Module 13: Checkpoint #2: Hypothesis Testing for a Population Proportion (all 4 attempts)

A researcher wants to find out if U.S. adults still support the death penalty at a proportion of 0.64 (as it was in 2003). This graph indicates the sampling distribution for the proportion of supporters in random samples of 25 adults. The standard deviation is approximately 0.10. What is the approximate test statistic for p̂ = 0.84? A. −2 B. −1 C. 0 D. 1 E. 2

2 Correct. 0.84 is in the right-hand tail of the distribution because it is 2 standard deviations above the mean using the formula z = (p̂ − p)/√[(p(1 − p)/n].

Does secondhand smoke increase the risk of a low birthweight? A baby is considered have low birthweight if he/she weighs less than 5.5 pounds at birth. According to the National Center of Health Statistics, about 7.8% of all babies born in the U.S. are categorized as low birthweight. Suspecting that the national percentage is higher than 7.8%, researchers randomly select 1,200 babies whose mothers had extensive exposure to secondhand smoke during pregnancy and find that 10.4% of the sampled babies are categorized as low birth weight. Let p be the proportion of all babies in the United States who are categorized as low birth weight. What are the appropriate null and alternative hypotheses for this research question? A. H0: p = 0.078 Ha: p ≠ 0.078 B. H0: p = 0.078 Ha: p > 0.078 C. H0: p = 0.104 Ha: p ≠ 0.104 D. H0: μ = 0.078 Ha: μ > 0.078

H0: p = 0.078 Ha: p > 0.078 Correct. The researchers want know if the proportion of low weight births for women exposed to secondhand smoke during their pregnancies is greater than 0.078.

Does going to a private university increase the chance that a student will graduate with student loan debt? A national poll by the Institute for College Access and Success showed that in 69% of college graduates from public and nonprofit colleges in 2013 had student loan debt. A researcher wanted to see if there was a significant increase in the proportion of student loan debt for public and nonprofit colleges in 2014. Suppose that the researcher surveyed 1500 graduates of public and nonprofit universities and found that 71% of graduates had student loan debt in 2014. Let p be the proportion of all graduates of public nonprofit universities that graduated with student loan debt. What are the appropriate null and alternative hypotheses for this research question? A. H0: p = 0.69 Ha: p ≠ 0.69 B. H0: p = 0.69 Ha: p > 0.69 C. H0: p = 0.71 Ha: p ≠ 0.71 D. H0: μ = 0.69 Ha: μ > 0.69

H0: p = 0.69 Ha: p > 0.69 Correct. The researchers want to compare the proportion of students graduating with student loan debt in 2014 is greater than 0.69.

According to a U.S. news poll, 38% of students in the class of 2013 had done an internship during their time as an undergraduate student. Dana is interested in finding out whether students at her university had an internship rate that was higher than the national average. She obtained a list of 25 randomly selected students from the population of all students at her university by requesting this information from the university's institutional research office. She collected the responses and calculated that the proportion for her university was 43%. Which one of the following statements about the z-test is correct? A. It is not safe to use the z-test for p, since n * (1 − po) is not large enough. B. It is safe to use the z-test for p. C. It is not safe to use the z-test for p, since the sample is not a random sample from the entire population (or cannot be considered as one). D. It is not safe to use the z-test for p, since n * po is not large enough.

It is not safe to use the z-test for p, since n * po is not large enough. Correct. 25 * 0.38 = 9.5. 9.5 is less than 10, so n is not large enough for this sample to behave as a normal sampling distribution of p-hat.

According to a U.S. news poll, 38% of students in the class of 2013 had done an internship during their time as an undergraduate student. Dana is interested in finding out whether students at her university had an internship rate that was higher than the national average. She decided to ask 100 architecture majors whether they had ever done an internship while at their university. She collected the responses and calculated that the proportion for her university was 43%. Which one of the following statements about the z-test is correct? A. It is safe to use the z-test for p. B. It is not safe to use the z-test for p, since n * (1 − po) is not large enough. C. It is not safe to use the z-test for p, since the sample is not a random sample from the entire population (or cannot be considered as one). D. It is not safe to use the z-test for p, since n * po is not large enough.

