10 There is a declining interest among teenagers to pursue a career in science a

Question

10

There is a declining interest among teenagers to pursue a career in science and health care (US News and World Report, May 23, 2011). Thirty college-bound students in Portland, Oregon, are asked about the field they would like to pursue in college. The choices offered in the questionnaire are science, business, and other. The gender information also is included in the questionnaire. The data is shown below. Use Table 1.

Construct the 95% confidence interval for the proportion of students who would like to pursue Science. (Round intermediate calculations to 4 decimal places, "z" value to 2 decimal places, and final answers to 3 decimal places.)

Construct the 95% confidence interval for the proportion of students who would like to pursue Business. (Round intermediate calculations to 4 decimal places, "z" value to 2 decimal places, and final answers to 3 decimal places.)

Construct and interpret the 90% confidence interval for the proportion of female students who are college bound. (Round intermediate calculations to 4 decimal places, "z" value to 2 decimal places, and final answers to 3 decimal places.)

There is a declining interest among teenagers to pursue a career in science and health care (US News and World Report, May 23, 2011). Thirty college-bound students in Portland, Oregon, are asked about the field they would like to pursue in college. The choices offered in the questionnaire are science, business, and other. The gender information also is included in the questionnaire. The data is shown below. Use Table 1.

Explanation / Answer

a1) 95% confidence interval for the proportion of students who would like to pursue Science x= 12 n= 30 p=x/n 0.4 CI = p +- z(a/2)*sqrt(p*(1-p)/n) lower =0.4-1.96*SQRT(0.4*(1-0.4)/30) 0.225 upper =0.4+1.96*SQRT(0.4*(1-0.4)/30) 0.575 a2) 95% confidence interval for the proportion of students who would like to pursue Business x= 10 n= 30 p=x/n 0.333333333 CI = p +- z(a/2)*sqrt(p*(1-p)/n) lower =0.33-1.96*SQRT(0.33*(1-0.33)/30) 0.162 upper =0.33+1.96*SQRT(0.33*(1-0.33)/30) 0.498 b) 90% confidence interval for the proportion of female students who are college bound x= 14 n= 30 p=x/n 0.467 z(a/2)= 1.645 CI = p +- z(a/2)*sqrt(p*(1-p)/n) lower =0.467-1.645*SQRT(0.467*(1-0.467)/30) 0.317 upper =0.467+1.645*SQRT(0.467*(1-0.467)/30) 0.617

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