningful.

c. Find the predicted value and residual for the United

States. Interpret the residual.

d. With UN data for all 23 OECD nations, the correlation

between these variables is 0.56. For women in

parliament, the mean is 26.5% and the standard

deviation is 9.8%. For female economic activity, the

mean is 76.8 and the standard deviation is 7.7. Find the

prediction equation, treating women in parliament as

the response variable.


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8. Predict Crime Using Poverty A recent analyses of data for

the 50 US states on y = violent crime rate (measured as

number of violent crimes per 100,000 people in the state)

and x = poverty rate (percent of people in the state living

at or below the poverty level) yielded the regression

equation = 209.9+ 25.5.a. Interpret the slope.

b. The state poverty rates ranged from 8.0 (for Hawaii)to 24.7 (for Mississippi). Over this range, find the

range of predicted values for the violent crime rate.

c. Would the correlation between these variables be

positive or negative? Why?

9. Football Discipline A large southern university had

problems with 17 football players being disciplined for

team rule violations, arrest charges, and possible NCAA

violations. The online Atlanta Journal Constitution ran a

poll with the question, “Has the football coach lost control

over his players?” having possible responses, “Yes, he’s

been too lenient,” and “No, he can’t control everything

teenagers do.”

a. Was there potential for bias in this study? If so, what

types of bias?

b. The poll results after two days were

Yes 6012 93%

No 487 7%

Does this large sample size guarantee that the results are

unbiased? Explain.

10. Video games mindless? “Playing video games not so

mindless.” This was the headline of a CNN news report

about a study that concluded that young adults whoregularly play video games demonstrated better visual

skills than young adults who do not play regularly. Sixteen

young men volunteered to take a series of tests that

measured their visual skills; those who had played video

games in the previous six months performed better on the

test than those who hadn’t played.

a. What are the explanatory and response variables?

b. Was this an observational study or an experiment?

Explain.

c. Specify a potential lurking variable. Explain.

11. Peyton Manning Completions As of the end of the 2010

NFL season, Indianapolis Colts quarterback Peyton

Manning, throughout his 13-year career, completed 65%

of all of his pass attempts. Suppose the probability each

pass attempted in the next season has probability 0.65 of

being completed.

a. Does this mean that if we watch Manning throw 100

times in the upcoming season, he would complete

exactly 65 passes? Explain.

b. Explain what this probability means in terms of

observing him over a longer period, say for 1000 passes

over the course of the next two seasons assuming

Manning is still at his typical playing level. Would it be

surprising if his completion percentage over a large

number of passes differed significantly from 65%?

12. Driver’s Exam Three 15-year-old friends with no particulabackground in driver’s education decide to take the written

part of the Georgia Driver’s Exam. Each exam was graded a

a pass (P) or a failure (F).

a. How many outcomes are possible for the grades

received by the three friends together? Using a tree

diagram, list the sample space.

b. If the outcomes in the sample space in part a are

equally likely, find the probability that all three pass the

exam.

c. In practice, the outcomes in part a are not equally likely

Suppose that statewide 70% of 15-year-olds pass the

exam. If these three friends were a random sample of

their age group, find the probability that all three pass.

d. In practice, explain why probabilities that apply to a

random sample are not likely to be valid for a sample of

three friends.

13. Grandparents Let X = the number of living grandparents

that a randomly selected adult American has. According to

recent General Social Surveys, its probability distribution is

approximately P(0) = 0.71, P(1) = 0.15, P(2) = 0.09, P(3) =

0.03, P(4) = 0.02.

a. Does this refer to a discrete or a continuous random

variable? Why?

b. Show that the probabilities satisfy the two conditionsfor a probability distribution.

c. Find the mean of this probability distribution.

14. Z-score and Tail Probability

a. Find the z-score for the number that is less than only 1%

of the values of a normal distribution. Sketch a graph to

show where this value is.

b. Find the z-scores corresponding to the (i) 90th

and (ii)

99th

percentiles of a normal distribution.

15. Cloning Butterflies The wingspans of recently cloned

monarch butterflies follow a normal distribution with mean

9 inches and standard deviation 0.75 inches. What

proportion of the butterflies has a wingspan

a. less than 8 inches?

b. wider than 10 inches?

c. between 8 and 10 inches?

d. ten percent of the butterflies have a wingspan wider

than how many inches?


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16. Exam Performance An exam consists of 50 multiple-

choice questions. Based on how much you studied, for any

given question you think you have a probability of p = 0.70

of getting the correct answer. Consider the sampling

distribution of the sample proportion of the 50 questions

on which you get the correct answer.

a. Find the mean and standard deviation of the sampling

distribution of this proportion.b. What do you expect for the shape of the sampling

distribution? Why?

c. If truly p = 0.70, would it be very surprising if you got

correct answers on only 60% of the questions? Justify

your answer by using the normal distribution to

approximate the probability of a sample proportion of

0.60 or less.

