Every problem in the chapter beside the book’s answer, a chapter at a time as you scroll.

Chapter 9: Hypothesis Testing with One Sample

60 problems · 2 figures

9-1 §9.1 source ↗

You are testing that the mean speed of your cable internet connection is more than three megabits per second. What is the random variable? Describe it in words.

Answer

The random variable is the mean Internet speed in megabits per second.

9-3 §9.1 source ↗

The American family has an average of two children. What is the random variable? Describe in words.

Answer

The random variable is the mean number of children an American family has.

9-5 §9.1 source ↗

A sociologist claims the probability that a person picked at random in Times Square in New York City is visiting the area is 0.83. You want to test to see if the proportion is actually less. What is the random variable? Describe in words.

Answer

The random variable is the proportion of people picked at random in Times Square visiting the city.

9-7 §9.1 source ↗

In a population of fish, approximately 42 percent are female. A test is conducted to see if, in fact, the proportion is less. State the null and alternative hypotheses.

Answer

a. H0: p = 0.42

b. Ha: p < 0.42

9-9 §9.1 source ↗

A random survey of 75 long-term marathon runners revealed that the mean length of time they've been running is 17.4 years with a standard deviation of 6.3 years. If you were conducting a hypothesis test to determine if the population mean time for these runners could likely be 15 years, what would the null and alternative hypotheses be?

a. H0: __________

b. Ha: __________

Answer

a. H0: μ = 15

b. Ha: μ ≠ 15

9-11 §9.2 source ↗

The mean price of mid-sized cars in a region is $32,000. A test is conducted to see if the claim is true. State the Type I and Type II errors in complete sentences.

Answer

Type I: The mean price of mid-sized cars is $32,000, but we conclude that it is not $32,000.

Type II: The mean price of mid-sized cars is not $32,000, but we conclude that it is $32,000.

9-13 §9.2 source ↗

For Exercise 9.12, what are α and β in words?

Answer

α = the probability that you think the bag cannot withstand –15 degrees F, when, in fact, it can.

β = the probability that you think the bag can withstand –15 degrees F, when, in fact, it cannot.

9-15 §9.2 source ↗

A group of doctors is deciding whether or not to perform an operation. Suppose the null hypothesis, H0, is: the surgical procedure will go well. State the Type I and Type II errors in complete sentences.

Answer

Type I: The procedure will go well, but the doctors think it will not.

Type II: The procedure will not go well, but the doctors think it will.

9-17 §9.2 source ↗

The power of a test is 0.981. What is the probability of a Type II error?

Answer

0.019

9-19 §9.2 source ↗

A microbiologist is testing a water sample for E. coli. Suppose the null hypothesis, H0, is the sample does not contain E. coli. The probability that the sample does not contain E. coli, but the microbiologist thinks it does is 0.012. The probability that the sample does contain E. coli, but the microbiologist thinks it does not is 0.002. What is the power of this test?

Answer

0.998

9-21 §9.3 source ↗

Which two distributions can you use for hypothesis testing for this chapter?

Answer

A normal distribution or a Student’s t-distribution

9-23 §9.3 source ↗

Which distribution do you use when the standard deviation is not known and you are testing one population mean? Assume sample size is large.

Answer

Use a Student’s t-distribution

9-25 §9.3 source ↗

A population has a mean of 25 and a standard deviation of five. The sample mean is 24, and the sample size is 108. What distribution should you use to perform a hypothesis test?

Answer

a normal distribution for a single population mean

9-27 §9.3 source ↗

You are performing a hypothesis test of a single population mean using a Student’s t-distribution. What must you assume about the distribution of the data?

Answer

It must be approximately normally distributed.

9-29 §9.3 source ↗

You are performing a hypothesis test of a single population proportion. What must be true about the quantities of np and nq?

Answer

They must both be greater than five.

9-31 §9.3 source ↗

You are performing a hypothesis test of a single population proportion. The data come from which distribution?

Answer

binomial distribution

9-33 §9.4 source ↗

The probability of winning the grand prize at a particular carnival game is 0.005. Is the outcome of winning very likely or very unlikely?

Answer

The outcome of winning is very unlikely.

9-35 §9.4 source ↗

It is believed that the mean height of high school students who play basketball on the school team is 73 inches with a standard deviation of 1.8 inches. A random sample of 40 players is chosen. The sample mean was 71 inches, and the sample standard deviation was 1.5 inches. Do the data support the claim that the mean height is less than 73 inches? The p-value is almost zero. State the null and alternative hypotheses and interpret the p-value.

Answer

H0: μ > = 73
Ha: μ < 73
The p-value is almost zero, which means there is sufficient data to conclude that the mean height of high school students who play basketball on the school team is less than 73 inches at the 5 percent level. The data do support the claim.

9-37 §9.4 source ↗

Does the shaded region represent a low or a high p-value compared to a level of significance of 1 percent?
Figure 9.22 in the book ↗

Answer

The shaded region shows a low p-value.

9-39 §9.4 source ↗

What should you do if α = p-value?

Answer

Do not reject H0.

9-41 §9.4 source ↗


Use the following information to answer the next seven exercises: Suppose that a recent article stated that the mean time students spend doing homework each week is 2.5 hours. A study was then done to see if the mean time has increased in the new century. A random sample of 26 students was taken. The mean length of time they did homework each week was three hours with a standard deviation of 1.8 hours. Suppose that it is somehow known that the population standard deviation is 1.5. Conduct a hypothesis test to determine if the mean length of time doing homework each week has increased. Assume the distribution of homework times is approximately normal.

