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
59 problems · 2 figures9-1 §9.1 source ↗
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a. H0: p = 0.42
b. Ha: p < 0.42
9-9 §9.1 source ↗
A random survey of 75 student loan recipients revealed that the mean length of time in repayment 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 in repayment 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 ↗
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 ↗
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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
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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.
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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% level. The data do support the claim.
9-37 §9.4 source ↗
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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 spent in jail by a first-time convicted burglar is 2.5 years. A study was then done to see if the mean time has increased in the new century. A random sample of 26 first-time convicted burglars in a recent year was picked. The mean length of time in jail from the survey was three years with a standard deviation of 1.8 years. 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 jail time has increased. Assume the distribution of the jail times is approximately normal.
Is this a test of means or proportions?
Answer
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 spent in jail by a first-time convicted burglar is 2.5 years. A study was then done to see if the mean time has increased in the new century. A random sample of 26 first-time convicted burglars in a recent year was picked. The mean length of time in jail from the survey was three years with a standard deviation of 1.8 years. 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 jail time has increased. Assume the distribution of the jail times is approximately normal.
In words, define the random variable for this test.
Answer
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 spent in jail by a first-time convicted burglar is 2.5 years. A study was then done to see if the mean time has increased in the new century. A random sample of 26 first-time convicted burglars in a recent year was picked. The mean length of time in jail from the survey was three years with a standard deviation of 1.8 years. 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 jail time has increased. Assume the distribution of the jail times is approximately normal.
Calculate the following:
a. _______
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 spent in jail by a first-time convicted burglar is 2.5 years. A study was then done to see if the mean time has increased in the new century. A random sample of 26 first-time convicted burglars in a recent year was picked. The mean length of time in jail from the survey was three years with a standard deviation of 1.8 years. 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 jail time has increased. Assume the distribution of the jail times is approximately normal.
State the distribution to use for the hypothesis test.
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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% 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 seniors get drunk each month.
e. Fewer than 5% 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 ten.
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% for females.
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% of Evergreen Valley College (EVC) students attended the opening night midnight showing of the latest Marvel Universe movie. The instructor surveys 84 students and finds that 11 attended the midnight showing. An appropriate alternative hypothesis is:
a. p = 0.20
b. p > 0.20
c. p < 0.20
d. p ≤ 0.20
Answer
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% 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 seniors get drunk each month.
e. Fewer than 5% of adults ride the bus to work in Los Angeles.
f. The mean number of cars a person owns in their lifetime is not more than ten.
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% for females.
j. Private universities 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% of Americans vote in presidential elections, when the actual percentage is at most 60%.Type II error: We conclude that at most 60% of Americans vote in presidential elections when, in fact, more than 60% 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 get drunk each month is not 29%, when it really is 29%. Type II error: We conclude that the proportion of high school seniors who get drunk each month is 29% when, in fact, it is not 29%.
e. Type I error: We conclude that fewer than 5% of adults ride the bus to work in Los Angeles, when the percentage that do is really 5% or more. Type II error: We conclude that 5% or more adults ride the bus to work in Los Angeles when, in fact, fewer that 5% do.
f. Type I error: We conclude that the mean number of cars a person owns in their 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 their 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%, when it is really at least 11%. Type II error: We conclude that the proportion of females who develop breast cancer is at least 11%, when in fact it is less than 11%.
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 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
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%, 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
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%, 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 ~ ________________
a.
b.
c. t22
d. t21
Answer
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.
NOTEIf 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, is the data highly inconsistent with the claim?
Answer
a. H0: μ ≥ 50,000
b. Ha: μ < 50,000
c. Let = the average lifespan of a brand of tires.
d. normal distribution
e. z = -2.315
f. p-value = 0.0103
g. Answers may vary.
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)
9-76 §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.
NOTEIf 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% level?
Answer
a. H0: μ = $1.00
b. Ha: μ ≠ $1.00
c. Let = the average cost of a daily newspaper.
d. normal distribution
e. z = –0.866
f. p-value = 0.3865
g. Answers may vary.
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.
NOTEIf 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 ten. 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 ten?
Answer
a. H0: μ = 10
b. Ha: μ ≠ 10
c. Let the mean number of sick days an employee takes per year.
d. Student’s t-distribution
e. t = –1.13
f. p-value = 0.1299
g. Answers may vary.
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% significance level, there is insufficient evidence to conclude that the mean number of sick days is not ten.
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.
NOTEIf 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% of the students who take their Elementary Statistics class go through life feeling more enriched. For some reason that the instructor 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 their past Elementary Statistics students and find that 34 feel more enriched as a result of taking the 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. 1.13
f. p-value = 0.1299
g. Answers may vary.
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 the 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.
NOTEIf 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 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.
NOTEIf 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 poll done for Newsweek found that 13% of Americans have seen or sensed the presence of an angel. A contingent doubts that the percent is really that high. It conducts its own survey. Out of 76 Americans surveyed, only two had seen or sensed the presence of an angel. As a result of the contingent’s survey, would you agree with the Newsweek poll? In complete sentences, also give three reasons why the two 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 seen or sensed angels
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 seen or sensed an angel is less than 13%.
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.
NOTEIf 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 twenty races given.