17. Aunt Erma’s Restaurant In Example 5 about Aunt Erma’s

Restaurant, the daily sales follow a probability distribution

that has a mean of µ = $900 and a standard deviation of σ

= $300. This past week the daily sales for the seven days

had a mean of $980 and a standard deviation of $276.

a. Identify the mean and standard deviation of the

population distribution.

b. Identify the mean and standard deviation of the data

distribution. What does the standard deviation

describe?

c. Identify the mean and the standard deviation of the

sampling distribution of the sample mean for samples

of seven daily sales. What does this standard

deviation describe?

18. Approval Rating for President Obama A July 2011 Gallup

poll based on the responses of 1500 adults indicated that46% of Americans approve of the job Barack Obama is

doing as president. One way to summarize the findings of

the poll is by saying, “It is estimated that 46% of American

approve of the job Barack Obama is doing as president.

This estimate has a margin of error of plus or minus 3%.”

How could you explain the meaning of this to someone

who has not taken a statistics course?

19. Vegetarianism Time magazine (July 15, 2002) quoted a

poll of 10,000 Americans in which only 4% said they were

vegetarian.

a. What has to be assumed about this sample to

construct a confidence interval for the populationproportion of vegetarians?

b. Construct a 99% confidence interval for the

population proportion. Explain why the interval is so

narrow, even though the confidence level is high.

c. In interpreting this confidence interval, can you

conclude that fewer than 10% of Americans are

vegetarians? Explain your reasoning.

20. Grandpas Using Email When the GSS asked in 2004,

“About how many hours per week do you spend sending

and answering email?” the eight males in the sample of age

at least 75 responded:

0, 1, 2, 2, 7, 10, 14, 15

a. The TI-83+/84 screen shot shows results of a statistical

analysis for finding a 90% confidence interval. Identify

the results shown and explain how to interpret them.b. Find and interpret a 90% confidence interval for the

population mean.

Explain why the population distribution may be skewed

right. If this is the case, is the interval you obtained in

part b useless, or is it still valid? Explain.

TInterval(2.34,10.41) = 6.38 = 6.02 = 8.00 21. US Popularity In 2007, a poll conducted for the BBC of

28,389 adults in 27 countries found that the United Stateshad fallen sharply in world esteem since 2001

(www.globescan.com). The United States was rated third

most negatively (after Israel and Iran), with 30% of those

polled saying they had a positive image of the United States

a. In Canada, for a random sample of 1008 adults, 56%

said the United States is mainly a negative influence in

the world. True or false: The 99% confidence interval of

(0.52, 0.60) means that we can be 99% confident that

between 52% and 60% of the population of all Canadian

adults have a negative image of the United States.

b. In Australia, for a random sample of 1004 people, 66%

said the United States is mainly a negative influence in

the world. True or false: The 95% confidence interval of(0.63, 0.69) means that for a random sample of 100

people, we can be 95% confident that between 63 and

69 people in the sample have a negative image of the

United States.

22. Driving After Drinking In December 2004, a report based o

the National Survey on Drug Use and Health estimated that

20% of all Americans of ages 16 to 20 drove under the

influence of drugs or alcohol in the previous year (AP,

December 30, 2004). A public health unit in Wellington, New

Zealand, plans a similar survey for young people of that age

in New Zealand. They want a 95% confidence interval tohave a margin of error of 0.04.

a. Find the necessary sample size if they expect results

similar to those in the United States.

b. Suppose that in determining the sample size, they use

the safe approach that sets = 0.50 in the formula forn. Then, how many records need to be sampled?

Compare this to the answer in part a. Explain why it is

better to make an educated guess about what to expect

for ̂, when possible.


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23. Mean property tax. A tax assessor wants to estimate the

mean property tax bill for all homeowners in Madison,

Wisconsin. A survey 10 years ago got a sample mean and

standard deviation of $1400 and $1000.

a. How many tax records should the tax assessor

randomly sample for 95% confidence interval for the

mean to have a margin of error equal to $100? Whatassumption does your solution make?

b. In reality, suppose that they’d now get a standard

deviation equal to $1500. Using the sample size you

derived in part a, without doing any calculation,

explain whether the margin of error for a 95%

confidence interval would be less than $100, equal to

$100 or more than $100.

c. Refer to part b. Would the probability that the sample

mean falls with $100 of the population mean be less

than 0.95, equal to 0.95 or greater than 0.95? Explain.

24. H0

or Ha? For each of the following hypothesis explain

whether it is a null hypothesis or alternative hypothesis:

a. For females, the population mean on the political

ideology scale is equal to 4.0.

b. For males, the population proportion who support the

death penalty is larger than 0.50.

c. The diet has an effect; the population mean can

change in weight being less than 0.

d. For all subway sandwich stores worldwide, the

difference between sales this month and in the

corresponding month last year has been a mean of 0.