Is this a test of means or proportions?

Answer

means

9-43 §9.4 source ↗


Use the following information to answer the next seven exercises: Suppose that a recent article stated that the mean time students spend doing homework each week is 2.5 hours. A study was then done to see if the mean time has increased in the new century. A random sample of 26 students was taken. The mean length of time they did homework each week was three hours with a standard deviation of 1.8 hours. Suppose that it is somehow known that the population standard deviation is 1.5. Conduct a hypothesis test to determine if the mean length of time doing homework each week has increased. Assume the distribution of homework times is approximately normal.

In words, define the random variable for this test.

Answer

the mean time spent on homework for 26 students

9-45 §9.4 source ↗


Use the following information to answer the next seven exercises: Suppose that a recent article stated that the mean time students spend doing homework each week is 2.5 hours. A study was then done to see if the mean time has increased in the new century. A random sample of 26 students was taken. The mean length of time they did homework each week was three hours with a standard deviation of 1.8 hours. Suppose that it is somehow known that the population standard deviation is 1.5. Conduct a hypothesis test to determine if the mean length of time doing homework each week has increased. Assume the distribution of homework times is approximately normal.

Calculate the following:

a. x ¯ _______

b. σ _______

c. sx _______

d. n _______

Answer

a. 3

b. 1.5

c. 1.8

d. 26

9-47 §9.4 source ↗


Use the following information to answer the next seven exercises: Suppose that a recent article stated that the mean time students spend doing homework each week is 2.5 hours. A study was then done to see if the mean time has increased in the new century. A random sample of 26 students was taken. The mean length of time they did homework each week was three hours with a standard deviation of 1.8 hours. Suppose that it is somehow known that the population standard deviation is 1.5. Conduct a hypothesis test to determine if the mean length of time doing homework each week has increased. Assume the distribution of homework times is approximately normal.

State the distribution to use for the hypothesis test.

Answer

X ¯ ~N( 2.5, 1.5 26 )

9-49 §9.5 source ↗

Assume H0: μ = 9 and Ha: μ < 9. Is this a left-tailed, right-tailed, or two-tailed test?

Answer

This is a left-tailed test.

9-51 §9.5 source ↗

Assume H0: p = 0.25 and Ha: p ≠ 0.25. Is this a left-tailed, right-tailed, or two-tailed test?

Answer

This is a two-tailed test.

9-53 §9.5 source ↗

Draw the graph of a two-tailed test.

Answer

Figure 9.23 in the book ↗

9-55 §9.5 source ↗

Your friend claims that his mean golf score is 63. You want to show that it is higher than that. What type of test would you use?

Answer

a right-tailed test

9-57 §9.5 source ↗

You flip a coin and record whether it shows heads or tails. You know the probability of getting heads is 50 percent, but you think it is less for this particular coin. What type of test would you use?

Answer

a left-tailed test

9-59 §9.5 source ↗

Assume the null hypothesis states that the mean is at least 18. Is this a left-tailed, right-tailed, or two-tailed test?

Answer

This is a left-tailed test.

9-61 §9.5 source ↗

Assume the null hypothesis states that the mean is equal to 88. The alternative hypothesis states that the mean is not equal to 88. Is this a left-tailed, right-tailed, or two-tailed test?

Answer

This is a two-tailed test.

9-62 §9.1 source ↗

Some of the following statements refer to the null hypothesis, some to the alternate hypothesis.

State the null hypothesis, H0, and the alternative hypothesis. Ha, in terms of the appropriate parameter (μ or p).

a. The mean number of years Americans work before retiring is 34.

b. At most 60 percent of Americans vote in presidential elections.

c. The mean starting salary for San Jose State University graduates is at least $100,000 per year.

d. Twenty-nine percent of high school students take physical education daily.

e. Less than 5 percent of adults ride the bus to work in Los Angeles.

f. The mean number of cars a person owns in her lifetime is not more than 10.

g. About half of Americans prefer to live away from cities, given the choice.

h. Europeans have a mean paid vacation each year of six weeks.

i. The chance of developing breast cancer is under 11 percent for women.

j. Private universities' mean tuition cost is more than $20,000 per year.

Answer

a. H0: μ = 34; Ha: μ ≠ 34

b. H0: p ≤ 0.60; Ha: p > 0.60

c. H0: μ ≥ 100,000; Ha: μ < 100,000

d. H0: p = 0.29; Ha: p ≠ 0.29

e. H0: p = 0.05; Ha: p < 0.05

f. H0: μ ≤ 10; Ha: μ > 10

g. H0: p = 0.50; Ha: p ≠ 0.50

h. H0: μ = 6; Ha: μ ≠ 6

i. H0: p ≥ 0.11; Ha: p < 0.11

j. H0: μ ≤ 20,000; Ha: μ > 20,000

9-64 §9.1 source ↗

A statistics instructor believes that fewer than 20 percent of Evergreen Valley College (EVC) students attended the opening night midnight showing of the latest Harry Potter movie. She surveys 84 of her students and finds that 11 attended the midnight showing. An appropriate alternative hypothesis is as follows:

a. p = 0.20

b. p > 0.20

c. p < 0.20

d. p ≤ 0.20

Answer

c

9-66 §9.2 source ↗

State the Type I and Type II errors in complete sentences given the following statements.

a. The mean number of years Americans work before retiring is 34.

b. At most 60 percent of Americans vote in presidential elections.

c. The mean starting salary for San Jose State University graduates is at least $100,000 per year.

d. 29 percent of high school students take physical education every day.

e. Less than 5 percent of adults ride the bus to work in Los Angeles.

f. The mean number of cars a person owns in his or her lifetime is not more than 10.

g. About half of Americans prefer to live away from cities, given the choice.

h. Europeans have a mean paid vacation each year of six weeks.

i. The chance of developing breast cancer is under 11 percent for women.

j. Private universitie' mean tuition cost is more than $20,000 per year.

