Question

9. An online site presented this​ question, "Would the recent norovirus outbreak deter you from taking a​ cruise?" Among the 34,592 people who​ responded, 65​% answered​ "yes." Use the sample data to construct a 95​% confidence interval estimate for the proportion of the population of all people who would respond​ "yes" to that question. Does the confidence interval provide a good estimate of the population​ proportion?

____<p<____

​(Round to three decimal places as​ needed.)

11. During a period of 11 years 697 of the people selected for grand jury duty were​ sampled, and 65​% of them were immigrants. Use the sample data to construct a​ 99% confidence interval estimate of the proportion of grand jury members who were immigrants. Given that among the people eligible for jury​ duty, 66.2​% of them were​ immigrants, does it appear that the jury selection process was somehow biased against​ immigrants?

___<p<____

​(Round to three decimal places as​ needed.)

Answer 1

9:

Here we have n 34592,p -0.65 Level of significance: Critical value: α-005 1.96 The standard deviation is: P)sqrt(0.65*(1-0.65)/34592) p(1-p) -0.0026 The requried confidence interval is: =0.65 ± (1.96*0.0026) -0.65 + 0.0051 (0.645,0.655) pt-P, Excel function for critical value: NORMSINV(1-(0.05/2)) Hence, the required confidence interval is (0.645,0.655)

The sample is not simple random sample so it may not represent the population correctly.

11:

Here we have n -697, p-0.65 Level of significance : Critical value: α-001 2.58 The standard deviation is: p(1-p) -sqrt(0.65*(1-0.65) 697) - 0.0181 The requried confidence interval is: ptz, 0.65+ (2.58 o.0181) 0.65+0.0467 (0.603,0.697) Excel function for critical value: NORMSINV(1-(0.01/2)) Hence, the required confidence interval is (0.603,0.697)

Since confidence interval contains 66.2% So it does not appear that the jury selection process was somehow biased against​ immigrants.