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Describe how the confidence level affects the width of the large-sample confidence interval for p.
As the confidence level increases, the width of the confidence interval also increases.
Describe how the sample size affects the width of the large-sample confidence interval for p.
As the sample size increases, the width of the confidence interval decreases.
Describe how the value of p^ affects the width of the large-sample confidence interval for p.
When p^ = 0.5, the margin of error (for a fixed sample size) is maximum and decreases symmetrically as p^ decreases toward 0 or increases toward 1. A larger margin of error results in a wider confidence interval.
The study "Digital Footprints" (Pew Internet & American Life Project, www.pewinternet.org, 2007) reported that 47% of Internet users have searched for information about themselves online. The 47% figure was based on a random sample of Internet users. Suppose that the sample size was n = 300 (the actual sample size was much larger). Answer the four key questions (QSTN) to confirm that the suggested method in this situation is a large-sample confidence interval for a population proportion.
Question type (Q): Estimation; Study type (S): Sample data: Type of data (T): One categorical variable; Number of samples or treatments (N): One sample.
The article "Kids Digital Day: Almost 8 Hours" (USA Today, January 20, 2010) summarized a national survey of 2,002 Americans ages 8 to 18. The sample was selected to be representative of Americans in this age group. Of those surveyed, 1,321 reported owning a cell phone. Use this information to construct and interpret a 90% confidence interval for the proportion of all Americans ages 8 to 18 who own a cell phone.
(0.6426, 0.6774). You can be 90% confident that the actual proportion of all Americans ages 8 to 18 who own a cell phone is somewhere between 0.6426 and 0.6774.
The article "Kids Digital Day: Almost 8 Hours" (USA Today, January 20, 2010) summarized a national survey of 2,002 Americans ages 8 to 18. The sample was selected to be representative of Americans in this age group. Of those surveyed, 1,522 reported owning an MP3 music player. Use this information to construct and interpret a 90% confidence interval for the proportion of all Americans ages 8 to 18 who own an MP3 music player.
(0.7443, 0.7757). You can be 90% confident that the actual proportion of all Americans ages 8 to 18 who own an MP3 music player is somewhere between 0.7443 and 0.7757.
The article "Kids Digital Day: Almost 8 Hours" (USA Today, January 20, 2010) summarized a national survey of 2,002 Americans ages 8 to 18. The sample was selected to be representative of Americans in this age group. Explain why the 90% confidence interval for the proportion of all Americans ages 8 to 18 who own a cell phone is narrower than the 90% confidence interval for the proportion of all Americans ages 8 to 18 who own an MP3 music player even though the confidence levels and the sample sizes used to compute the two intervals were the same.
The interval of all Americans ages 8 to 18 who own a cell phone is narrower because p^ of 0.76 is farther from 0.5 than p^ of 0.66. For a fixed sample size and confidence level, the confidence interval is widest when p^ 0.5 and decreases symmetrically as p^ moves closer to 0 or 1.
Explain why the statement p^ = 0.40 is not a legitimate hypothesis.
p^ is a sample statistic. Hypotheses are about population characteristics.
CareerBuilder.com conducted a survey to learn about the proportion of employers who had ever sent an employee home because they were dressed inappropriately (June 17, 2008, www.careerbuilder.com). Suppose you are interested in determining if the resulting data provide strong evidence in support of the claim that more than one-third of employers have sent an employee home to change clothes. To answer this question, what null and alternative hypotheses should you test?
H0: p = 1/3 and Ha: p > 1/3
The report “How Teens Use Media” (Nielsen, June 2009) says that 67% of U.S. teens have a Facebook page that they update at least once a week. Suppose you plan to select a random sample of 400 students at the local high school. You will ask each student in the sample if he or she has a Facebook page and if it is updated at least once per week. You plan to use the resulting data to decide if there is evidence that the proportion of students at the high school who have a Facebook page that they update at least once a week differs from the national figure given in the Nielsen report. What hypotheses should you test?
H0: p = 0.67 and Ha: p ≠ 0.67
The article “Poll Finds Most Oppose Return to Draft, Wouldn’t Encourage Children to Enlist” (Associated Press, December 18, 2005) reports that in a random sample of 1,000 American adults, 430 answered yes to the following question: “If the military draft were reinstated, would you favor drafting women as well as men?” The data were used to test H0: p=0.5 versus Ha: p<0.5, and the null hypothesis was rejected. Based on the result of the hypothesis test, what can you conclude about the proportion of American adults who favor drafting women if a military draft were reinstated?
There is convincing evidence that the proportion of American adults who favor drafting women is less than 0.5.
The article “Poll Finds Most Oppose Return to Draft, Wouldn’t Encourage Children to Enlist” (Associated Press, December 18, 2005) reports that in a random sample of 1,000 American adults, 430 answered yes to the following question: “If the military draft were reinstated, would you favor drafting women as well as men?” The data were used to test H0: p=0.5 versus Ha: p<0.5, and the null hypothesis was rejected. Is it reasonable to say that the data provide strong support for the alternative hypothesis?
Yes
The article “Poll Finds Most Oppose Return to Draft, Wouldn’t Encourage Children to Enlist” (Associated Press, December 18, 2005) reports that in a random sample of 1,000 American adults, 430 answered yes to the following question: “If the military draft were reinstated, would you favor drafting women as well as men?” The data were used to test H0: p=0.5 versus Ha: p<0.5, and the null hypothesis was rejected. Is it reasonable to say that the data provide strong evidence against the null hypothesis?
Yes
Do state laws that allow private citizens to carry concealed weapons result in a reduced crime rate? The author of a study carried out by the Brookings Institution is reported as saying, “The strongest thing I could say is that I don’t see any strong evidence that they are reducing crime” (San Luis Obispo Tribune, January 23, 2003). Is this conclusion consistent with testing H0: concealed weapons laws reduce crime versus Ha: concealed weapons laws do not reduce crime or with testing H0: concealed weapons laws do not reduce crime versus Ha: concealed weapons laws reduce crime Explain.
The conclusion is consistent with testing H0; concealed weapons laws do not reduce crime versus Ha; concealed weapons laws reduce crime.