25. ESP A person who claims to possess extrasensory

perception (ESP) says she can guess more often than notthe outcome of a flip of a balanced coin. Out of 20 flips,

she guesses correctly 12 times. Would you conclude that

she truly has ESP? Answer by reporting all five steps of a

significance test of the hypothesis that each of her guesses

has probability 0.50 of being correct against the

alternative that corresponds to her having ESP.

26. Jurors and gender A jury list contains the names of all

individuals who may be called for jury duty. The

proportion of the available jurors on the list who are

women is 0.53. If 40 people are selected to serve as

candidates for being picked on the jury, show all steps of

significance test of the hypothesis that the selections are

random with respect to gender.

a. Set up notation and hypotheses, and specify

assumptions.

b. 5 of the 40 selected were women. Find the test

statistic.

c. Report the P-value, and interpret.

d. Explain how to make a decision using a significance

level or 0.01.

27. Type I and Type II errors Refer to the previous exercise.

a. Explain what type I and Type II errors mean in the

context of that exercise.

b. If you made an error with the decisions in part d, is it a

Type I or Type II error?

28. Tennis balls in control? When it is operating correctly amachine for manufacturing tennis balls produces balls with a

mean weight of 57.6 grams. The last eight balls

manufactured had weights

57.3, 57.4, 57.2, 57.5, 57.4, 57.1, 57.3, 57.0

a. Using a calculator or software, find the test statistic and

P-value for a test of whether the process is in control

against the alternative that the true mean of the

process now differs from 57.6.

b. For significance level of 0.05, explain what you would

conclude. Express your conclusion so it would be

understood by someone who never studied statistics.

c. If your decision in part b is in error, what type of error

have you made?

29. Wage claim false? Management claims that the mean

income for all senior-level assembly-line workers in a large

company equals $500 per week. An employee decides to

test the claim, believing that it is actually less than $500. For

a random sample of nine employees, the incomes are:

430, 450, 450, 440, 460, 420, 430, 450, 440.

Conduct a significance test of whether the population mea

income equals $500 per week against the alternative that i

less. Include all assumptions, the hypotheses, test statisticsP-value, and interpret the results in context.

30. Legal trial errors Consider the analogy discussed in Section

9.4 between making a decision about null hypothesis in a

significance test and making a decision about the innocence

or guilt of a defendant in a criminal trial.

a. Explain the difference between Type I and Type II errors

in the trial setting.

b. In this context, explain intuitively why decreasing the

chance of Type I error increases the chance of Type II

error.


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31. Gender and belief in afterlife. This table shows results from

the 2008 General Social Survey on gender and whether or

not on believes in an afterlife.

Belief in Afterlife

Gender Yes No Total

Female 599 111 710

Male 425 168 593

a. Denote the population proportion who believe in an

afterlife by p1 for females and by p2 for males.

Estimate p1, p2 and ( p1 – p2 ).

b. Find the standard error for the estimate of ( p1 – p2 ).

Interpret.

c. Construct a 95% confidence interval for ( p1 – p2 ). Can

you conclude which of and is larger? Explain.d. Suppose that, unknown to us, p1 = 0.81 and p2 = 0.72.

Does the confidence interval in part c contain the

parameter it is designed to estimate? Explain.

32. Belief depend on gender? Refer to the previous exercise.

a. Find the standard error of ̂ ̂ for a test of: = .b. For two-sided test, find the test statistic and P-value,

and make a decision using significance level 0.05.

Interpret.

c. Suppose that actually p1 = 0.81 and p2 = 0.72. Was the

decision in part b in error?

d. State the assumption on which the methods in this

exercise are based.

33. Heavier horseshoe crabs more likely to mate? A study of a

sample of horseshoe crabs on a Florida island (J. Brockman,Ethology , vol. 102, 1996, pp. 1-21) investigated the factors

that were associated with whether or not female crabs had a

male crab mate. Basic statistics, including five-number

summary on weight (kg) for the 111 female crabs who had a

male crab nearby and for the 62 female crabs who did not

have a male nearby, are given in the table. Assume that

these horseshoe crabs have the properties of random

sample of all such crabs.

Summary Statistics for Weights of Horseshoe Crabs

# Mean Std. Dev. Min Q1 Med Q3 Max

Mate 111 2.6 0.6 1.5 2.2 2.6 3.0 5.2

No Mate 62 2.1 0.4 1.2 1.8 2.1 2.4 3.2

34. Sex roles A study of the effect of the gender of the tester

the sex-role differentiation scores16

in Manhattan gave a

random sample of preschool children the Occupational

Preference Test. Children were asked to give three choices

what they wanted to be when they grew up. Each occupat

was rated on a scale from 1 (traditionally feminine) to 5

(traditionally masculine), and a child’s score was the meanthe three selections. When the tester was male, the 50 gir

had = 2.9 and s = 1.4, whereas when the tester wasfemale, the 90 girls had = 3.2 and s = 1.2. Show all stepa test of the hypothesis that the population mean is the sa

for the female and male testers, against the alternative th

they differ. Report the P-value and interpret.

