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Exam 19: Comparing Two Proportions

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Question 31

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A university wishes to compare the proportions of female students in the science and social science programs.A random sample of 120 science students contained 43 females.A random sample of 200 social science students contained 132 females.Construct a 99% confidence interval for the difference in the proportions of female students in the science and social science programs at this university.

A) (-0.44,-0.16)
B) (-0.66,0.30)
C) (-0.11,0.28)
D) (-0.57,0.25)
E) (0.30,0.66)

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Question 32

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A university found it retained 25 students out of 352 in 2012 and 36 students out of 334 in 2013.

A) P-value = 0.0455; If there is no difference in the proportions,there is about a 4.55% chance of seeing the observed difference or larger by natural sampling variation.
B) P-value = 0.9545; If there is no difference in the proportions,there is about a 95.45% chance of seeing the observed difference or larger by natural sampling variation.
C) P-value = 0.0455; There is about a 4.55% chance that the two proportions are equal.
D) P-value = 0.091; There is about a 9.1% chance that the two proportions are equal.
E) P-value = 0.091; If there is no difference in the proportions,there is about a 9.1% chance of seeing the observed difference or larger by natural sampling variation.

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Question 33

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A study investigated the smoking status of people in two age groups.Let $\mathrm { P } 1$ represent the proportion of people aged 20-24 who are smokers.Let $\mathrm { P } 2$ represent the proportion of people aged 25-29 who are smokers.A 95% confidence interval for the difference in smoking rates for $\mathrm { P } 1$ - $\mathrm { P } 2$ was found to be (0.045,0.112) .Give an interpretation of this confidence interval.

A university found it retained 25 students out of 352 in 2012 and 26 students out of 334 in 2013.Construct a 90% confidence interval for the difference in the proportions of students retained in 2012 and 2013.

Suppose that you perform a two-proportion z-test and reject the null hypothesis that there is no difference in the two proportions.If that conclusion is actually wrong,which type of error did you commit?

A new manager,hired at a large warehouse,was told to reduce the employee sick leave.The manager introduced a new incentive program for employees with perfect attendance.Suppose the proportion of the employees sick prior to the new program is $p _ { p }$ and the employees sick after the new program is $\mathrm { Pa }$ .A study found a 98% confidence interval for $p _ { p }$ - $\mathrm { Pa }$ is (-0.0668,0.1125) .Give an interpretation of this confidence interval.

You wish to construct a 95% confidence interval to compare two proportions.If we wish to obtain a margin of error of at most 0.04,what sample size should we take from each group? Assume equal sample sizes.

A researcher finds that of 1,000 people who said that they attend a religious service at least once a week,31 stopped to help a person with car trouble.Of 1,200 people interviewed who had not attended a religious service at least once a month,22 stopped to help a person with car trouble.At the 0.05 significance level,do the data provide sufficient evidence to conclude that the two proportions are different?

A marketing survey involves product recognition in Ontario and British Columbia.Of 558 Ontario residents surveyed,193 knew the product while 196 out of 614 British Columbia residents knew the product.At the 0.05 significance level,do the data provide sufficient evidence to conclude that the recognition rate in Ontario differs from the recognition rate in British Columbia?

Suppose the proportion of first year students at a particular university who purchased used textbooks in the past year is $\mathrm { P } 1$ and the proportion of second year students at the university who purchased used textbooks in the past year is p2.A study found a 95% confidence interval for p1 - p2 is $( 0.232,0.421 )$ Give an interpretation of this confidence interval.

Use the given sample data to test the claim that p1 > p2.Use a significance level of 0.01. $\begin{array} { ll } \underline{Sample 1} & \underline{Sample 2}\\\mathrm{n}_{1}=85 & \mathrm{n}_{2}=90\\x _ { 1 } = 38 & x _ { 2 } = 23 \\\end{array}$

A Member of Parliament wants to measure the level of support in his area for campaign finance reform and determine if there is a gender gap among voters with respect to this issue.One aide suggests that they find separate confidence intervals for the percent of men and the percent of women who favour reform and then see if the intervals overlap.Another aide suggests that they find a confidence interval for the difference in the proportions of men and women who favour reform.Which is the correct approach? Why?

In a random sample of 300 Canadian women,68% watch CBC News.In a random sample of 200 Canadian men,56% watch CBC News.Construct a 98% confidence interval for the difference in the proportions of women and men who watch CBC News.