Do state laws that allow private citizens to carry concealed weapons result in a reduced crime rate? The author of a study carried out by the Brookings Institution is reported as saying, “The strongest thing I could say is that I don’t see any strong evidence that they are reducing crime” (San Luis Obispo Tribune, January 23, 2003). Does the stated conclusion indicate that the null hypothesis was rejected or not rejected? Explain.
The null hypothesis was not rejected because no evidence was found that the laws were reducing crime.
In a hypothesis test, what does it mean to say that the null hypothesis was rejected?
The sample data provide convincing evidence against the null hypothesis. If the null hypothesis were true, the sample data would be very unlikely.
Is p = 0.65 a legitimate hypotheses?
legitimate
Is p^ = 0.90 a legitimate hypotheses?
not legitimate
Is p^ = 0.10 a legitimate hypotheses?
not legitimate
Is p = 0.5 a legitimate hypotheses?
legitimate
Is p > 4.30 a legitimate hypotheses?
not legitimate
A college has decided to introduce the use of plus and minus with letter grades, as long as there is convincing evidence that more than 60% of the faculty favor the change. A random sample of faculty will be selected, and the resulting data will be used to test the relevant hypotheses. If p represents the proportion of all faculty who favor a change to plus–minus grading, which of the following pairs of hypotheses should be tested? H0: p = 0.6 verses Ha: p < 0.6 or H0: p = 0.6 verses Ha: p > 0.6. Explain your choice.
H0: p = 0.6 versus Ha: p > 0.6. In order to make the change, the university requires evidence that more than 60% of the faculty are in favor of the change.
One type of error in a hypothesis test is rejecting the null hypothesis when it is true. What is the other type of error that might occur when a hypothesis test is carried out?
Not rejecting the null hypothesis when it is not true
Suppose that for a particular hypothesis test, the consequences of a Type I error are very serious. Would you want to carry out the test using a small significance level α (such as 0.01) or a larger significance level (such as 0.10)? Explain the reason for your choice.
A small significance level because α is the probability of a Type I error.
Occasionally, warning flares of the type contained in most automobile emergency kits fail to ignite. A consumer group wants to investigate a claim that the proportion of defective flares made by a particular manufacturer is much higher than the advertised value of 0.1. A large number of flares will be tested, and the results will be used to decide between H0: H0: p = 0.01 and Ha: p > 0.01, where p represents the actual proportion of defective flares made by this manufacturer. If H0 is rejected, charges of false advertising will be filed against the manufacturer. Explain why the alternative hypothesis was chosen to be Ha: p > 0.01.
Before filing charges of false advertising against the company, the consumer advocacy group would require convincing evidence that more than 10% of the flares are defective.
Occasionally, warning flares of the type contained in most automobile emergency kits fail to ignite. A consumer group wants to investigate a claim that the proportion of defective flares made by a particular manufacturer is much higher than the advertised value of 0.1. A large number of flares will be tested, and the results will be used to decide between H0: H0: p = 0.01 and Ha: p > 0.01, where p represents the actual proportion of defective flares made by this manufacturer. If H0 is rejected, charges of false advertising will be filed against the manufacturer. Give a consequence of a Type I and Type II error.
A Type I error is thinking that more than 10% of the flares are defective when in fact 10% (or fewer) of the flares are defective. This would result in the expensive and time-consuming process of filing charges of false advertising against the company when the company advertising is not false. A Type II error is not thinking that more than 10% of the flares are defective when in fact more than 10% of the flares are defective. This would result in the consumer advocacy group not filing charges when the company advertising was false.
A television manufacturer states that at least 90% of its TV sets will not need service during the first 3 years of operation. A consumer group wants to investigate this statement. A random sample of n = 100 purchasers is selected and each person is asked if the set purchased needed repair during the first 3 years. Let p denote the proportion of all sets made by this manufacturer that will not need service in the first 3 years. The consumer group does not want to claim false advertising unless there is strong evidence that p < 0.09. The appropriate hypotheses are then H0: p = 0.09 versus Ha: p < 0.09. In the context of this problem, describe Type I and Type II errors, and discuss the possible consequences of each.
A Type I error would be thinking that less than 90% of the TV sets need no repair when in fact (at least) 90% need no repair. The consumer agency might take action against the manufacturer when the manufacturer is not at fault. A Type II error would be not thinking that less than 90% of the TV sets need no repair when in fact less than 90% need no repair. The consumer agency would not take action against the manufacturer when the manufacturer is making untrue claims about the reliability of the TV sets.
A television manufacturer states that at least 90% of its TV sets will not need service during the first 3 years of operation. A consumer group wants to investigate this statement. A random sample of n = 100 purchasers is selected and each person is asked if the set purchased needed repair during the first 3 years. Let p denote the proportion of all sets made by this manufacturer that will not need service in the first 3 years. The consumer group does not want to claim false advertising unless there is strong evidence that p < 0.09. The appropriate hypotheses are then H0: p = 0.09 versus Ha: p < 0.09. Would you recommend a test procedure that uses α = 0.01 or one that uses α = 0.10? Explain.
Taking action against the manufacturer when the manufacturer is not at fault could involve large and unnecessary legal costs to the consumer agency. α = 0.01 should be recommended.
Describe the two types of errors that might be made when a hypothesis test is carried out.
A Type I error is rejecting a true null hypothesis, and a Type II error is not rejecting a false null hypothesis
The article “Fewer Parolees Land Back Behind Bars” (Associated Press, April 11, 2006) includes the following statement: “Just over 38% of all felons who were released from prison in 2003 landed back behind bars by the end of the following year, the lowest rate since 1979.” Explain why it would not be necessary to carry out a hypothesis test to determine if the proportion of felons released in 2003 who landed back behind bars by the end of the following year was less than 0.40.
The “38%” value given in the article is the proportion of all felons; in other words, it is a population proportion. Therefore, you know that the population proportion is less than 0.4, and there is no need for a hypothesis test.
The article “Breast-Feeding Rates Up Early” (USA Today, Sept. 14, 2010) summarizes a survey of mothers whose babies were born in 2009. The Center for Disease Control sets goals for the proportion of mothers who will still be breast-feeding their babies at various ages. The goal for 12 months after birth is 0.25 or more. Suppose that the survey used a random sample of 1,200 mothers and that you want to use the survey data to decide if there is evidence that the goal is not being met. Let p denote the proportion of all mothers of babies born in 2009 who were still breast-feeding at 12 months. Describe the shape, center, and spread of the sampling distribution of p^ for random samples of size 1,200 if the null hypothesis H0: p = 0.25 is true.