It is not safe to use the z-test for p, since the sample is not a random sample from the entire population (or cannot be considered as one). Correct. Her question was about whether students at her entire university had a different internship rate than the national average, but she only asked architecture students. This was not a random sample from the entire population, which would have included a variety of majors. Perhaps architecture majors have a higher rate (or a lower rate) of internships than other majors on campus.

Dr. Gray would like to do a survey on whether the proportion of people ages 18 to 22 who have seen a healthcare professional (doctor, nurse, hospital, etc.) in the past year is higher than the national average of 82%. She plans to give her survey to 80 student athletes at your college by distributing surveys after sports practice. Is this survey valid or not valid for testing the hypothesis that the proportion of people ages 18 to 22 who have seen a healthcare professional in the past year is higher than the national average? A. Valid B. Not valid

Not Valid Correct. Student athletes probably see health care professionals more than average because of the need for physical exams related to sports. This is also not a random sample of people ages 18 to 22.

Suppose that a major polling organization wanted to test the hypothesis that there was a change in the president's "approval rating" since last month. Last month, 35% of the representative sample of registered voters approved of the president. For this month, the null hypothesis was that the approval rating equals 35% and the alternative hypothesis is that the approval rating does not equal 35%. The significance level for this test was 0.05. The results of the hypothesis test of the new survey showed a p-value of 0.008. Which of the following statements is correct? Check all that apply. A. The results were statistically significant. B. The results were not statistically significant. C. The null hypothesis should be rejected. D. The null hypothesis should be accepted. E. The null hypothesis cannot be rejected.

The results were statistically significant. The null hypothesis should be rejected. Correct. The results were statistically significant because the p-value of 0.008 was less than the significance level of 0.05. The null hypothesis should be rejected because the p-value of 0.008 was less than the significance level of 0.05.

A national poll by the Institute for College Access and Success showed that in 69% of college graduates from public and nonprofit colleges in 2013 had student loan debt. Dr. Blackman wanted to find out if her public nonprofit university had a lower proportion of students who graduated with student loan debt in 2013. For this survey, the null hypothesis was that the proportion of students with graduated with student loan debt equals 69% and the alternative hypothesis is that the proportion with student loan debt does not equal 69%. The significance level for this test was 0.05. The results of the hypothesis test of the new survey showed a p-value of 0.039. Which of the following statements is correct? Check all that apply. A. The results were statistically significant. B. The results were not statistically significant. C. The null hypothesis should be rejected. D. The null hypothesis should be accepted. E. The null hypothesis cannot be rejected.

The results were statistically significant. The null hypothesis should be rejected. Correct. The results were statistically significant because the p-value of 0.039 was less than the significance level of 0.05. The null hypothesis should be rejected because the p-value of 0.008 was less than the significance level of 0.05.

According to a Pew Research Center, in May 2011, 35% of all American adults had a smartphone (one which the user can use to read email and surf the Internet). A communications professor at a university believes this percentage is higher among community college students. She selects 300 community college students at random and finds that 120 of them have a smartphone. In testing the hypotheses H0, P = 0.35, versus Ha, p > 0.35, she calculates the test statistic as Z = 1.82. Use the normal table to identify the appropriate p-value for this Z score. Click here to access the normal table. Given these results, which of the following is an appropriate conclusion? A. There is enough evidence to show that more than 35% of community college students own a smartphone (p-value = 0.034). B. There is enough evidence to show that more than 35% of community college students own a smartphone (p-value = 0.068). C. There is not enough evidence to show that more than 35% of community college students own a smartphone (p-value = 0.966). D. There is not enough evidence to show that more than 35% of community college students own a smartphone (p-value = 0.034).

There is enough evidence to show that more than 35% of community college students own a smartphone (p-value = 0.034). Correct. The area to the right of Z = 1.82 on the standard normal curve is about 0.034. This is smaller than α = 0.10 so we reject the null hypothesis and accept the alternative.