Answer

a. Type I error: We conclude that the mean is not 34 years, when it really is 34 years. Type II error: We conclude that the mean is 34 years, when in fact it really is not 34 years.

b. Type I error: We conclude that more than 60 percent of Americans vote in presidential elections, when the actual percentage is at most 60 percent.Type II error: We conclude that at most 60 percent of Americans vote in presidential elections when, in fact, more than 60 percent do.

c. Type I error: We conclude that the mean starting salary is less than $100,000, when it really is at least $100,000. Type II error: We conclude that the mean starting salary is at least $100,000 when, in fact, it is less than $100,000.

d. Type I error: We conclude that the proportion of high school seniors who take physical education daily is not 29%, when it really is 29%. Type II error: We conclude that the proportion of high school seniors who take physical education daily is 29% when, in fact, it is not 29%.

e. Type I error: We conclude that fewer than 5 percent of adults ride the bus to work in Los Angeles, when the percentage that do is really 29%. Type II error: We conclude that 29%. or more adults ride the bus to work in Los Angeles when, in fact, fewer that 29% do.

f. Type I error: We conclude that the mean number of cars a person owns in his or her lifetime is more than 10, when in reality it is not more than 10. Type II error: We conclude that the mean number of cars a person owns in his or her lifetime is not more than 10 when, in fact, it is more than 10.

g. Type I error: We conclude that the proportion of Americans who prefer to live away from cities is not about half, though the actual proportion is about half. Type II error: We conclude that the proportion of Americans who prefer to live away from cities is half when, in fact, it is not half.

h. Type I error: We conclude that the duration of paid vacations each year for Europeans is not six weeks, when in fact it is six weeks. Type II error: We conclude that the duration of paid vacations each year for Europeans is six weeks when, in fact, it is not.

i. Type I error: We conclude that the proportion is less than 11 percent, when it is really at least 11 percent. Type II error: We conclude that the proportion of women who develop breast cancer is at least 11 percent, when in fact it is less than 11 percent.

j. Type I error: We conclude that the average tuition cost at private universities is more than $20,000, though in reality it is at most $20,000. Type II error: We conclude that the average tuition cost at private universities is at most $20,000 when, in fact, it is more than $20,000.

9-68 §9.2 source ↗

When a new drug is created, the pharmaceutical company must subject it to testing before receiving the necessary permission from the U.S. Food and Drug Administration (FDA) to market the drug. Suppose the null hypothesis is the drug is unsafe. What is the Type II error?

a. To conclude the drug is safe when, in fact, it is unsafe.

b. Not to conclude the drug is safe when, in fact, it is safe.

c. To conclude the drug is safe when, in fact, it is safe.

d. Not to conclude the drug is unsafe when, in fact, it is unsafe.

Answer

b

9-70 §9.2 source ↗

It is believed that Lake Tahoe Community College (LTCC) Intermediate Algebra students get less than seven hours of sleep per night, on average. A survey of 22 LTCC Intermediate Algebra students generated a mean of 7.24 hours with a standard deviation of 1.93 hours. At a level of significance of 5 percent, do LTCC Intermediate Algebra students get less than seven hours of sleep per night, on average?

The Type II error is not to reject that the mean number of hours of sleep LTCC students get per night is at least seven when, in fact, the mean number of hours

a. is more than seven hours.

b. is at most seven hours.

c. is at least seven hours.

d. is less than seven hours.

Answer

d

9-72 §9.3 source ↗

It is believed that Lake Tahoe Community College (LTCC) Intermediate Algebra students get less than seven hours of sleep per night, on average. A survey of 22 LTCC Intermediate Algebra students generated a mean of 7.24 hours with a standard deviation of 1.93 hours. At a level of significance of 5 percent, do LTCC Intermediate Algebra students get less than seven hours of sleep per night, on average? The distribution to be used for this test is X ¯ ~ ________________

a. N(7.24, 1.93 22 )

b. N(7.24,1.93)

c. t22

d. t21

Answer

d

9-74 §9.5 source ↗

For each of the word problems, use a solution sheet to do the hypothesis test. The solution sheet is found in Appendix E, Solution Sheets. Please feel free to make copies of the solution sheets. For the online version of the book, it is suggested that you copy the .doc or the .pdf files.

Note

If you are using a Student's-t-distribution for one of the following homework problems, you may assume that the underlying population is normally distributed. In general, you must first prove that assumption, however.

A particular brand of tires claims that its deluxe tire averages at least 50,000 miles before it needs to be replaced. From past studies of this tire, the standard deviation is known to be 8,000. A survey of owners of that tire design is conducted. From the 28 tires surveyed, the mean lifespan was 46,500 miles with a standard deviation of 9,800 miles. Using alpha = 0.05, are the data highly inconsistent with the claim?