35. Internet book prices Anna’s project for her introductory

statistics course was to compare the selling prices of

textbooks at two Internet bookstores. She first took a

random sample of 10 textbooks used that term in courses

her college, based on the list of texts compiled by the coll

bookstore. The prices of those textbooks at two Internet

sites were

Site A: $115, $79, $43, $140, $99, $30, $80, $99, $119, $6

Site B: $110, $79, $40, $129, $99, $30, $69, $99, $109, $6

a. Are these independent samples or dependent sample

Justify your answer.

b. Find the mean for each sample. Find the mean of the

difference scores. Compare, and interpret.

c. Using software or a calculator, construct a 90%

confidence interval comparing the population mean

prices of all the textbooks used that term at her colle

Interpret.

36. Comparing book prices 2 For the data in the previous

exercise, use software or a calculator to perform a

significance test comparing the population mean prices.

Show all steps of the test, and indicate whether you woul

conclude that the mean price is lower at one or the two

Internet bookstores.

a. Sketch box plots for the weight distributions of the two

groups. Interpret by comparing the groups with respect

to the shape, center, and variability.

b. Estimate the difference between the mean weights of

the female crabs who have mates and those who don't.

c. Find the standard error for the estimate in part b.

d. Construct a 90% confidence interval for the difference

between the population mean weights, and interpret.


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37. Down syndrome diagnostic test The table shown, from

Example 8 in Chapter 5, cross-tabulates whether a fetus

has Down syndrome by whether or not the triple blood

diagnostic test for Down syndrome is positive (that is,

indicates that the fetus has Down syndrome).

a. Tabulate the conditional distributions for the blood

test result given the true Down syndrome status.

b. For the Down cases, what percentage was diagnosed

as positive by the diagnostic test? For the unaffected

cases, what percentage got a negative test result?

Does the diagnostic test appear to be a good one?

c. Construct the conditional distribution on Down

syndrome status, for those who have a positive test

result. (Hint : You condition on the first column total

and find proportions in that column.) Of those cases,

what percentage truly have Down syndrome? Is the

result surprising? Explain why this probability is

small.

Blood Test Result

Down Syndrome Status Positive Negative Total

D (Down) 48 6 54

D0

(unaffected) 1307 3921 5228

Total 1355 3927 5282

40. Tanning experiment Suppose the tanning experiment

described in Examples 1 and 2 used only four participants,

two for each treatment.

a. Show the six possible ways the four ranks could be

allocated, two to each treatment, with no ties.

b. For each possible sample, find the mean rank for each

treatment and the difference between the mean ranks.c. Presuming H0 is true of identical treatment effects,

construct the sampling distribution of the difference

between the sample mean ranks for the two

treatments.

41. Test for tanning experiment Refer to the previous exercise

For the actual experiment, suppose the participants using

the tanning studio got ranks 1 and 2 and the participants

using the tanning lotion got ranks 3 and 4.

a. Find and interpret the P-value for the alternative

hypothesis that the tanning studio tends to give better

tans than the tanning lotion.

b. Find and interpret the P-value for the alternative

hypothesis that the treatments have different effects.

c. Explain why it is a waste of time to conduct this

experiment if you plan to use a 0.05 significance level to

make a decision.

38. Down and chi-squared For the data in the previousexercise, = 114.4. Show all steps of the chi-squaredtest of independence.

39. Gender gap? Exercise 11.1 showed a 2 x 3 table relating

gender and political party identification, shown againhere. The chi-squared statistic for these data equals 8.294.

Conduct all five steps of chi-squared test.

Political Party

Sex Dem Indep Repub

F 422 381 273

M 299 365 232


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Answers

1.

UW Student Survey

a. The population is the entire UW student body of 40,858. The sample is the 100 students who were asked to complete thequestionnaire.

b. This value would not necessarily equal the value of the entire population of UW students. It is quite possible that the sample of 100 is not exacrepresentative of the whole student body. This percentage is only an estimate of the percentage of all students who would respond this way. I

unlikely that any single sample of 100 would have a percentage that was exactly the percentage of the entire population.

c.

The numerical summery is the sample statistics because it only summarizes for a sample, not for a population.

2. Median versus mean sales price of new homes

We would expect the mean sales price to have been higher due to the distribution being skewed to the right. A few very expensive

homes will greatly affect the mean, but not the median sales price.

3.

Female Heights

a. According to the Empirical Rule, 95% of the scores in a bell-shaped distribution fall within two standard deviation of the mean.

̅ 2 = 65 2(3.5) = 58 ̅ + 2 = 65 + 2(3.5) = 72

Thus, 95% of the heights likely fall between 58 and 72 inches.b. The height for a woman who is three standard deviations below the mean is 54.5.

