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Chapter 11: Q. 11.116 (page 477)

Ballistic Fingerprinting. Refer to Exercise 11.110 and find and interpret a $98 %$ confidence interval for the difference between the percentages of women and men who favor ballistic fingerprinting.

Short Answer

Expert verified

There is $98 %$ interval for the difference between the proportion is (0.147,0.273)

Step by step solution

01

Given Information

The given values are,

$x_{1} = 446, n_{1} = 537, x_{2} = 307, n_{2} = 495, 伪 = 0.02$

02

Explanation

The formula for ${\tilde{p}}_{1}$ is given by,

${\tilde{p}}_{1} = \frac{x_{1}}{n_{1}}$

The formula for ${\tilde{p}}_{2}$ is given by,

${\tilde{p}}_{2} = \frac{x_{2}}{n_{2}}$

The formula for $z$ is given by,

$z = \frac{{\tilde{p}}_{1} - {\tilde{p}}_{2}}{\sqrt{{\tilde{p}}_{p} (1 - {\tilde{p}}_{p})} \sqrt{\frac{1}{m_{1}} + \frac{1}{m_{2}}}}$

The value of ${\tilde{p}}_{1}$ is calculated as,

${\tilde{p}}_{1} = \frac{x_{1}}{n_{1}} = \frac{446}{537} = 0.83$

The value of ${\tilde{p}}_{2}$ is calculated as,

${\tilde{p}}_{2} = \frac{x_{2}}{n_{2}} = \frac{307}{495} = 0.62$

Calculate the value of $z_{\frac{a}{2}}$ with $98 %$ confidence level is.

$z \frac{a}{2} = Z 0.2 / 2$

$= Z 0.01 = 2.326$

The $98 %$ confidence interval is.

$({\tilde{p}}_{1} - {\tilde{p}}_{2}) 卤 z_{\frac{a}{2}} \sqrt{\frac{{\tilde{p}}_{1} (1 - {\tilde{p}}_{1})}{n_{1}} + \frac{{\tilde{p}}_{1} (1 - {\tilde{p}}_{1})}{n_{2}}} = (0.83 - 0.62) 卤 2.326 \sqrt{\frac{0.83 (1 - 0.83)}{537} + \frac{0.62 (1 - 0.62)}{495}}$

$= 0.21 卤 0.063$

The interval for the difference between the proportion is $(0.147, 0.273)$

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Most popular questions from this chapter

Obtain a sample size that will ensure a margin of error of at most the one specified.

Margin of error $= 0.01$

Confidence level $= 90 %$

Getting a Job. The National Association of Colleges and Employers sponsors the Graduating Student and Alumni Survey. Part f the survey gauges student optimism in landing a job after graduation. According to one year's survey results, published in American Demographics, the $1218$ respondents, $733$ said that they expected difficulty finding a job. Note: In this problem and the next, round your proportion answers to four decimal places.

a. Use these data to find and interpret a $95 %$ confidence interval for the proportion of students who expect difficulty finding a job.

b. Find the margin of error for the estimate of $p$ .

c. Express the confidence interval in the form point estimate $卤$ margin of error.

Obtain a sample size that will ensure a margin of error of at most the one specified.

Margin of error $= 0.02$

Confidence level $= 95 %$

$x_{1} = 18, n_{1} = 30, x_{2} = 10, n_{2} = 20; 95 %$ confidence interval

Women in the Labor Force. The Organization for Economic Cooperation and Development (OFCD) summarizes data on labor-force participation rates in O E C D in Figures. Independent simple random samples were taken of 300 U.S. women and 250 Canadian women. Of the U.S. women, 215 were found to be in the labor force; of the Canadian women. 186 were found to be in the labor force.

a. At the $5 %$ significance level, do the data suggest that there is a difference between the labor-force participation rates of U.S. and Canadian women?

b. Find and interpret a $95 %$ confidence interval for the difference between the labor-force participation rates of U.S. and Canadian women.

See all solutions

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