The sampling distribution of p^ will be approximately normal with a mean of 0.25 and a standard deviation of 0.0125.
The article “Breast-Feeding Rates Up Early” (USA Today, Sept. 14, 2010) summarizes a survey of mothers whose babies were born in 2009. The Center for Disease Control sets goals for the proportion of mothers who will still be breast-feeding their babies at various ages. The goal for 12 months after birth is 0.25 or more. Suppose that the survey used a random sample of 1,200 mothers and that you want to use the survey data to decide if there is evidence that the goal is not being met. Let p denote the proportion of all mothers of babies born in 2009 who were still breast-feeding at 12 months. Would you be surprised to observe a sample proportion of p^ = 0.24 for a sample of size 1,200 if the null hypothesis H0: p = 0.25 were true? Explain why or why not.
A sample proportion of p^ = 0.24 would not be surprising if p = 0.25, because it is less than one standard deviation below 0.25. This is not unusual for a normal distribution.
The article “Breast-Feeding Rates Up Early” (USA Today, Sept. 14, 2010) summarizes a survey of mothers whose babies were born in 2009. The Center for Disease Control sets goals for the proportion of mothers who will still be breast-feeding their babies at various ages. The goal for 12 months after birth is 0.25 or more. Suppose that the survey used a random sample of 1,200 mothers and that you want to use the survey data to decide if there is evidence that the goal is not being met. Let p denote the proportion of all mothers of babies born in 2009 who were still breast-feeding at 12 months. Would you be surprised to observe a sample proportion as small as p^ = 0.20 for a sample of size 1,200 if the null hypothesis H0: p = 0.25  were true? Explain why or why not.
A sample proportion of p^ = 0.20 would be surprising if p = 0.25, because it is 4 standard deviations below 0.25. This would be unusual for a normal distribution.
The article “Breast-Feeding Rates Up Early” (USA Today, Sept. 14, 2010) summarizes a survey of mothers whose babies were born in 2009. The Center for Disease Control sets goals for the proportion of mothers who will still be breast-feeding their babies at various ages. The goal for 12 months after birth is 0.25 or more. Suppose that the survey used a random sample of 1,200 mothers and that you want to use the survey data to decide if there is evidence that the goal is not being met. Let p denote the proportion of all mothers of babies born in 2009 who were still breast-feeding at 12 months. The actual sample proportion observed in the study was p^ = 0.22. Based on this sample proportion, is there convincing evidence that the goal is not being met, or is the observed sample proportion consistent with what you would expect to see when the null hypothesis is true? Support your answer with a probability calculation.
If p = 0.25, P(p^≤ 0.22) = P(z ≤ -2.4) = 0.0082. Because this probability is small, there is convincing evidence that the goal is not being met.
The article “The Benefits of Facebook Friends: Social Capital and College Students’ Use of Online Social Network Sites” (Journal of Computer-Mediated Communication [2007]: 1143–1168) describes a study of n = 286 undergraduate students at Michigan State University. Suppose that it is reasonable to regard this sample as a random sample of undergraduates at Michigan State. You want to use the survey data to decide if there is evidence that more than 75% of the students at this university have a Facebook page that includes a photo of themselves. Let p denote the proportion of all Michigan State undergraduates who have such a page. Describe the shape, center, and spread of the sampling distribution of p^ for random samples of size 286 if the null hypothesis H0: p = 0.75 is true.
The sampling distribution of p^ will be approximately normal with a mean of 0.75 and a standard deviation of 0.0256.
The article “The Benefits of Facebook Friends: Social Capital and College Students’ Use of Online Social Network Sites” (Journal of Computer-Mediated Communication [2007]: 1143–1168) describes a study of n = 286 undergraduate students at Michigan State University. Suppose that it is reasonable to regard this sample as a random sample of undergraduates at Michigan State. You want to use the survey data to decide if there is evidence that more than 75% of the students at this university have a Facebook page that includes a photo of themselves. Let p denote the proportion of all Michigan State undergraduates who have such a page. Would you be surprised to observe a sample proportion of p^ = 0.83 for a sample of size 286 if the null hypothesis H0: p = 0.75  were true? Explain why or why not.
A sample proportion of p^ = 0.83 would be surprising if p = 0.75, because it is more than 3 standard deviations above 0.75. This would be unusual for a normal distribution.
The article “The Benefits of Facebook Friends: Social Capital and College Students’ Use of Online Social Network Sites” (Journal of Computer-Mediated Communication [2007]: 1143–1168) describes a study of n = 286 undergraduate students at Michigan State University. Suppose that it is reasonable to regard this sample as a random sample of undergraduates at Michigan State. You want to use the survey data to decide if there is evidence that more than 75% of the students at this university have a Facebook page that includes a photo of themselves. Let p denote the proportion of all Michigan State undergraduates who have such a page. Would you be surprised to observe a sample proportion of p^ = 0.79 for a sample of size 286 if the null hypothesis H0: p = 0.75  were true? Explain why or why not.
A sample proportion of p^ = 0.79 would not be surprising if p = 0.75, because it is about 1.5 standard deviations above 0.75. This is not unusual for a normal distribution.
The article “The Benefits of Facebook Friends: Social Capital and College Students’ Use of Online Social Network Sites” (Journal of Computer-Mediated Communication [2007]: 1143–1168) describes a study of n = 286 undergraduate students at Michigan State University. Suppose that it is reasonable to regard this sample as a random sample of undergraduates at Michigan State. You want to use the survey data to decide if there is evidence that more than 75% of the students at this university have a Facebook page that includes a photo of themselves. Let p denote the proportion of all Michigan State undergraduates who have such a page. The actual sample proportion observed in the study was p^ = 0.80. Based on this sample proportion, is there convincing evidence that the null hypothesis H0: p = 0.75  is not true, or is p^ consistent with what you would expect to see when the null hypothesis is true? Support your answer with a probability calculation.
If p = 0.75, P(p^ ≥ 0.80) = P(z ≥ 1.95) = 0.0256. Because this probability is small, there is convincing evidence that the null hypothesis is not true.