Answer

a. H0: μ ≥ 50,000

b. Ha: μ < 50,000

c. Let X ¯ = the average lifespan of a brand of tires.

d. normal distribution

e. z = -2.315

f. p-value = 0.0103

g. Check student’s solution.

h.

i. Alpha: 0.05

ii. Decision: Reject the null hypothesis.

iii. Reason for decision: The p-value is less than 0.05.

iv. Conclusion: There is sufficient evidence to conclude that the mean lifespan of the tires is less than 50,000 miles.

i. (43,537, 49,463)

9-75 §9.5 source ↗

For each of the word problems, use a solution sheet to do the hypothesis test. The solution sheet is found in Appendix E, Solution Sheets. Please feel free to make copies of the solution sheets. For the online version of the book, it is suggested that you copy the .doc or the .pdf files.

Note

If you are using a Student's-t-distribution for one of the following homework problems, you may assume that the underlying population is normally distributed. In general, you must first prove that assumption, however.

In 2009, President Barack Obama announced a new national fuel economy and emissions policy for cars and light trucks. It stated that the combined fleet fuel economy for an auto manufacturer of cars and light trucks will have to average 35.5 mpg or better by 2016. From past studies on fuel economy, it is known that the standard deviation of a typical fleet is 7.6 mpg. An auto manufacturer selects a random sample of 55 cars and light trucks and finds the sample mean fuel economy to be 34.6 mpg with a standard deviation of 10.3 mpg. Can the manufacturer claim that their fleet meets the fuel economy standard in the 2016 policy at the 5 percent level?

Answer

a. H0: μ ≥ 35.5

b. Ha: μ < 35.5

c. Let x ¯ = the average mpg for the sample of cars and trucks in the fleet

d. normal distribution

e. z = -0.648

f. p-value = 0.2578

g. Check student’s solution.

h.

i. Alpha: 0.05

ii. Decision: Do not reject the null hypothesis.

iii. Reason for decision: The p-value is greater than 0.05.

iv. Conclusion: There is sufficient evidence to support the claim that the manufacturer’s fleet meets the fuel economy standards in the 2016 policy.

i. (31.88 mpg, 37.32 mpg)

9-76 §9.5 source ↗

For each of the word problems, use a solution sheet to do the hypothesis test. The solution sheet is found in Appendix E, Solution Sheets. Please feel free to make copies of the solution sheets. For the online version of the book, it is suggested that you copy the .doc or the .pdf files.

Note

If you are using a Student's-t-distribution for one of the following homework problems, you may assume that the underlying population is normally distributed. In general, you must first prove that assumption, however.

The cost of a daily newspaper varies from city to city. However, the variation among prices remains steady with a standard deviation of 20¢. A study was done to test the claim that the mean cost of a daily newspaper is $1.00. Twelve costs yield a mean cost of 95¢ with a standard deviation of 18¢. Do the data support the claim at the 1 percent level?

Answer

a. H0: μ = $1.00

b. Ha: μ ≠ $1.00

c. Let x ¯ = the average cost of a daily newspaper.

d. normal distribution

e. z = –0.866

f. p-value = 0.3865

g. Check student’s solution.

h.

i. Alpha: 0.01

ii. Decision: Do not reject the null hypothesis.

iii. Reason for decision: The p-value is greater than 0.01.

iv. Conclusion: There is sufficient evidence to support the claim that the mean cost of daily papers is $1. The mean cost could be $1.

i. ($0.84, $1.06)

9-78 §9.5 source ↗

For each of the word problems, use a solution sheet to do the hypothesis test. The solution sheet is found in Appendix E, Solution Sheets. Please feel free to make copies of the solution sheets. For the online version of the book, it is suggested that you copy the .doc or the .pdf files.

Note

If you are using a Student's-t-distribution for one of the following homework problems, you may assume that the underlying population is normally distributed. In general, you must first prove that assumption, however.

The mean number of sick days an employee takes per year is believed to be about 10. Members of a personnel department do not believe this figure. They randomly survey eight employees. The number of sick days they took for the past year are as follows: 12; 4; 15; 3; 11; 8; 6; 8. Let x = the number of sick days they took for the past year. Should the personnel team believe that the mean number is 10?

Answer

a. H0: μ = 10

b. Ha: μ ≠ 10

c. Let X ¯ = the mean number of sick days an employee takes per year.

d. Student’s t-distribution

e. t = –1.12

f. p-value = 0.300

g. Check student’s solution.

h.

i. Alpha: 0.05

ii. Decision: Do not reject the null hypothesis.

iii. Reason for decision: The p-value is greater than 0.05.

iv. Conclusion: At the 5 percent significance level, there is insufficient evidence to conclude that the mean number of sick days is not 10.

i. (4.9443, 11.806)

9-80 §9.5 source ↗

For each of the word problems, use a solution sheet to do the hypothesis test. The solution sheet is found in Appendix E, Solution Sheets. Please feel free to make copies of the solution sheets. For the online version of the book, it is suggested that you copy the .doc or the .pdf files.

Note

If you are using a Student's-t-distribution for one of the following homework problems, you may assume that the underlying population is normally distributed. In general, you must first prove that assumption, however.

Your statistics instructor claims that 60 percent of the students who take her Elementary Statistics class go through life feeling more enriched. For some reason that she can't quite figure out, most people don't believe her. You decide to check this out on your own. You randomly survey 64 of her past Elementary Statistics students and find that 34 feel more enriched as a result of her class. Now, what do you think?