̅ 3 = 65 3(3.5) = 54.5 This is on the cusp of what would be considered an outlier according to the z-score criterion. Scores that are beyond three standa

deviations from the mean are considered to be potential outliers. So, yes, this height is bordering on unusual.

4. High School Graduation Rates

a.

The range is the difference between the lowest and the highest scores: 92.3 78.3 = 14. The interquartile range (IQR) is the difference between scores at the 25

th and 75

th percentiles: IQR=Q3Q1= 88.8 83.6 = 5.2.

b. 1.5(IQR)= 7.8 from Q1 or Q3; this criterion suggests that potential outliers would be scores less than 75.8 and greater than 96.6.There are no scores beyond these values, so it would not indicate any potential outliers.

5.

Blood Pressure

a.

= − ̅ = − = 1.19 A score of 1.19 indicates that a person with a blood pressure of 140, the cutoff having a high blood pressure, falls 1.19standard deviations above the mean.

b. About 95% of all the values in a bell-shaped distribution fall within two standard deviation of the meanin this case 32. Subtractingtwo times the standard deviation from the mean and adding two times the standard deviation to the mean tells us about 95% ofsystolic blood pressures fall between 89 and 153.

6. Life After Death for Males and Females

Using 2008 data:

a. Opinion about life after death

Gender Yes No Total

Male 0.77 0.23 808

Female 0.85 0.15 979

b.

Overall, both male and women are more likely to believe in life after death than not, but women are somewhat more likely to do

so.


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7.

Women in Government and Economic Life

a.

The correlation between women in parliament and female economic activity is 0.745. This correlation is supported by the positive

linear trend evident in the scatterplot, but note this is largely driven by the point (for Japan) having female economic activity very low

(65).

b. The regression equation is = 48.91 + 0.1986 . Since the –intercept correspond to an value of 0, the –intercept is not

meaningful in this case. (Female economic activity=0 is outside the range of the observed data).c.

The predicted value for U.S. is 48.91 +0.1986(81) = 25.5 with 15.0 25.5 = 10.5 as the corresponding residual. Theregression equation underestimates the percentage of women in parliament by 10.5% for the U.S.

d. = 0.56 .. = 0.7127 and = 26.5 0.7127(76.8) = 28.24. Thus the prediction equation is = 28.24 +0.7127.8.

Predict Crime Using Poverty

a. The slope of 25.5 indicates that for each percentage increase in the poverty, the predicted violent crime rate increases by 25.5 crime

per 100, 000people statewide.

b. The range predicted values runs from 413.9 to 839.8 crimes per 100, 000 people.

= 209.9 +25.5(8.0) = 413.9 = 209.9 + 25.5(24.7) = 839.75 (rounds to 839.8)

c. The correlation would be positive we know this because the slope is positive and the slope and the correlation have always the same

sign.

9. Football Discipline

a. Because this is a volunteer sample, there is the potential for sampling bias, both because of the sample is not selected randomly

(those who have responded might have been those felt the most strongly) and because of the undercover age (anyone without

Internet access would not have been able to participate). There also is potential for response biases because of the statements are

leading.

b. If the sample is biased due to undercoverage and lack of random sampling, it does not matter how big the sample. It is almost alway

better to have a small random sample than a large volunteer sample.


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10.

Video games mindless?

a. The explanatory is the history playing video games, and the response variable is visual skills.

b. This was an observation study because the men were not randomly assigned to treatment (played video games versus hadn’t played

those who already were in these groups were observed.

c.

One possible lurking variable is reaction time. Excellent reaction times might make it easier and therefore more fun, to play video

games, leading young people to be more likely to play. Excellent reaction times also might lead young men to perform better on task

measuring visual skills. These young men might have performed better on tasks measuring visual skills regardless of whether they

played video games.11.

Peyton Manning Completions

a. No. What it means that in 100 passes we expect to see about 65 completions, but actually the number may vary somewhat.

b. If Manning is still at his typical paying level, it would be quite surprising if his completion percentage over a large number of passes

differed significantly from 0.6. The more passes he throws, the closer the observed percentage should be to 0.65.

12.

Driver’s Exam

a. There are 2 × 2 × 2 = 8 possible outcomes. Construct the tree yourself. An alternative way of presenting the outcomes:let P=”pass” and F=”failure”, the outcomes are: PPP, PPF, PFP, FPP, PFF, FPF, FFP, FFF.

b. If the eight outcomes are equally likely, all three pass the exam is = 0.125. This also be calculated by multiplying the probability th

the first would pass(0.5), by the probability that the second would pass(0.5), by the probability that the third would pass(0.5),

0.5 × 0.5 ×0.5 = 0.125.

c.

If the three friends were a random sample of their age group, the probability that all would pass 0.7 ×0.7 ×0.7 = 0.343. d. The probabilities that apply to a random sample are not likely to be valid for a sample of three friends because the three friends are

likely to be similar on many characteristics that might affect the performance on such a test ( e.g., IQ) . In addition it is possible that

they studied together.