Every year on Groundhog Day (February 2), the famous groundhog Punxsutawney Phil tries to predict whether there will be 6 more weeks of winter. The article “Groundhog Has Been Off Target” (USA Today, Feb. 1, 2011) states that “based on weather data, there is no predictive skill for the groundhog.” Suppose that you plan to take a random sample of 20 years and use weather data to determine the proportion of these years the groundhog’s prediction was correct. Describe the shape, center, and spread of the sampling distribution of p^ for samples of size 20 if the groundhog has only a 50–50 chance of making a correct prediction.
The sampling distribution of p^ will be approximately normal with a mean of 0.5 and a standard deviation of 0.1118.
Every year on Groundhog Day (February 2), the famous groundhog Punxsutawney Phil tries to predict whether there will be 6 more weeks of winter. The article “Groundhog Has Been Off Target” (USA Today, Feb. 1, 2011) states that “based on weather data, there is no predictive skill for the groundhog.” Suppose that you plan to take a random sample of 20 years and use weather data to determine the proportion of these years the groundhog’s prediction was correct. What sample proportion values would convince you that the groundhog’s predictions have a better than 50–50 chance of being correct?
One reasonable answer would be sample proportions greater than 0.7236 (these are sample proportions that are more than two standard deviations above 0.5).
Use the definition of the P-value to explain why H0 would be rejected if P-value = 0.0003.
A P-value of 0.0003 means that it is very unlikely (probability = 0.0003), assuming that H0 is true, that you would get a sample result at least as inconsistent with H0 as the one obtained in the study. H0 would be rejected.
Use the definition of the P-value to explain why H0 would not be rejected if P-value = 0.350.
A P-value of 0.350 means that it is not particularly unlikely (probability = 0.350), assuming that H0 is true, that you would get a sample result at least as inconsistent with H0 as the one obtained in the study. There is no reason to reject H0.
The article “Poll Finds Most Oppose Return to Draft, Wouldn’t Encourage Children to Enlist” (Associated Press, December 18, 2005) reports that in a random sample of 1,000 American adults, 700 indicated that they oppose the reinstatement of a military draft. Suppose you want to use this information to decide if there is convincing evidence that the proportion of American adults who oppose reinstatement of the draft is greater than two-thirds. What hypotheses should be tested in order to answer this question?
H0: p = 2/3 and Ha: p > 2/3
The article “Poll Finds Most Oppose Return to Draft, Wouldn’t Encourage Children to Enlist” (Associated Press, December 18, 2005) reports that in a random sample of 1,000 American adults, 700 indicated that they oppose the reinstatement of a military draft. Suppose you want to use this information to decide if there is convincing evidence that the proportion of American adults who oppose reinstatement of the draft is greater than two-thirds. The P-value for this test is 0.013. What conclusion would you reach if α = 0.05?
The null hypothesis would be rejected because the P-value (0.013) is less than α=0.05.
Explain why a P-value of 0.0002 would be interpreted as strong evidence against the null hypothesis.
A P-value of 0.0002 means that it is very unlikely (probability = 0.0002), assuming that H0 is true, that you would get a sample result at least as inconsistent with H0 as the one obtained in the study. This is strong evidence against the null hypothesis.
Answer the following four key questions (Q: Estimating or hypothesis testing? S: Sample data or experiment data? T: One variable or two? Categorical or numerical? N: How many samples or treatments?) on the following scenario and indicate whether the method that you would consider would be a large-sample hypothesis test for a population proportion: The paper “College Students’ Social Networking Experiences on Facebook” (Journal of Applied Developmental Psychology [2009]: 227–238) summarized a study in which 92 students at a private university were asked how much time they spent on Facebook on a typical weekday. The researchers were interested in estimating the average time spent on Facebook by students at this university.
Estimation, sample data, one numerical variable, one sample. A hypothesis test for a population proportion would not be appropriate.
Answer the following four key questions (Q: Estimating or hypothesis testing? S: Sample data or experiment data? T: One variable or two? Categorical or numerical? N: How many samples or treatments?) on the following scenario and indicate whether the method that you would consider would be a large-sample hypothesis test for a population proportion: The article “iPhone Can Be Addicting, Says New Survey” (msnbc.com, March 8, 2010) described a survey administered to 200 college students who owned an iPhone. One of the questions on the survey asked students if they slept with their iPhone in bed with them. You would like to use the data from this survey to determine if there is convincing evidence that a majority of college students with iPhones sleep with their phones.
Hypothesis testing, sample data, one categorical variable, one sample. A hypothesis test for a population proportion would be appropriate.
Answer the following four key questions (Q: Estimating or hypothesis testing? S: Sample data or experiment data? T: One variable or two? Categorical or numerical? N: How many samples or treatments?) on the following scenario and indicate whether the method that you would consider would be a large-sample hypothesis test for a population proportion: USA Today (Feb. 17, 2011) reported that 10% of 1,008 American adults surveyed about their use of e-mail said that they had ended a relationship by e-mail. You would like to use this information to estimate the proportion of all adult Americans who have used e-mail to end a relationship.
Estimation, sample data, one categorical variable, one sample. A hypothesis test for a population proportion would not be appropriate.
Assuming a random sample from a large population, is the following null hypotheses and sample sizes is the large-sample z test appropriate? H0: p = 0.2, n = 25
Large-sample z test is not appropriate.
Assuming a random sample from a large population, is the following null hypotheses and sample sizes is the large-sample z test appropriate? H0: p = 0.6, n = 200
Large-sample z test is appropriate.
Assuming a random sample from a large population, is the following null hypotheses and sample sizes is the large-sample z test appropriate? H0: p = 0.9, n = 100
Large-sample z test is appropriate.
Assuming a random sample from a large population, is the following null hypotheses and sample sizes is the large-sample z test appropriate? H0: p = 0.05, n = 75
Large-sample z test is not appropriate.
Let p denote the proportion of students at a large university who plan to purchase a campus meal plan in the next academic year. For a large-sample z test of H0: p = 0.20 versus Ha: p =<0.20 , find the P-value associated with a z test statistic of –0.55.
0.2912
Let p denote the proportion of students at a large university who plan to purchase a campus meal plan in the next academic year. For a large-sample z test of H0: p = 0.20 versus Ha: p =<0.20 , find the P-value associated with a z test statistic of –0.92.
