Question

People were polled on how many books they read the previous year. Initial survey results indicate thats= 19.6books. Complete parts ​(a) through ​(d) below.

​(a) How many subjects are needed to estimate the mean number of books read the previous year within four books with 95​% ​confidence?

(b) How many subjects are needed to estimate the mean number of books read the previous year within one book with 90​% ​confidence?

(c) What effect does doubling the required accuracy have on the sample​ size?

(d) How many subjects are needed to estimate the mean number of books read the previous year within two books with 99​% ​confidence? Compare this result to part ​(a). How does increasing the level of confidence in the estimate affect sample​ size? Why is this​ reasonable?

Answer 1

Given, s = 19.6

a) With 95% confidence interval the critical value is z0.025 = 1.96

Margin of error = 4

=> z0.025 * s/sqrt(n) = 4

=> 1.96 * 19.6/sqrt(n) = 4

=> sqrt(n) = 1.96 * 19.6 / 4

=> sqrt(n) = 9.604

=> n = 92.24 = 92

b) Margin of error = 1

With 90% confidence interval the critical value is z0.05 = 1.645

=> z0.05 * s/sqrt(n) = 1

=> 1.645 * 19.6/sqrt(n) = 1

=> sqrt(n) = 1.645 * 19.6 / 1

=> sqrt(n) = 32.242

=> n = 1039.55 = 1040

c) Doubling the required accuracy nearly quadruples the sample size

d) With 99% confidence interval the critical value is z0.005 = 2.58

Margin of error = 2

=> z0.005 * s/sqrt(n) = 2

=> 2.58 * 19.6/sqrt(n) = 2

=> sqrt(n) = 2.58 * 19.6 / 2

=> sqrt(n) = 25.284

=> n = 639.28 = 639

Comparing to part (a),

Increasing the level of confidence increases the sample size required. For a fixed margin of error, greater confidence can be achieved with a larger sample size