13. Grandparents

a. This refers to a discrete random variable because there can only be whole numbers of grandparents. One can’t have 1.78

grandparents.

b. The probabilities satisfy the two conditions for a probability distribution because they each fall between 0 and 1, and the sum of the

probabilities of all possible values is 1.

c. The mean of this probability is 0(0.71) + 1(0.15) + 2(0.09) + 3(0.03) + 4(0.02) ≈ 0.5. 14. Z-score and Tail Probability

a.

The z- score that is less than only 1% of the values would be greater than 99% of the values. If we look up 0.99 on Table A, we see th

z-score is 2.33.

b.

(i) The z-score that is above 0.90 is 1.28.

(ii) The z-score that is above 0.99 is 2.33.

15. Cloning Butterflies

a. = − . = 1.33; according to Table A, 0.092 of the butterflies have a wingspan less than 8 inches.b. = −. = 1.33; according to Table A, 0.092 of the butterflies have a wingspan wider than 10 inches.c. From parts a and b, 1 2(0.092) = 0.816 of the butterflies have wingspans between 8 and 10 inches.d.

From Table A, the 90th

percentile is 1.28. 1.28 ×0.75 + 9 = 9.96 inches. Thus 10% of the butterflies have wingspan wider than 9.96inches.

16.

Exam Performance

a. Mean= = 0.70, standard error= (−) = .(−.) = 0.0648.b. Since = 50, by the Central Limit Theorem, we would expect the shape of the sampling distribution to be approximately normal

with mean=0.70 and standard deviation≈ 0.0648.c. The z-score for 0.60 is

.−.. = 1.54 giving a cumulative probability of 0.06. It would not be surprising to only get 60% of the

answers correct.


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17.

Aunt Erma’s Restaurant

a. The population distribution has a mean $900 and a standard deviation of $300.

b. The population distribution has a mean $980 and a standard deviation of $276. The standard deviation of the data distribution

describes the spread of the daily sales values for this past week.

c.

The mean of the sampling distribution of the sample mean = =$900; standard error= √ =

√ ≈ 113.4 dollars. The standarderror describes the spread of the sample means based on sample of seven days sales.

18.

Approval Rating for President Obama

We could tell to someone who hadn’t taken a statistics course that we do not know the exact percentage of the population who

approve of the job Barack Obama is doing as president, but we are quite sure that it is within 3% of 46%, that is, between 43% and

49%.

19. Vegetarianism

a. We must assume that the data were obtained randomly.

b. = (−) = .(−.), ≈ 0.002.

The confidence interval is ̂ ± () Lower limit: 0.04 2.58(0.002) = 0.035. Upper limit: 0.04 + 2.58(0.002) = 0.045. The interval is so narrow, even though the confidence level is high, mainly because of the very large sample size. The very largesample size contributes to a small standard error by providing a very large denominator for the standard error calculation.

c. We can conclude that fewer than 10% of Americans are vegetarians because 10% falls above the highest believable value in the

confidence interval.

20. Grandpas Using Email

a. The first result is a 90% confidence interval for the mean hours spent per week sending and answering e-mail for males at least age

75. The sample mean, ̅, is listed as 6.38 hours. Thus, the estimated mean spent per week sending and answering e-mail for males atleast age 75 is 6.38 hours. The sample standard deviation is 6.02. This quantity estimates the population standard deviation which

tells us how far we can expect a typical observation to vary from the mean. These estimates are based on a sample of size 8.

b.

The confidence interval is 2.34 to 10.41. We can be 90% confident that the populations mean numbers of hours spent per weeksending and answering e-mail for males at least age 75 is 2.34 to 10.41 hours.

c. Since there are likely to be a lot of men over the age of 75 who do not use email but also some who use e-mail regularly, this

distribution is likely skewed right. Since the t-distribution is robust to violations of normal assumption, the interval is still valid.

21. US Popularity

a. True

b. False

22. Driving After Drinking

a. = [(−)] = [.(−.)](.)

(.) = 385. b.

= [(−)]

= [.(−.)](.)

(.) = 601. This is larger than the answer in a. If we can make an educated guess about what to expect for the proportion, we can use a smaller

sample size, saving possibly unnecessary time and money.

23. Mean property tax

a. = = ()(.)

() = 385. The solution makes the assumption that the standard deviation will be similar now.

b. The margin of error would be more than $100 because the standard error would be larger than predicted.

c. With a larger margin of error, the 95% confidence interval is wider; thus, the probability the sample mean is within $100(which is les

than the margin of error from b) of the population mean is less than 0.95.


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24.

H0 or Ha?

a.

null hypothesis

b. alt alternative hypothesis

c. alternative hypothesis

d. null hypothesis

25. ESP

1.