0.1788
Let p denote the proportion of students at a large university who plan to purchase a campus meal plan in the next academic year. For a large-sample z test of H0: p = 0.20 versus Ha: p =<0.20 , find the P-value associated with a z test statistic of –1.99.
0.0233
Let p denote the proportion of students at a large university who plan to purchase a campus meal plan in the next academic year. For a large-sample z test of H0: p = 0.20 versus Ha: p =<0.20 , find the P-value associated with a z test statistic of –2.24.
0.0125
Let p denote the proportion of students at a large university who plan to purchase a campus meal plan in the next academic year. For a large-sample z test of H0: p = 0.20 versus Ha: p =<0.20 , find the P-value associated with a z test statistic of 1.40.
0.9192
The paper “Debt Literacy, Financial Experiences and Over-Indebtedness” (Social Science Research Network, Working paper W14808, 2008) included data from a survey of 1,000 Americans. One question on the survey was: “You owe $3,000 on your credit card. You pay a minimum payment of $30 each month. At an Annual Percentage Rate of 12% (or 1% per month), how many years would it take to eliminate your credit card debt if you made no additional charges?” Answer options for this question were: (a) less than 5 years; (b) between 5 and 10 years; (c) between 10 and 15 years; (d) never—you will continue to be in debt; (e) don’t know; and (f) prefer not to answer. Only 354 of the 1,000 respondents chose the correct answer of never. Assume that the sample is representative of adult Americans. Is there convincing evidence that the proportion of adult Americans who can answer this question correctly is less than 0.40 (40%)? Use the five-step process for hypothesis testing (HMCCC) described in this section and α = 0.05 to test the appropriate hypotheses.
H0: p = 0.4, Ha: p < 0.4, z = -2.969, P-value = 0.001, reject H0. There is convincing evidence that the proportion of all adult Americans who would answer the question correctly is less than 0.4.
The paper “Debt Literacy, Financial Experiences and Over-Indebtedness” (Social Science Research Network, Working paper W14808, 2008) included data from a survey of 1,000 Americans. One question on the survey was: “You owe $3,000 on your credit card. You pay a minimum payment of $30 each month. At an Annual Percentage Rate of 12% (or 1% per month), how many years would it take to eliminate your credit card debt if you made no additional charges?” Answer options for this question were: (a) less than 5 years; (b) between 5 and 10 years; (c) between 10 and 15 years; (d) never—you will continue to be in debt; (e) don’t know; and (f) prefer not to answer. The paper also reported that 37.8% of those in the sample chose one of the wrong answers (a, b, or c) as their response to this question. Is it reasonable to conclude that more than one-third of adult Americans would select a wrong answer to this question? Use α = 0.05.
H0: p = 1/3, Ha: p < 1/3, z = 2.996, P-value = 0.001, reject H0. There is convincing evidence that more than one-third of adult Americans would select a wrong answer.
The paper “Teens and Distracted Driving” (Pew Internet & American Life Project, 2009) reported that in a representative sample of 283 American teens ages 16 to 17, there were 74 who indicated that they had sent a text message while driving. For purposes of this exercise, assume that this sample is a random sample of 16- to 17-year-old Americans. Do these data provide convincing evidence that more than a quarter of Americans ages 16 to 17 have sent a text message while driving? Test the appropriate hypotheses using a significance level of 0.01.
H0: p = 0.25, Ha: p > 0.25, z = 0.45, P-value = 0.328, reject H0, fail to reject H0. There is no convincing evidence that more than 25% of Americans ages 16 to 17 have sent a text message while driving.
The article “Theaters Losing Out to Living Rooms” (San Luis Obispo Tribune, June 17, 2005) states that movie attendance declined in 2005. The Associated Press found that 730 of 1,000 randomly selected adult Americans prefer to watch movies at home rather than at a movie theater. Is there convincing evidence that a majority of adult Americans prefer to watch movies at home? Test the relevant hypotheses using a 0.05 significance level.
H0: p = 0.5, Ha: p > 0.5, z = 14.546, P-value ≈ 0, reject H0. There is convincing evidence that a majority of adult Americans prefer to watch movies at home.
The article referenced in the previous exercise also reported that 470 of 1,000 randomly selected adult Americans thought that the quality of movies being produced is getting worse. Is there convincing evidence that less than half of adult Americans believe that movie quality is getting worse? Use a significance level of 0.05.
z = -1.897, P-value = 0.0289, reject H0 
The article referenced in the previous exercise also reported that 470 of 1,000 randomly selected adult Americans thought that the quality of movies being produced is getting worse. Suppose that the sample size had been 100 instead of 1,000, and that 47 thought that movie quality was getting worse (so that the sample proportion is still 0.47). Based on this sample of 100, is there convincing evidence that less than half of adult Americans believe that movie quality is getting worse? Use a significance level of 0.05.
z = -0.6, P-value = 0.274, fail to reject H0 
In a survey of 1,005 adult Americans, 46% indicated that they were somewhat interested or very interested in having Web access in their cars (USA Today, May 1, 2009). Suppose that the marketing manager of a car manufacturer claims that the 46% is based only on a sample and that 46% is close to half, so there is no reason to believe that the proportion of all adult Americans who want car Web access is less than 0.50. Is the marketing manager correct in his claim? Provide statistical evidence to support your answer. For purposes of this exercise, assume that the sample can be considered representative of adult Americans.
 H0: p = 0.5, Ha: = p < 0.5, z = -2.536, P-value = 0.006, reject H0. There is convincing evidence that the proportion of all adult Americans who want car Web access is less than 0.5. The marketing manager is not correct in his claim.
In a survey of 1,000 women ages 22 to 35 who work full-time, 540 indicated that they would be willing to give up some personal time in order to make more money (USA Today, March 4, 2010). The sample was selected to be representative of women in the targeted age group. Do the sample data provide convincing evidence that a majority of women ages 22 to 35 who work full-time would be willing to give up some personal time for more money? Test the relevant hypotheses using α = 0.01.
z = 2.530, P-value = 0.0057, reject H0
In a survey of 1,000 women ages 22 to 35 who work full-time, 540 indicated that they would be willing to give up some personal time in order to make more money (USA Today, March 4, 2010). The sample was selected to be representative of women in the targeted age group. Would it be reasonable to generalize the conclusion of a significance test to all working women? Explain why or why not.