Answer

a. H0: p ≥ 0.6

b. Ha: p < 0.6

c. Let P′ = the proportion of students who feel more enriched as a result of taking elementary statistics.

d. normal for a single proportion

e. 1.12

f. p-value = 0.1308

g. Check student’s solution.

h.

i. Alpha: 0.05

ii. Decision: Do not reject the null hypothesis.

iii. Reason for decision: The p-value is greater than 0.05.

iv. Conclusion: There is insufficient evidence to conclude that less than 60 percent of her students feel more enriched.

i. Confidence interval: (0.409, 0.654)
The “plus-4s” confidence interval is (0.411, 0.648)

9-82 §9.5 source ↗

For each of the word problems, use a solution sheet to do the hypothesis test. The solution sheet is found in Appendix E, Solution Sheets. Please feel free to make copies of the solution sheets. For the online version of the book, it is suggested that you copy the .doc or the .pdf files.

Note

If you are using a Student's-t-distribution for one of the following homework problems, you may assume that the underlying population is normally distributed. In general, you must first prove that assumption, however.

Refer to Exercise 9.81. Conduct a hypothesis test to see if your decision and conclusion would change if your belief were that the brown trout’s mean I.Q. is not four.

Answer

a. H0: μ = 4

b. Ha: μ ≠ 4

c. Let X ¯ the average I.Q. of a set of brown trout.

d. two-tailed Student's t-test

e. t = 1.95

f. p-value = 0.076

g. Check student’s solution.

h.

i. Alpha: 0.05

ii. Decision: Reject the null hypothesis.

iii. Reason for decision: The p-value is greater than 0.05

iv. Conclusion: There is insufficient evidence to conclude that the average IQ of brown trout is not four.

i. (3.8865, 5.9468)

9-84 §9.5 source ↗

For each of the word problems, use a solution sheet to do the hypothesis test. The solution sheet is found in Appendix E, Solution Sheets. Please feel free to make copies of the solution sheets. For the online version of the book, it is suggested that you copy the .doc or the .pdf files.

Note

If you are using a Student's-t-distribution for one of the following homework problems, you may assume that the underlying population is normally distributed. In general, you must first prove that assumption, however.

A group of researchers research a common contagious disease. A newspaper found that 13 percent of Americans have been diagnosed with the disease in the last year. The researchers doubt that the percentage is really that high. It conducts its own survey. Out of 76 Americans surveyed, only two had been diagnosed with the disease. Would you agree with the newspaper's poll? In complete sentences, give three reasons why polls might give different results.

Answer

a. H0: p ≥ 0.13

b. Ha: p < 0.13

c. Let P′ = the proportion of Americans who have the disease

d. normal for a single proportion

e. –2.688

f. p-value = 0.0036

g. Check student’s solution.

h.

i. Alpha: 0.05

ii. Decision: Reject the null hypothesis.

iii. Reason for decision: The p-value is less than 0.05.

iv. Conclusion: There is sufficient evidence to conclude that the percentage of Americans who have been diagnosed with the disease is less than 13 percent.

i. (0, 0.0623).
The plus-4s confidence interval is (0.0022, 0.0978)

9-86 §9.5 source ↗

For each of the word problems, use a solution sheet to do the hypothesis test. The solution sheet is found in Appendix E, Solution Sheets. Please feel free to make copies of the solution sheets. For the online version of the book, it is suggested that you copy the .doc or the .pdf files.

Note

If you are using a Student's-t-distribution for one of the following homework problems, you may assume that the underlying population is normally distributed. In general, you must first prove that assumption, however.

Use the Lap time data for Lap 4 (see Appendix C: Data Sets) to test the claim that Terri finishes Lap 4, on average, in less than 129 seconds. Use all 20 races given.

Answer

a. H0: μ ≥ 129

b. Ha: μ < 129

c. Let X ¯ = the average time in seconds that Terri finishes Lap 4.

d. Student's t-distribution

e. t = 1.209

f. 0.8792

g. Check student’s solution.

h.

i. Alpha: 0.05

ii. Decision: Do not reject the null hypothesis.

iii. Reason for decision: The p-value is greater than 0.05.

iv. Conclusion: There is insufficient evidence to conclude that Terri’s mean lap time is less than 129 seconds.

i. (128.63, 130.37)

9-88 §9.5 source ↗

Note

The following questions were written by past students. They are excellent problems!

"Asian Family Reunion," by Chau Nguyen

Every two years it comes around.

We all get together from different towns.

In my honest opinion,

It's not a typical family reunion.

Not forty, or fifty, or sixty,

But how about seventy companions!

The kids would play, scream, and shout

One minute they're happy, another they'll pout.

The teenagers would look, stare, and compare

From how they look to what they wear.

The men would chat about their business

That they make more, but never less.

Money is always their subject

And there's always talk of more new projects.

The women get tired from all of the chats

They head to the kitchen to set out the mats.

Some would sit and some would stand

Eating and talking with plates in their hands.

Then come the games and the songs

And suddenly, everyone gets along!

With all that laughter, it's sad to say

That it always ends in the same old way.

They hug and kiss and say "good-bye"

And then they all begin to cry!

I say that 60 percent shed their tears

But my mom counted 35 people this year.

She said that boys and men will always have their pride,

So we won't ever see them cry.

I myself don't think she's correct,

So could you please try this problem to see if you object?

Answer

a. H0: p = 0.60

b. Ha: p < 0.60

c. Let P′ = the proportion of family members who shed tears at a reunion.

d. normal for a single proportion

e. –1.71

f. 0.0438

g. Check student’s solution.

h.

i. Alpha: 0.05

ii. Decision: Reject the null hypothesis.

iii. Reason for decision: p-value < alpha

iv. Conclusion: At the 5 percent significance level, there is sufficient evidence to conclude that the proportion of family members who shed tears at a reunion is less than 0.60. However, the test is weak because the p-value and alpha are quite close, so other tests should be done.

i. We are 95 percent confident that between 38.29 percent and 61.71 percent of family members will shed tears at a family reunion. (0.3829, 0.6171). The plus-4s confidence interval (see chapter 8) is (0.3861, 0.6139)

Note that here the large-sample 1 – PropZTest provides the approximate p-value of 0.0438. Whenever a p-value based on a normal approximation is close to the level of significance, the exact p-value based on binomial probabilities should be calculated whenever possible. This is beyond the scope of this course.