Assumptions: The data are categorical (correct vs. incorrect guesses) and are obtained randomly. The expected successes and failureare less than 15 under null hypothesis : = (20)(0.5) = 10 < 15 and (1 ) = (20)(0.5) = 10 < 15, so this test isapproximate.

2. Hypotheses: : = 0.5; : > 0.5 3. Test statistic: = .−. .(−.)/ = 0.89 4. P-value:0.19

5. Conclusion: If the null hypothesis were true, the probability would be 0.19 of getting a test statistic at least as extreme as the value

observed. There is no strong evidence that the population proportion correct guesses is higher than 0.50.

26. Jurors and gender

a. The proportion of jurors who are women

Hypotheses: : = 0.53; : ≠ 0.53 b. Test statistic: = .−. .(−.)/ = 5.1 c. P-value is 0.000. If the null hypothesis were true, the probability would be almost 0 of getting a test statistic at least as extreme as

the value observed.

d. This P-value is more extreme than the significance level of 0.01. We can reject the null hypothesis; we have strong evidence that

women are not being selected in numbers proportionate to their representation in their jury pool.

27. Type I and Type II errors

a. In the previous exercises, a Type I error would have occurred if we had rejected the null hypothesis, concluding that the women wer

passed over jury duty, when they really were not. A Type II error would have occurred if we had failed to reject the null hypothesis,

but women were picked disproportionate to their representation in their jury pool.

b.

If we made an error, it was a Type I error.

28. Tennis balls in control?

a. Software indicates a test statistic of 5.5 and a P-value of 0.001.b.

For a significance level of 0.05, we conclude that the process is not in control. The machine is producing tennis balls that weight less

than they are supposed to.

c. If we rejected the null hypothesis when in fact it is true, we have made a Type I error and conclude that the process is not in control

when it actually is.

d.

29. Wage claim false?

1. Assumptions: The data are quantitative. The data seem to have been produced using randomization. We also assume an

approximately normal population distribution.

2. Hypotheses: : = 500; : < 500 3.

The sample mean is 441.11, and the standard deviation is 12.69.

Test statistic: = .−.√

= 13.9 4.

P-value:0.000

5. Conclusion: If the null hypothesis were true, the probability would be almost 0 of getting a test statistic at least as extreme as the

value observed. There is extremely strong evidence that the population mean is less than 500; we can conclude that the mean incom

is less than $500 per week.


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30.

Legal trial errors

a. A Type I error in a trail setting would occur if we convicted a defendant who was not guilty. A Type II error would occur if we failed to

convict a guilty defendant.

b. To decrease the chance of a Type I error, we would decrease the significance level. In doing this, it is more difficult to reject the null

hypothesis (i.e., find someone guilty) . Thus, there will be more guilty people who are not found guilty, a Type II error.

31. Gender and belief in afterlife.

a. The sample proportions who report that they believe in afterlife for females is = 0.8437, for males = 0.7167, and for the

difference between females and males is 0.8437 0.7167 = 0.127.b. The standard error for the estimate of ( ) is

= (−) + (−)

= .(−.)

+ .(−.) = 0.0230

This is the standard deviation for the difference between males and females for the sample of these sizes.

c. The confidence interval is ̂ ̂ ± (); lower endpoint: 0.1271.96(0.0230) = 0.0819; upper end point:0.127+1.96(0.0230) = 0.172. The confidence interval is (0.0819,0.172).Because 0 is less than all plausible values given in the confidence interval, we can conclude that the proportion of females who repo

that they believe in afterlife is higher than the proportion of males.

d.

The difference between these population proportions, 0.09, is in the confidence interval. The confidence interval in part c) contains

the parameter it is designated to estimate.

32. Belief depend on gender

a.

= ̂(1 ̂) + = 0.7859(1 0.7859)

+ = 0.0228.

b. = (−)− = .. = 5.57; P-value≈ 0. If the null hypothesis were true, the probability would be approximately 0 of getting a test statistic at least as extreme as the value

observed. Therefore, we reject the null hypothesis, and conclude that the population proportions believing in afterlife are different

for females and males.

c.

If the population difference were 0.810.72=0.09, our decision would have been correct.d.

The assumptions on which the methods in this exercises are based are independent random samples for the two groups and that we

had at least 5 successes and 5 failures.

33.

Heavier horseshoe crabs more likely to mate?

a. Construct yourself two boxplots of weight for female carbs who have mates and who do not have mates on the same scale. Then

conclude that the female crabs have a higher median and a bigger spread if they had a mate than if they did not have a mate. The

distribution for female crabs with a mate is right-skewed, whereas the distribution for female crabs without a mate is symmetrical.

b. The estimated difference between the mean weights of female crabs who have mates and do not have mates is 2.62.1=0.5.

c. = + =

. + . = 0.076

d. (̅ ̅) ± .() Because and are large, we will approximate t with the normal distribution using z=1.645.0.5 1.645(0.076) = 0.375 0.5 + 1.645(0.076) = 0.625 (0.375, 0.625)

We can be 90% confidence the difference between the population mean weights of female crabs with and without a mate is betwee

0.375 and 0.625. Because 0 does not fall in this interval, we can conclude that female crabs with a met weight more than do female

crabs without a mate.