Answer
a. H0: μ ≥ 129
b. Ha: μ < 129
c. Let = 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 ↗
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% 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% confident that between 38.29% and 61.71% of family members will shed tears at a family reunion. (0.3829, 0.6171). The“plus-4s” confidence interval (see Confidence Intervals) 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.
9-90 §9.5 source ↗
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 = 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)
9-92 §9.5 source ↗
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 = 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% 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)
9-94 §9.5 source ↗
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%.
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% 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).
9-96 §9.5 source ↗
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% of his friends have pierced ears.
i. Confidence Interval: (0.245, 0.515): The “plus-4s” confidence interval is (0.259, 0.519).
9-98 §9.5 source ↗
The following questions were written by past students. They are excellent problems!
Toastmasters International cites a report by Gallup Poll that 40% of Americans fear public speaking. A student believes that less than 40% of students at their school fear public speaking. The student randomly surveys 361 schoolmates and finds that 135 report they fear public speaking. Conduct a hypothesis test to determine if the percent at their 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% 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).
9-100 §9.5 source ↗
The following questions were written by past students. They are excellent problems!
According to New York City's Community Health Survey, the most recent adult smoking rate is 12%. Suppose that a survey is conducted to determine this year’s rate. Nine out of 70 randomly chosen N.Y. City residents reply that they smoke. Conduct a hypothesis test to determine if the rate is still 12% or if it has decreased.
Answer
a. H0: p = 0.12
b. Ha: p < 0.12
c. Let the proportion of NYC residents that smoke.
d. Normal for a single proportion
e. 0.2207
f. p-value = 0.5873
g. Answers may vary
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% significance level, there is insufficient evidence to conclude that the proportion of NYC residents who smoke is less than 0.14.
i. Confidence interval: Confidence Interval: (0.0502, 0.2070): The “plus-4s” confidence interval (see Confidence Intervals) is (0.0676, 0.2297).
9-102 §9.5 source ↗
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: p = 0.12
b. Ha: p < 0.12
c. Let the proportion of NYC residents that smoke.
d. Normal for a single proportion
e. 0.2207
f. p-value = 0.5873
g. Answers may vary
h.
i. alpha 0.05
ii. Decision: Do not reject the null hypothesis.
iii. Reason for decision: The p-value is less than 0.05.
iv. Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the mean salary of California registered nurses exceeds $69,110.
i. ($68,757, $73,485)
9-104 §9.5 source ↗
The following questions were written by past students. They are excellent problems!
Over the past few decades, public health officials have examined the link between weight concerns and teen girls' smoking. Researchers surveyed a group of 273 randomly selected teen girls living in Massachusetts (between 12 and 15 years old). After four years the girls were surveyed again. Sixty-three said they smoked to stay thin. Is there good evidence that more than 30% of the teen girls smoke to stay thin?
After conducting the test, your decision and conclusion are
a. Reject H0: There is sufficient evidence to conclude that more than 30% of teen girls smoke to stay thin.
b. Do not reject H0: There is not sufficient evidence to conclude that less than 30% of teen girls smoke to stay thin.
c. Do not reject H0: There is not sufficient evidence to conclude that more than 30% of teen girls smoke to stay thin.
d. Reject H0: There is sufficient evidence to conclude that less than 30% of teen girls smoke to stay thin.
Answer
9-106 §9.5 source ↗
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
9-108 §9.5 source ↗
Instructions: For the following ten exercises,
Hypothesis testing: For the following ten exercises, answer each question.
A recent survey in the N.Y. Times Almanac indicated that 48.8% of families own stock. A broker wanted to determine if this survey could be valid. They 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. There is enough evidence to conclude that the percent of families owning stocks is not 48.8%.
f. The survey does not appear to be accurate.
9-110 §9.5 source ↗
Instructions: For the following ten exercises,
Hypothesis testing: For the following ten exercises, answer each question.
The US Department of Energy reported that 51.7% 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 in Kentucky? 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% significance level, there is not enough evidence to conclude that the proportion of homes in Kentucky that are heated by natural gas is different from the proportion of homes heated by natural gas across the country.
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.
9-112 §9.5 source ↗
Instructions: For the following ten exercises,
Hypothesis testing: For the following ten exercises, answer each question.
The Weather Underground reported that the mean amount of summer rainfall for the northeastern US 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% 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% because the p-value is almost 0.
9-114 §9.5 source ↗
Instructions: For the following ten exercises,
Hypothesis testing: For the following ten 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% 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.
9-116 §9.5 source ↗
Instructions: For the following ten exercises,
Hypothesis testing: For the following ten exercises, answer each question.
The student academic group on a college campus claims that first-year 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 first-year 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% significance level, there is not enough evidence to conclude that first-year 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 Introductory Statistics 2e by Barbara Illowsky and Susan Dean, OpenStax, and are used under the CC BY-NC-SA 4.0 licence. Access for free at https://openstax.org/books/introductory-statistics-2e/pages/1-introduction. Illustrations: Copyright Rice University, OpenStax, under CC BY-NC-SA 4.0 license. This adaptation is released under the same licence, is non-commercial, and is neither affiliated with nor endorsed by OpenStax or Rice University.