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1 Sample Practice problems - chapter 12-1 and 2 proportions for inference - Z Distributions Name MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. Provide an appropriate response. 1) A state university wants to increase its retention rate of 4% for graduating students from the previous year. After implementing several new programs during the last two years, the university reevaluated its retention rate using a random sample of 352 students and found the retention rate at 5%. Test an appropriate hypothesis and state your conclusion. Be sure the appropriate assumptions and conditions are satisfied before you proceed. A) H0: p = 0.04; HA: p < 0.04; z = -1.07; P-value = This data shows that more than 4% of students are retained; the university should continue with the new programs. B) H0: p = 0.04; HA: p > 0.04; z = 0.96; P-value = This data does not show that more than 4% of students are retained; the university should not continue with the new programs. C) H0: p = 0.04; HA: p > 0.04; z = -1.07; P-value = This data does not show that more than 4% of students are retained; the university should not continue with the new programs. D) H0: p = 0.04; HA: p < 0.04; z = 1.07; P-value = This data shows that more than 4% of students are retained; therefore, the university should continue with the new programs. E) H0: p = 0.04; HA: p 0.04; z = 1.07; P-value = This data does not show that more than 4% of students are retained; the university should not continue with the new programs. 1) 2) A statistics professor wants to see if more than 80% of her students enjoyed taking her class. At the end of the term, she takes a random sample of students from her large class and asks, in an anonymous survey, if the students enjoyed taking her class. Which set of hypotheses should she test? A) H0: p < 0.80 HA: p > 0.80 B) H0: p > 0.80 HA: p = 0.80 C) H0: p = 0.80 HA: p < 0.80 D) H0: p = 0.80 HA: p > 0.80 E) H0: p < 0.80 HA: p ) 1

2 Write the null and alternative hypotheses you would use to test the following situation. 3) The federal guideline for smog is 12% pollutants per 10,000 volume of air. A metropolitan city is trying to bring its smog level into federal guidelines. The city comes up with a new policy where city employees are to use city transportation to and from work. A local environmental group does not think the city is doing enough and no real decrease will occur. An independent agency, hired by the city, runs a test on the air. What are the null and alternative hypotheses? A) H0: p = 0.12 HA: p < 0.12 B) H0: p 0.12 HA: p = 0.12 C) H0: p = 0.12 HA: p > 0.12 D) H0: p < 0.12 HA: p > 0.12 E) H0: p = 0.12 HA: p ) Provide an appropriate response. 4) A state university wants to increase its retention rate of 4% for graduating students from the previous year. After implementing several new programs during the last two years, the university reevaluated its retention rate using a random sample of 352 students and retained 18 students. Should the university continue its new programs? Test an appropriate hypothesis using α = 0.10 and state your conclusion. Be sure the appropriate assumptions and conditions are satisfied before you proceed. A) z = 1.07; P-value = The change is statistically significant. A 95% confidence interval is (3.1%, 67.2%). This is clearly lower than 4%. The chance of observing 18 or more retained students of 352 is only 28.46% if the dropout rate is really 4%. B) z = -1.07; P-value = The university should continue with the new programs. There is an 85.77% chance of having 18 or more of 352 students in a random sample be retained if in fact 4% are retained. C) z = 1.07; P-value = The university should not continue with the new programs. There is a 14.23% chance of having 18 or more of 352 students in a random sample be retained if in fact 4% are retained. The P-value of is greater than the alpha level of D) z = -1.07; P-value = The change is statistically significant. A 98% confidence interval is (2.7%, 7.5%). This is clearly lower than 4%. The chance of observing 18 or more retained students of 352 is only 14.23% if the dropout rate is really 4%. E) z = 1.07; P-value = The change is statistically significant. A 90% confidence interval is (3.4%, 6.8%). This is clearly higher than 4%. The chance of observing 18 or more retained students of 352 is only 85.77% if the dropout rate is really 4%. 4) SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question. Decide whether or not the conditions and assumptions for inference with the two-proportion z-test are satisfied. Explain your answer. 5) A poll of randomly selected Americans between the ages of 20 and 29 reports that 35 of 5) 410 men and 59 of 398 women suffered from insomnia at least once a week during the past year. 2