No. The survey only included women ages 22 to 35.
In a representative sample of 2,013 American adults, 1,590 indicated that lack of respect and courtesy in American society is a serious problem (Associated Press, April 3, 2002). Is there convincing evidence that more than three-quarters of American adults believe that lack of respect and courtesy is a serious problem? Test the relevant hypotheses using a significance level of 0.05.
H0: p = 0.75, Ha: p > 0.75, z = 4.13, P-value = 0.0057, reject H0. There is convincing evidence that more than three-quarters of American adults believe that lack of respect and courtesy is a serious problem.
A number of initiatives on the topic of legalized gambling have appeared on state ballots. A political candidate has decided to support legalization of casino gambling if he is convinced that more than two-thirds of American adults approve of casino gambling. Suppose that 1,035 of the people in a random sample of 1,523 American adults said they approved of casino gambling. Is there convincing evidence that more than two-thirds approve?
z = 1.069, P-value = 0.143, fail to reject H0
The survey described in the previous exercise also asked the individuals in the sample what they thought was their best chance to obtain more than $500,000 in their lifetimes. Twenty-eight percent responded “win a lottery or sweepstakes.” Does this provide convincing evidence that more than one-fourth of U.S. adults see a lottery or sweepstakes win as their best chance of accumulating $500,000? Carry out a test using a significance level of 0.01.
H0: p = 0.25, Ha: p > 0.25, z = 2.202, P-value = 0.014, reject H0. There is not convincing evidence that more than one-fourth of American adults see a lottery or sweepstakes win as their best chance of accumulating $500,000.
The article “Americans Seek Spiritual Guidance on Web” (San Luis Obispo Tribune, October 12, 2002) reported that 68% of the U.S. population belongs to a religious community. In a survey on Internet use, 84% of “religion surfers” (defined as those who seek spiritual help online or who have used the Web to search for prayer and devotional resources) belong to a religious community. Suppose that this result was based on a representative sample of 512 religion surfers. Is there convincing evidence that the proportion of religion surfers who belong to a religious community is different from 0.68, the proportion for the U.S. population? Use α = 0.05.
H0: p = 0.68, Ha: p ≠ 0.25, z = 7.761, P-value ≈ 0, reject H0. There is not convincing evidence that more than one-fourth of American adults see a lottery or sweepstakes win as their best chance of accumulating $500,000. There is convincing evidence that the proportion of religion surfers who belong to a religious community is different from 0.68.
Suppose you plan to test H0: p = 0.40 versus Ha: p > 0.40 . Assuming the same significance level, would the power of the test be greater if n = 50 or if n = 100? Explain your choice.
n = 100; the power is greater for a larger sample size.
Suppose you plan to test H0: p = 0.25 versus Ha: p ≠ 0.25 . Assuming the same sample size and significance level, would the power of the test be smaller if the actual value of the population proportion were 0.15 or 0.30? Explain your choice.
p = 0.30; the power is smaller when the actual value of p is closer to the hypothesized value.
Suppose you plan to test H0: p = 0.82 versus Ha: p > 0.82. Assuming the same sample size, would the power of the test be greater for a significance level of α = 0.01 or α = 0.05? Explain your choice.
α = 0.05. The power is greater when the significance level is larger.
The city council in a large city has become concerned about the trend toward exclusion of renters with children in apartments within the city. The housing coordinator has decided to select a random sample of 125 apartments and determine for each whether children are permitted. Let p be the proportion of all apartments that prohibit children. If the city council is convinced that p is greater than 0.75, it will consider appropriate legislation. If 102 of the 125 sampled apartments exclude renters with children, would a test with α=0.05 lead you to the conclusion that more than 75% of all apartments exclude children?
Testing H0: p = 0.75 against Ha: p > 0.75, z = 1.704, P-value = 0.044. H0 is rejected, there is convincing evidence that more than 75% of apartments exclude children.
The city council in a large city has become concerned about the trend toward exclusion of renters with children in apartments within the city. The housing coordinator has decided to select a random sample of 125 apartments and determine for each whether children are permitted. Let p be the proportion of all apartments that prohibit children. If the city council is convinced that p is greater than 0.75, it will consider appropriate legislation. What is the power of the test when p=0.8 and α=0.05?
0.351
The report “How Teens Use Media” (Nielsen, June 2009) says that 37% of U.S. teens access the Internet from a mobile phone. Suppose you plan to select a random sample of students at the local high school and will ask each student in the sample if he or she accesses the Internet from a mobile phone. You want to determine if there is evidence that the proportion of students at the high school who access the Internet using a mobile phone differs from the national figure of 0.37 given in the Nielsen report. What hypotheses should you test?
H0: p = 0.37, Ha: p ≠ 0.37
Explain why failing to reject the null hypothesis in a hypothesis test does not mean there is convincing evidence that the null hypothesis is true.
Failing to reject the null hypothesis means that you are not convinced that the null hypothesis is false. This is not the same as being convinced that it is true.
The National Cancer Institute conducted a 2-year study to determine whether cancer death rates for areas near nuclear power plants are higher than for areas without nuclear facilities (San Luis Obispo Telegram-Tribune, September 17, 1990). A spokesperson for the Cancer Institute said, “From the data at hand, there was no convincing evidence of any increased risk of death from any of the cancers surveyed due to living near nuclear facilities. However, no study can prove the absence of an effect.” Suppose p denotes the true proportion of the population in areas near nuclear power plants who die of cancer during a given year. The researchers at the Cancer Institute might have considered two rival hypotheses of the form H0: p is equal to the corresponding value for areas without nuclear facilities ha: p is greater than the corresponding value for areas without nuclear facilities Did the researchers reject H0 or fail to reject H0?
The researchers failed to reject H0 
The National Cancer Institute conducted a 2-year study to determine whether cancer death rates for areas near nuclear power plants are higher than for areas without nuclear facilities (San Luis Obispo Telegram-Tribune, September 17, 1990). A spokesperson for the Cancer Institute said, “From the data at hand, there was no convincing evidence of any increased risk of death from any of the cancers surveyed due to living near nuclear facilities. However, no study can prove the absence of an effect.” If the Cancer Institute researchers are incorrect in their conclusion that there is no evidence of increased risk of death from cancer associated with living near a nuclear power plant, are they making a Type I or a Type II error? Explain.