According to a 2015 survey by mobile loyalty company SessionM, 47% of smartphone users preferred online shopping to in-store shopping. A professor at a local community college believes that the percentage of smartphone users who prefer online shopping may be higher among community college students. She randomly selects 150 community college students with smartphones and she finds that 84 of the students surveyed prefer online shopping. In testing the hypotheses: H0: p = 0.47 versus Ha: p > 0.47, she calculates the test statistic as Z = 2.21. The alpha level for this problem is 0.05. Use the Normal Table to identify the appropriate p-value for this Z score. Click here to access the normal table. Given these results, which of the following is an appropriate conclusion? A. There is enough evidence to show that more than 47% of community college with smartphones prefer online shopping (p-value = 0.0136). B. There is enough evidence to show that more than 47% of community college with smartphones prefer online shopping (p-value = 0.0272). C. There is not enough evidence to show that more than 47% of community college with smartphones prefer online shopping (p-value = 0.9864). D. There is not enough evidence to show that more than 47% of community college with smartphones prefer online shopping (p-value = 0.0136).

There is enough evidence to show that more than 47% of community college with smartphones prefer online shopping (p-value = 0.0136). Correct. The area to the right of Z = 2.21 on the standard normal curve is about 0.0136. This is smaller than α = 0.05 so we reject the null hypothesis and accept the alternative.

The proportion of college football players who have had at least one concussion is estimated to be 34% in the United States. We wanted to know if football players at our university were less likely to have suffered a concussion, so we surveyed a random sample of 100 past and present football players at our university. Is this survey valid or not valid for testing the hypothesis that the proportion of college football players at our university with at least one concussion is less than the national average? A. Valid B. Not valid

Valid Correct. The population of interest (football players at our university) was randomly sampled. There were also enough people in the sample to have a normal sampling distribution of p-hat.

A manufacturer of t-shirts marks a shirt as "irregular" when it has defects such as crooked seams, stains, rips, or holes. A small number of irregular t-shirts are expected as part of the manufacturing process, but if more than 8% of the t-shirts manufactured at a plant are classified as irregular, the manager has to do an investigation to try to find the source of the increased mistakes in the manufacturing process. In order to test whether his plant is making a higher than expected number of irregular t-shirts, the manager of a plant randomly selects 100 t-shirts and finds that 12 are irregular. He plans to test the hypotheses: H0, P = 0.08, versus Ha, p > 0.08 (where p is the true proportion of irregular t-shirts). What is the test statistic? A. Z = −1.23 B. Z = 1.47 C. Z = −1.47 D. Z = 5

Z = 1.47 Correct. The sample proportion is about 1.47 standard errors above the p = 0.08 using the formula z = (p̂ − p)/√[(p(1 − p)/n].

A quality control engineer at a potato chip company tests the bag-filling machine by weighing bags of potato chips. Not every bag contains exactly the same weight. But if more than 15% of bags are overfilled, then they stop production to fix the machine. They define overfilled to be more than 1 ounce above the weight on the package. The engineer weighs 100 bags and finds that 21 of them are overfilled. He plans to test the hypotheses: H0: p = 0.15 versus Ha: p > 0.15 (where p is the true proportion of overfilled bags). What is the test statistic? A. Z = −1.47 B. Z = 1.68 C. Z = 4 D. Z = −1.68

Z = 1.68 Correct. The sample proportion is about 1.68 standard errors above the p = 0.15 using the formula z = (p-hat − p)/√[(p(1 − p)/n].

A researcher wants to find out if U.S. adults still support the death penalty at a proportion of 0.64 (as it was in 2003). This graph indicates the sampling distribution for the proportion of supporters in random samples of 25 adults. The standard deviation is approximately 0.10. What is the approximate test statistic for p̂ = 0.54? A. −2 B. −1 C. 0 D. 1 E. 2

−1 Correct. The p̂ of 0.54 is less than the population proportion of 0.64, so the test statistic should be a negative number using the formula z = (p̂ − p)/√[(p(1 − p)/n]. Using the formula, 0.54 is 1 standard deviation below the mean.