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34.

Sex roles

1. Assumptions: The data are quantitative (Child’s score); the samples are independent and we will assume that they were obtaine d

randomly; we will assume that the populations scores distribution are approximately normal for each group.

2. Hypotheses: : = ; : ≠ where group 1 represents with the group with male tester group 2 represents the group witfemale tester.

3. = . + .

= 0.2349, = .−.. = 1.28 4.

P-value:0.2055. If the null hypothesis were true, the probability would be 0.205 of getting a test statistic at least as extreme as the value observed.

Since the P-value is quite large, there is not much evidence of a difference in the population mean of the children’s scores when the

tester is male than female.

35. Internet book prices

a. The samples are dependent because they are prices of the same ten books at two different internet sites

b. Let group 1 be the prices from Site A and group 2 from Site B. Then, ̅ = $87.30,̅ = $83.00,̅ = $4.30. Sample mean price forthe books from Site A is higher than the sample mean price for the book from Site B. Thus, the sample mean of the difference

between the prices from the two sites is positive.

c. A 90% confidence interval for is given by (1.53, 7.03). Since 0 is less than the values in the confidence interval, we canconclude that the prices for textbooks used at her college are more expensive at Site A than at Site B.

36.

Comparing book prices 21. Assumptions: the differences in prices are a random sample from a population that is approximately normal.

2. : = 0; : ≠ 0

3. = −/√ = .

./√ = 2.88

4. P-value:0.02

5.

If the null hypothesis is true, the probability obtaining a difference in sample means extreme as the value observed is 0.02. We

would reject the null hypothesis and conclude there is a significant difference in the prices of textbooks used at her college between

the two sites for =0.05, or 0.10, but not for =0.01.37. Down syndrome diagnostic test

a. BLOOD TEST RESULT

STATUS Positive Negative Total

D(Down) 0.89 0.11 54

(Unafected) 0.25 0.75 5228b.

For the Down cases, 89% were correctly diagnosed. For the unaffected cases, 75% get a negative result. The test seems fairly good,

but there are a good number of false positives and false negatives.

c. BLOOD TEST RESULT

STATUS Positive Negative

D(Down) 0.035 0.002

(Unafected) 0.965 0.998Total 1355 3927

Of the positive cases, only 0.035 truly have Down syndrome. This result is not surprising because there are so few cases overall. The

fairly large numbers of false positives will overwhelm the much smaller number of actual cases.

38.

Down and chi-squared

1. The assumptions are there are two categorical variables (Down syndrome status and blood test result), that randomization was used

to obtain the data and that the expected count was at least five in all cells.

2. : Down syndrome status and blood test result are independent : Down syndrome status and blood test result are dependent

3. = 114.4, df=14.

P-value:0.000


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5.

If the null hypothesis were true, the probability would be almost 0 of getting a test statistic at least as extreme as the value

observed. There is very strong evidence of an association between test result and actual status.

39. Gender gap?

1. The assumptions are there are two categorical variables (party identification and gender), that randomization was used to obt ain th

data and that the expected count was at least five in all cells.

2. : Party identification and gender are independent : Party identification and gender are dependent

3. = 8.294, df=24. P-value:0.016

5. If the null hypothesis were true, the probability would be 0.016 of getting a test statistic at least as extreme as the value observed.

There is very strong evidence that party identification depends on gender.

40. Tanning experiment

a.

Treatments Ranks

Lotion (1, 2) (1, 3) (1, 4) (2, 3) (2, 4) (3, 4)

Studio (3, 4) (2, 4) (2, 3) (1, 4) (1, 3) (1, 2)

b.

Lotion mean rank 1.5 2.0 2.5 2.5 3.0 3.5

Studio mean rank 3.5 3.0 2.5 2.5 2.0 1.5

Difference of mean

ranks

-2.0 -1.0 0.0 0.0 1.0 2.0

c.

Difference between mean ranks probability

-2.0 1/6

-1.0 1/6

0.0 2/6

1.0 1/6

2.0 1/6

41.

Test for tanning experiment

a. The P-value is 1/6 =0.17; if the treatments had identical effects, the probability would be 0.17 of getting sample like we observed, or

even more extreme, in this direction. It is plausible that the null hypothesis is correct, and that the studio does not lead to better

results than the lotion.

b.

The P-value is 2/6 =0.33; if the treatments had identical effects, the probability would be 0.33 of getting sample like we observed, or

even more extreme, in this direction. It is plausible that the null hypothesis is correct, and that the treatments do not lead to differe

results.

c. It is a waste of time to conduct this experiment if we plan to use a 0.05 significance level because the smallest possible P-value is 0.1

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