The poll introduced in Exercise 5.13 also asked whether re- spondents expected t

Question

The poll introduced in Exercise 5.13 also asked whether re- spondents expected to see a female president in their lifetime. 78% of the 1,835 respondents said “yes”. (a) Construct a 90% confidence interval for the proportion of Americans who expect to see a female president in their lifetime, and interpret this interval in context of the data. (b) How would you expect the width of a 98% confidence interval to compare to the interval you calculated in part (a)? Explain your reasoning. (c) Now construct the 98% confidence interval, and comment on whether your answer to part (b) is confirmed. The poll introduced in Exercise 5.13 also asked whether re- spondents expected to see a female president in their lifetime. 78% of the 1,835 respondents said “yes”. (a) Construct a 90% confidence interval for the proportion of Americans who expect to see a female president in their lifetime, and interpret this interval in context of the data. (b) How would you expect the width of a 98% confidence interval to compare to the interval you calculated in part (a)? Explain your reasoning. (c) Now construct the 98% confidence interval, and comment on whether your answer to part (b) is confirmed. The poll introduced in Exercise 5.13 also asked whether re- spondents expected to see a female president in their lifetime. 78% of the 1,835 respondents said “yes”. (a) Construct a 90% confidence interval for the proportion of Americans who expect to see a female president in their lifetime, and interpret this interval in context of the data. (b) How would you expect the width of a 98% confidence interval to compare to the interval you calculated in part (a)? Explain your reasoning. (c) Now construct the 98% confidence interval, and comment on whether your answer to part (b) is confirmed.

Explanation / Answer

P = 0.78

n = 1835

A) At 90% confidence interval the critical value is 1.645

The confidence interval is

P +/- z* *  sqrt(P * (1 - P)/n)

= 0.78 +/- 1.645 * sqrt(0.78 * 0.22/1835)

= 0.78 +/- 0.0159

= 0.7641, 0.7959

B) The width of 98% confidence interval will be more wider than the width of 90% confidence interval.

C) At 98% confidence interval the critical value is 2.33

Confidence interval is

P +/- z* *  sqrt(P * (1 - P)/n)

= 0.78 +/- 2.33 * sqrt(0.78 * 0.22/1835)

= 0.78 +/- 0.0225

= 0.7575, 0.8025

Yes the answer to part(b) is confirmed .

The width of 98% confidence interval is more wider than the width of 90% confidence interval.

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