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Note

The following questions were written by past students. They are excellent problems!

"Blowing Bubbles," by Sondra Prull

Studying stats just made me tense,

I had to find some sane defense.

Some light and lifting simple play

To float my math anxiety away.

Blowing bubbles lifts me high

Takes my troubles to the sky.

POIK! They're gone, with all my stress

Bubble therapy is the best.

The label said each time I blew

The average number of bubbles would be at least 22.

I blew and blew and this I found

From 64 blows, they all are round!

But the number of bubbles in 64 blows

Varied widely, this I know.

20 per blow became the mean

They deviated by 6, and not 16.

From counting bubbles, I sure did relax

But now I give to you your task.

Was 22 a reasonable guess?

Find the answer and pass this test!

Answer

a. H0: μ ≥ 22

b. Ha: μ < 22

c. Let X ¯ = the mean number of bubbles per blow.

d. Student's t-distribution

e. –2.667

f. p-value = 0.00486

g. Check student’s solution.

h.

i. Alpha: 0.05

ii. Decision: Reject the null hypothesis.

iii. Reason for decision: The p-value is less than 0.05.

iv. Conclusion: There is sufficient evidence to conclude that the mean number of bubbles per blow is less than 22.

i. (18.501, 21.499)

9-91 §9.5 source ↗

Note

The following questions were written by past students. They are excellent problems!

Macaroni and Cheese, please!! by Nedda Misherghi and Rachelle Hall

As a poor starving student I don't have much money to spend for even the bare necessities. So my favorite and main staple food is macaroni and cheese. It's high in taste and low in cost and nutritional value.

One day, as I sat down to determine the meaning of life, I got a serious craving for this, oh, so important, food of my life. So I went down the street to Greatway to get a box of macaroni and cheese, but it was SO expensive! $2.02 !!! Can you believe it? It made me stop and think. The world is changing fast. I had thought that the mean cost of a box (the normal size, not some super-gigantic-family-value-pack) was at most $1, but now I wasn't so sure. However, I was determined to find out. I went to 53 of the closest grocery stores and surveyed the prices of macaroni and cheese. Here are the data I wrote in my notebook:

Price per box of Mac and Cheese

• 5 stores @ $2.02

• 15 stores @ $0.25

• 3 stores @ $1.29

• 6 stores @ $0.35

• 4 stores @ $2.27

• 7 stores @ $1.50

• 5 stores @ $1.89

• 8 stores @ $0.75

I could see that the cost varied but I had to sit down to figure out whether or not I was right. If it does turn out that this mouth-watering dish is at most $1, then I'll throw a big cheesy party in our next statistics lab, with enough macaroni and cheese for just me. After all, as a poor starving student I can't be expected to feed our class of animals!

Answer

a. H0: μ ≤ 1

b. Ha: μ > 1

c. Let X ¯ = the mean cost in dollars of macaroni and cheese in a certain town.

d. Student's t-distribution

e. t = 0.340

f. p-value = 0.36756

g. Check student’s solution.

h.

i. Alpha: 0.05

ii. Decision: Do not reject the null hypothesis.

iii. Reason for decision: The p-value is greater than 0.05

iv. Conclusion: The mean cost could be $1, or less. At the 5 percent significance level, there is insufficient evidence to conclude that the mean price of a box of macaroni and cheese is more than $1.

i. (0.8291, 1.241)

9-93 §9.5 source ↗

Note

The following questions were written by past students. They are excellent problems!

"Untitled," by Stephen Chen

I've often wondered how software is released and sold to the public. Ironically, I work for a company that sells products with known problems. Unfortunately, most of the problems are difficult to create, which makes them difficult to fix. I usually use the test program X, which tests the product, to try to create a specific problem. When the test program is run to make an error occur, the likelihood of generating an error is 1 percent.

So, armed with this knowledge, I wrote a new test program Y that will generate the same error that test program X creates, but more often. To find out if my test program is better than the original, so that I can convince the management that I'm right, I ran my test program to find out how often I can generate the same error. When I ran my test program 50 times, I generated the error twice. While this may not seem much better, I think that I can convince the management to use my test program instead of the original test program. Am I right?

Answer

a. H0: p = 0.01

b. Ha: p > 0.01

c. Let P′ = the proportion of errors generated

d. Normal for a single proportion

e. 2.13

f. 0.0165

g. Check student’s solution.

h.

i. Alpha: 0.05

ii. Decision: Reject the null hypothesis.

iii. Reason for decision: The p-value is less than 0.05.

iv. Conclusion: At the 5 percent significance level, there is sufficient evidence to conclude that the proportion of errors generated is more than 0.01.

i. Confidence interval: (0, 0.094).
The plus-4s confidence interval is (0.004, 0.144).

9-95 §9.5 source ↗

Note

The following questions were written by past students. They are excellent problems!

"Phillip’s Wish," by Suzanne Osorio

My nephew likes to play

Chasing the girls makes his day.

He asked his mother

If it is okay

To get his ear pierced.

She said, “No way!”