Type II error
The National Cancer Institute conducted a 2-year study to determine whether cancer death rates for areas near nuclear power plants are higher than for areas without nuclear facilities (San Luis Obispo Telegram-Tribune, September 17, 1990). A spokesperson for the Cancer Institute said, “From the data at hand, there was no convincing evidence of any increased risk of death from any of the cancers surveyed due to living near nuclear facilities. However, no study can prove the absence of an effect.” Comment on the spokesperson’s last statement that no study can prove the absence of an effect. Do you agree with this statement?
Yes
The article “Cops Get Screened for Digital Dirt” (USA Today, Nov. 12, 2010) summarizes a report on law enforcement agency use of social media to screen applicants for employment. The report was based on a survey of 728 law enforcement agencies. One question on the survey asked if the agency routinely reviewed applicants’ social media activity during background checks. For purposes of this exercise, suppose that the 728 agencies were selected at random and that you want to use the survey data to decide if there is convincing evidence that more than 25% of law enforcement agencies review applicants’ social media activity as part of routine background checks. Describe the shape, center, and spread of the sampling distribution of p^ for samples of size 728 if the null hypothesis H0: p = 0.25 is true.
The sampling distribution of p^ will be approximately normal with a mean of 0.25 and a standard deviation of 0.0160.
The article “Cops Get Screened for Digital Dirt” (USA Today, Nov. 12, 2010) summarizes a report on law enforcement agency use of social media to screen applicants for employment. The report was based on a survey of 728 law enforcement agencies. One question on the survey asked if the agency routinely reviewed applicants’ social media activity during background checks. For purposes of this exercise, suppose that the 728 agencies were selected at random and that you want to use the survey data to decide if there is convincing evidence that more than 25% of law enforcement agencies review applicants’ social media activity as part of routine background checks. Would you be surprised to observe a sample proportion of p^ = 0.27 for a sample of size 728 if the null hypothesis H0: p = 0.25 were true? Explain why or why not.
A sample proportion of p^ = 0.27 would not be surprising if p = 0.25, because it is only 1.25 standard deviations above 0.25. This is not unusual for a normal distribution.
The article “Cops Get Screened for Digital Dirt” (USA Today, Nov. 12, 2010) summarizes a report on law enforcement agency use of social media to screen applicants for employment. The report was based on a survey of 728 law enforcement agencies. One question on the survey asked if the agency routinely reviewed applicants’ social media activity during background checks. For purposes of this exercise, suppose that the 728 agencies were selected at random and that you want to use the survey data to decide if there is convincing evidence that more than 25% of law enforcement agencies review applicants’ social media activity as part of routine background checks. Would you be surprised to observe a sample proportion of p^ = 0.31 for a sample of size 728 if the null hypothesis H0: p = 0.25 were true? Explain why or why not.
A sample proportion of p^ = 0.31 would be surprising if p = 0.25, because it is 3.75 standard deviations above 0.25. This would be unusual for a normal distribution.
The article “Cops Get Screened for Digital Dirt” (USA Today, Nov. 12, 2010) summarizes a report on law enforcement agency use of social media to screen applicants for employment. The report was based on a survey of 728 law enforcement agencies. One question on the survey asked if the agency routinely reviewed applicants’ social media activity during background checks. For purposes of this exercise, suppose that the 728 agencies were selected at random and that you want to use the survey data to decide if there is convincing evidence that more than 25% of law enforcement agencies review applicants’ social media activity as part of routine background checks. The actual sample proportion observed in the study was p^ = 0.33. Based on this sample proportion, is there convincing evidence that more than 25% of law enforcement agencies review social media activity as part of background checks, or is this sample proportion consistent with what you would expect to see when the null hypothesis is true?
If p = 0.25, P(p^ ≥ 0.33) = P(z ≥ 5) ≈ 0. Because this probability is approximately 0, there is convincing evidence that more than 25% of law enforcement agencies review social media activity.
The report “Credit Card Statistics, Industry Facts, Debt Statistics” (creditcards.com) summarizes a survey of college undergraduates. Each person in a sample of students was asked about his or her credit card balance, and the average balance was calculated to be $3,173. You are interested in determining if there is evidence that the mean credit card debt for undergraduates is greater than the known average from the previous year.
Hypothesis testing, sample data, one numerical variable, one sample. A hypothesis test for a population proportion would not be appropriate.
Suppose that the sample of 899 college students described in the previous exercise can be regarded as representative of college students in the United States. What hypotheses would you test to answer the question posed in the previous exercise?
H0: p = 0.25, Ha: p > 0.25
Suppose that the sample of 899 college students described in the previous exercise can be regarded as representative of college students in the United States. Is the sample size large enough for a large-sample test for a population proportion to be appropriate?
Yes
Suppose that the sample of 899 college students described in the previous exercise can be regarded as representative of college students in the United States. What is the value of the test statistic and the associated P-value for this test?
z = 3.33, P-value ≈ 0
Suppose that the sample of 899 college students described in the previous exercise can be regarded as representative of college students in the United States. If a significance level of 0.05 were chosen, would you reject the null hypothesis or fail to reject the null hypothesis?
reject H0
Past experience is that when individuals are approached with a request to fill out and return a particular questionnaire in a provided stamped and addressed envelope, the response rate is 40%. An investigator believes that if the person distributing the questionnaire were stigmatized in some obvious way, potential respondents would feel sorry for the distributor and thus tend to respond at a rate higher than 40%. To test this theory, a distributor wore an eye patch. Of the 200 questionnaires distributed by this individual, 109 were returned. Does this provide evidence that the response rate in this situation is greater than the previous rate of 40%? State and test the appropriate hypotheses using a significance level of 0.05.
H0: p = 0.40, Ha: p > 0.40, z = 4.186, P-value ≈ 0, reject H0. There is convincing evidence that the response rate is greater than 40%.
USA Today (February 16, 2012) reported that the percentage of U.S. residents living in poverty was 12.5% for men and 15.1% for women. These percentages were estimates based on data from large representative samples of men and women. Suppose that the sample sizes were 1,200 for men and 1,000 for women. You would like to use the survey data to estimate the difference in the proportion living in poverty for men and women. Answer the four key questions (QSTN) for this problem. What method would you consider based on the answers to these questions?