3 Use a two proportion z-test to perform the required hypothesis test. State the conclusion. 6) A marketing survey involves product recognition in New York and California. Of 558 New Yorkers surveyed, 193 knew the product while 196 out of 614 Californians knew the product. At the 0.05 significance level, do the data provide sufficient evidence to conclude that the recognition rate in New York differs from the recognition rate in California? 6) MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. A two-sample z-test for two population proportions is to be performed using the P-value approach. The null hypothesis is H 0 : p 1 = p 2 and the alternative is H a : p 1 p 2. Use the given sample data to find the P-value for the hypothesis test. Give an interpretation of the p-value. 7) A poll reported that 3 out of 50 college seniors surveyed did not have jobs, while 7 out of 50 college juniors did not have jobs during the academic year. A) P-value = ; If there is no difference in the proportions, there is about a 6.13% chance of seeing the exact observed difference by natural sampling variation. B) P-value = ; If there is no difference in the proportions, there is about a 0.72% chance of seeing the observed difference or larger by natural sampling variation. C) P-value = ; There is about a 18.24% chance that the two proportions are equal. D) P-value = ; If there is no difference in the proportions, there is about a 18.24% chance of seeing the observed difference or larger by natural sampling variation. E) P-value = ; There is about a 6.13% chance that the two proportions are equal. 7) SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question. Provide an appropriate response. 8) Suppose the proportion of sophomores at a particular college who purchased used textbooks in the past year is ps and the proportion of freshmen at the college who purchased used textbooks in the past year is pf. A study found a 98% confidence interval for ps - pf is (0.236, 0.422). Does this interval suggest that sophomores are more likely than freshmen to buy used textbooks? Explain. 8) 9) A random sample of 150 men found that 88 of the men exercise regularly, while a random sample of 200 women found that 130 of the women exercise regularly. Based on the results, construct and interpret a 95% confidence interval for the difference in the proportions of women and men who exercise regularly. A friend says that she believes that a higher proportion of women than men exercise regularly. Does your confidence interval support this conclusion? Explain. 9) 3

4 10) A city council must decide whether to fund a new "welfare-to-work" program to assist long-time unemployed people in finding jobs. This program would help clients fill out job applications and give them advice about dealing with job interviews. A six-month trial has just ended. At the start of this trial a number of unemployed residents were randomly divided into two groups; one group went through the help program and the other group did not. Data about employment at the end of this trial are shown in the table. Should the city council fund this program? Test an appropriate hypothesis and state your conclusion. 10) Current job status Employed Unemployed Group 1 (Help program) Group 2 (No help)

5 Answer Key Testname: UNTITLED1 1) B 2) D 3) A 4) C 5) The assumptions and conditions necessary for inference are satisfied. The samples are both random. Each sample contains less than 10% of the population. The samples are independent of each other. There are at least 10 successes and at least 10 failures in each sample. 6) H 0 : p1 - p2 = 0 H A : p1 - p2 0 Test statistic: z = 0.97 P-value = Fail to reject the null hypothesis. There is not sufficient evidence to conclude that the recognition rate in New York differs from the the recognition rate in California. 7) D 8) Yes. Since 0 is not in the interval, there is evidence that sophomores are more likely than freshmen to buy used textbooks. 9) Conditions: * Randomization Condition: We are told that we have random samples. * 10% Condition: We have less than 10% of all men and less than 10% of all women. * Independent samples condition: The two groups are clearly independent of each other. * Success/Failure Condition: Of the men, 88 exercise regularly and 62 do not; of the women, 130 exercise regularly and 70 do not. The observed number of both successes and failures in both groups is at least 10. With the conditions satisfied, the sampling distribution of the difference in proportions is approximately Normal with a mean of pm - pw, the true difference between the population proportions. We can find a two-proportion z-interval. We know: nm = 150, p^m = = 0.587, n w = 200, p^w = = We estimate SD ( p^m - p^w) as SE( p^m - p^w) = p^mq^m nm + p^wq^w nw = (0.587)(0.413) + (0.65)(0.35) = ME = z * SE( p^m - p^w) = 1.96(0.0525) = The observed difference in sample proportions = p^m - p^w = = , so the 95% confidence interval is ± , or -16.6% to 4.0%. We are 95% confident that the proportion of women who exercise regularly is between 4.0% lower and 16.6% higher than the proportion of men who exercise regularly. Since zero is contained in my confidence interval, I cannot say that a higher proportion of women than men exercise regularly. My confidence interval does not support my friend's claim. 5

6 Answer Key Testname: UNTITLED1 10) H0: p1 - p2 = 0 HA: p1 - p2 > 0 People were randomly assigned to groups, we assume the groups are independent, and 20, 34, 13, 33 are all > 10. OK to do a 2-proportion z-test. p^1 = 0.370, p^2 = 0.283, p^ = = 0.33, z = ( ) - 0 (0.33)(0.67) 54 + (0.33)(0.67) 46 = 0.93 P = P(p^1 -p^2 > 0.087) = P(z > 0.93) = We fail to reject the null because P is very large. We do not have evidence that the help program is beneficial, so it should not be funded. 6

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