To poke a hole through your ear,

Is not what I want for you, dear.

He argued his point quite well,

Says even my macho pal, Mel,

Has gotten this done.

It’s all just for fun.

C’mon please, mom, please, what the hell.

Again Phillip complained to his mother,

Saying half his friends (including their brothers)

Are piercing their ears

And they have no fears

He wants to be like the others.

She said, “I think it’s much less.

We must do a hypothesis test.

And if you are right,

I won’t put up a fight.

But, if not, then my case will rest.”

We proceeded to call fifty guys

To see whose prediction would fly.

Nineteen of the fifty

Said piercing was nifty

And earrings they’d occasionally buy.

Then there’s the other thirty-one,

Who said they’d never have this done.

So now this poem’s finished.

Will his hopes be diminished,

Or will my nephew have his fun?

Answer

a. H0: p = 0.50

b. Ha: p < 0.50

c. Let P′ = the proportion of friends that has a pierced ear.

d. normal for a single proportion

e. –1.70

f. p-value = 0.0448

g. Check student’s solution.

h.

i. Alpha: 0.05

ii. Decision: Reject the null hypothesis.

iii. Reason for decision: The p-value is less than 0.05. (However, they are very close.)

iv. Conclusion: There is sufficient evidence to support the claim that less than 50 percent of his friends have pierced ears.

i. Confidence interval: (0.245, 0.515): The plus-4s confidence interval is (0.259, 0.519).

9-97 §9.5 source ↗

Note

The following questions were written by past students. They are excellent problems!

Toastmasters International cites a report by Gallup Poll that 40 percent of Americans fear public speaking. A student believes that less than 40 percent of students at her school fear public speaking. She randomly surveys 361 schoolmates and finds that 135 report they fear public speaking. Conduct a hypothesis test to determine if the percentage at her school is less than 40.

Answer

a. H0: p = 0.40

b. Ha: p < 0.40

c. Let P′ = the proportion of schoolmates who fear public speaking.

d. normal for a single proportion

e. –1.01

f. p-value = 0.1563

g. Check student’s solution.

h.

i. Alpha: 0.05

ii. Decision: Do not reject the null hypothesis.

iii. Reason for decision: The p-value is greater than 0.05.

iv. Conclusion: There is insufficient evidence to support the claim that less than 40 percent of students at the school fear public speaking.

i. Confidence interval: (0.3241, 0.4240): The plus-4s confidence interval is (0.3257, 0.4250).

9-99 §9.5 source ↗

Note

The following questions were written by past students. They are excellent problems!

According to an article in a local poll, a city found that 14 percent of its residents walk for exercise. Suppose that a survey is conducted to determine this year’s rate. Nine out of 70 randomly chosen city residents replied that they walk for exercise. Conduct a hypothesis test to determine if the rate is still 14 percent or if it has decreased.

Answer

a. H0: p = 0.14

b. Ha: p < 0.14

c. Let P′ = the proportion of nursing home residents that have the disease.

d. normal for a single proportion

e. –0.2756

f. p-value = 0.3914

g. Check student’s solution.

h.

i. Alpha: 0.05

ii. Decision: Do not reject the null hypothesis.

iii. Reason for decision: The p-value is greater than 0.05.

iv. At the 5 percent significance level, there is insufficient evidence to conclude that the proportion of nursing home residents that have the disease is less than 0.14.

i. Confidence interval: (0.0502, 0.2070): The plus-4s confidence interval (see chapter 8) is (0.0676, 0.2297).

9-101 §9.5 source ↗

Note

The following questions were written by past students. They are excellent problems!

Registered nurses earned an average annual salary of $69,110. For that same year, a survey was conducted of 41 California registered nurses to determine if the annual salary is higher than $69,110 for California nurses. The sample average was $71,121 with a sample standard deviation of $7,489. Conduct a hypothesis test.

Answer

a. H0: μ = 69,110

b. Ha: μ > 69,110

c. Let X ¯ = the mean salary in dollars for California registered nurses.

d. Student's t-distribution

e. t = 1.719

f. p-value: 0.0466

g. Check student’s solution.

h.

i. Alpha: 0.05

ii. Decision: Reject the null hypothesis.

iii. Reason for decision: The p-value is less than 0.05.

iv. Conclusion: At the 5 percent significance level, there is sufficient evidence to conclude that the mean salary of California registered nurses exceeds $69,110.

i. ($68,757, $73,485)

9-103 §9.5 source ↗

Note

The following questions were written by past students. They are excellent problems!

Harley Davidson motorcycles are the largest selling motorcycle in the United States, with 14 percent of all motorcycles sold in 2012. Interestingly, a random sample of 1,945 stolen motorcycles was selected, and it was found that just 8 percent of them were Harleys. Is there good evidence that the proportion of Harleys among stolen motorcycles is significantly less than their share of all motorcycles? After conducting the test, what decision and conclusion would you make?

a. Reject H0: There is sufficient evidence to conclude that the proportion of Harleys stolen is significantly less than their share of all motorcycles

b. Do not reject H0: There is not sufficient evidence to conclude that the proportion of Harleys stolen is significantly less than their share of all motorcycles

c. Do not reject H0: There is sufficient evidence to conclude that the proportion of Harleys stolen is significantly more than their share of all motorcycles

d. Reject H0: There is sufficient evidence to conclude that the proportion of Harleys stolen is significantly more than their share of all motorcycles

Answer

a. H0: p ≥ 0.14, Ha: p < 0.14

b. p-value < 0.0002

c. Alpha: 0.05

d. Reject the null hypothesis.

e. At the 5 percent significance level, there is sufficient evidence to conclude that the proportion of Harleys stolen is significantly less than their share of all motorcycles. (conclusion a)

9-105 §9.5 source ↗

Note

The following questions were written by past students. They are excellent problems!