Estimation, sample data, one categorical variable, two samples. A large-sample confidence interval for a difference in proportions should be considered.
USA Today (February 16, 2012) reported that the percentage of U.S. residents living in poverty was 12.5% for men and 15.1% for women. These percentages were estimates based on data from large representative samples of men and women. Suppose that the sample sizes were 1,200 for men and 1,000 for women. You would like to use the survey data to estimate the difference in the proportion living in poverty for men and women. Use the five-step process for estimation problems (EMCCC) to compute and interpret a 90% large-sample confidence interval for the difference in the proportion living in poverty for men and women.
(-0.050, -0.002). You can be 90% confident that the actual difference between the proportion of male U.S. residents living in poverty and this proportion for females is between -0.050 and -0.002. Because both endpoints of this interval are negative, you would estimate that the proportion of men living in poverty is smaller than the proportion of women living in poverty by somewhere between 0.002 and 0.050.
The article "Portable MP3 Player Ownership Reaches New High" (Ipsos Insight, June 29, 2006) reported that in 2006, 20% of those in a random sample of 1,112 Americans ages 12 and older indicated that they owned an MP3 player. In a similar survey conducted in 2005, only 15% reported owning an MP3 player. Suppose that the 2005 figure was also based on a random sample of size 1,112. Are the sample sizes large enough to use the large-sample confidence interval for a difference in population proportions?
Yes. p^1 = 0.20 and p^2 = 0.15. Then n1*p^1 = 222.4, n1(1 - p^1) = 889.6, n2*p^2 =166.8, n2(1 - p^2) = 945.2 are all greater than 10.
The article "Portable MP3 Player Ownership Reaches New High" (Ipsos Insight, June 29, 2006) reported that in 2006, 20% of those in a random sample of 1,112 Americans ages 12 and older indicated that they owned an MP3 player. In a similar survey conducted in 2005, only 15% reported owning an MP3 player. Suppose that the 2005 figure was also based on a random sample of size 1,112. Estimate the difference in the proportion of Americans ages 12 and older who owned an MP3 player in 2006 and the corresponding proportion for 2005 using a 95% confidence interval.
(0.018, 0.082)
The article "Portable MP3 Player Ownership Reaches New High" (Ipsos Insight, June 29, 2006) reported that in 2006, 20% of those in a random sample of 1,112 Americans ages 12 and older indicated that they owned an MP3 player. In a similar survey conducted in 2005, only 15% reported owning an MP3 player. Suppose that the 2005 figure was also based on a random sample of size 1,112. Is zero included in the 95% confidence interval? What does this suggest about the change in this proportion from 2005 to 2006?
Zero is not included in the confidence interval. This means that you can be confident that the proportion of Americans ages 12 and older who owned an MP3 player was greater in 2006 than in 2005.
The article "Portable MP3 Player Ownership Reaches New High" (Ipsos Insight, June 29, 2006) reported that in 2006, 20% of those in a random sample of 1,112 Americans ages 12 and older indicated that they owned an MP3 player. In a similar survey conducted in 2005, only 15% reported owning an MP3 player. Suppose that the 2005 figure was also based on a random sample of size 1,112. Interpret the 95% confidence interval in the context of this problem.
You can be 95% confident that the proportion of Americans ages 12 and older who owned an MP3 player was greater in 2006 than in 2005 by somewhere between 0.018 and 0.082.
"Mountain Biking May Reduce Fertility in Men. Study Says" was the headline of an article appearing in the San Luis Obispo Tribune (December 3, 2002). This conclusion was based on an Austrian study that compared sperm counts of avid mountain bikers (those who ride at least 12 hours per week) and nonbikers. Ninety percent of the avid mountain bikers studied had low sperm count compared to 26% of the nonbikers. Suppose that these percentages were based on independent samples of 100 avid mountain bikers and 100 nonbikers and that these samples are representative of avid mountain bikers and nonbikers. Using a confidence level of 95%, estimate the difference between the proportion of avid mountain bikers with low sperm count and the proportion for nonbikers.
(0.536, 0.744). You can be 95% confident that the proportion of avid mountain bikers who have low sperm count is higher than the proportion for nonbikers by somewhere between 0.536 and 0.744.
"Mountain Biking May Reduce Fertility in Men. Study Says" was the headline of an article appearing in the San Luis Obispo Tribune (December 3, 2002). This conclusion was based on an Austrian study that compared sperm counts of avid mountain bikers (those who ride at least 12 hours per week) and nonbikers. Ninety percent of the avid mountain bikers studied had low sperm count compared to 26% of the nonbikers. Suppose that these percentages were based on independent samples of 100 avid mountain bikers and 100 nonbikers and that these samples are representative of avid mountain bikers and nonbikers. Is it reasonable to conclude that mountain biking 12 hours per week or more causes low sperm count based on a95% confidence interval? Explain.
No, because this was an observational study, and it is not a good idea to draw cause-and-effect conclusions from an observational study.
To learn about the proportion of college students who are registered to vote, surveys were conducted in 2010 and in 2011. Suppose that in 2010, 78% of the people in a representative sample of 2,300 college students were registered to vote. In 2011, 70% of the people in a representative sample of 2,294 college students were registered to vote. Use the five-step process for estimation problems (EMC³) to construct and interpret a 99% large-sample confidence interval for the difference in the proportion of college students who were registered to vote in 2010 and this proportion in 2011.
(0.047, 0.113). You can be 99% confident that the proportion of high school students who were registered to vote in 2010 was greater than this proportion in 2011 by somewhere between 0.047 and 0.114.
The article referenced in the previous exercise also reported that 5.7% of high school graduates were unemployed in 2008 and 9.7% of high school graduates were unemployed in 2009. Suppose that the reported percentages were based on independently selected representative samples of 400 high school graduates in each of these 2 years. Construct and interpret a 99% large-sample confidence interval for the difference in the proportions of high school graduates who were unemployed in these 2 years.
(-0.089, 0.009). Because 0 is included in this interval, there may be no difference in the proportion of high school graduates who were unemployed in 2008 and the proportion who were unemployed in 2009.