Previously, an organization reported that teenagers spent 4.5 hours per week, on average, on the phone. The organization thinks that, currently, the mean is higher. Fifteen randomly chosen teenagers were asked how many hours per week they spend on the phone. The sample mean was 4.75 hours with a sample standard deviation of 2.0. Conduct a hypothesis test.
At a significance level of a = 0.05, what is the correct conclusion?

a. There is enough evidence to conclude that the mean number of hours is more than 4.75.

b. There is enough evidence to conclude that the mean number of hours is more than 4.5.

c. There is not enough evidence to conclude that the mean number of hours is more than 4.5.

d. There is not enough evidence to conclude that the mean number of hours is more than 4.75.

Answer

c

9-107 §9.5 source ↗

Hypothesis testing: For the following 10 exercises, answer each question.

A recent survey in the New York Times Almanac indicated that 48.8 percent of families own stock. A broker wanted to determine if this survey could be valid. He surveyed a random sample of 250 families and found that 142 owned some type of stock. At the 0.05 significance level, can the survey be considered to be accurate?

Answer

a. H0: p = 0.488 Ha: p ≠ 0.488

b. p-value = 0.0114

c. alpha = 0.05

d. Reject the null hypothesis.

e. At the 5 percent level of significance, there is enough evidence to conclude that 48.8 percent of families own stocks.

f. The survey does not appear to be accurate.

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Hypothesis testing: For the following 10 exercises, answer each question.

The U.S. Department of Energy reported that 51.7 percent of homes were heated by natural gas. A random sample of 221 homes in Kentucky found that 115 were heated by natural gas. Does the evidence support the claim for Kentucky at the α = 0.05 level? Are the results applicable across the country? Why?

Answer

a. H0: p = 0.517 Ha: p ≠ 0.517

b. p-value = 0.9203.

c. alpha = 0.05.

d. Do not reject the null hypothesis.

e. At the 5 percent significance level, there is not enough evidence to conclude that the proportion of homes in Kentucky that are heated by natural gas is 0.517.

f. However, we cannot generalize this result to the entire nation. First, the sample’s population is only the state of Kentucky. Second, it is reasonable to assume that homes in the extreme north and south will have extreme high usage and low usage, respectively. We would need to expand our sample base to include these possibilities if we wanted to generalize this claim to the entire nation.

9-111 §9.5 source ↗

Hypothesis testing: For the following 10 exercises, answer each question.

The Weather Underground reported that the mean amount of summer rainfall for the northeastern United States is at least 11.52 inches. Ten cities in the northeast are randomly selected and the mean rainfall amount is calculated to be 7.42 inches with a standard deviation of 1.3 inches. At the α = 0.05 level, can it be concluded that the mean rainfall was below the reported average? What if α = 0.01? Assume the amount of summer rainfall follows a normal distribution.

Answer

a. H0: µ ≥ 11.52 Ha: µ < 11.52

b. p-value = 0.000002 which is almost 0.

c. alpha = 0.05.

d. Reject the null hypothesis.

e. At the 5 percent significance level, there is enough evidence to conclude that the mean amount of summer rain in the northeaster US is less than 11.52 inches, on average.

f. We would make the same conclusion if alpha was 1 percent because the p-value is almost 0.

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Hypothesis testing: For the following 10 exercises, answer each question.

A report by the Gallup Poll found that a woman visits her doctor, on average, at most 5.8 times each year. A random sample of 20 women results in these yearly visit totals:
3; 2; 1; 3; 7; 2; 9; 4; 6; 6; 8; 0; 5; 6; 4; 2; 1; 3; 4; 1
At the α = 0.05 level, can it be concluded that the sample mean is higher than 5.8 visits per year?

Answer

a. H0: µ ≤ 5.8 Ha: µ > 5.8

b. p-value = 0.9987

c. alpha = 0.05

d. Do not reject the null hypothesis.

e. At the 5 percent level of significance, there is not enough evidence to conclude that a woman visits her doctor, on average, more than 5.8 times a year.

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Hypothesis testing: For the following 10 exercises, answer each question.

The student academic group on a college campus claims that freshman students study at least 2.5 hours per day, on average. One Introduction to Statistics class was skeptical. The class took a random sample of 30 freshman students and found a mean study time of 137 minutes with a standard deviation of 45 minutes. At α = 0.01 level, is the student academic group’s claim correct?

Answer

a. H0: µ ≥ 150 Ha: µ < 150

b. p-value = 0.0622

c. alpha = 0.01

d. Do not reject the null hypothesis.

e. At the 1 percent significance level, there is not enough evidence to conclude that freshmen students study less than 2.5 hours per day, on average.

f. The student academic group’s claim appears to be correct.

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Problems and answers are adapted from Statistics by Barbara Illowsky and Susan Dean, attributed to the Texas Education Agency (TEA) and published by OpenStax, and are used under the CC BY 4.0 licence. Access for free at https://openstax.org/books/statistics/pages/1-introduction. The original material is available at https://www.texasgateway.org/book/tea-statistics. Changes were made to the original material, including updates to art, structure, and other content updates. Art is credited in its own caption wherever the book credits it. This adaptation is neither affiliated with nor endorsed by OpenStax, Rice University or the Texas Education Agency (TEA).