Question

4.10 A book publishing company has recently produced a batch of several thousand copies of a hardcover book. They have hired a team of engineers to inspect the machines they use because too many of the books have been assembled with missing pages. p = the proportion of defective books in the batch. 96 of 160 randomly selected books from the batch are defective. Use this data to find a 90% confidence interval for p.

4.11 Refer to the previous problem. The book publishing company decides that the error margin in the confidence interval is too big, and they would like a 90% confidence interval for p to have an error margin no larger than .01. What sample size is necessary to achieve this? Use the estimate of p from the original sample (96 defective books out of 160).

4.12 Refer to the previous problem. What if the book publishing company had no prior estimate of p? What sample size would be necessary to achieve an error margin no larger than .01 at 90% confidence?

Answer 1

Solution :

4.10) Given that,

n = 160

x = 96

Point estimate = sample proportion = = x / n = 96 / 160 = 0.60

1 - = 1 - 0.60 = 0.40

At 90% confidence level

= 1 - 90%

=1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z0.05 = 1.645

Margin of error = E = Z / 2 * (( * (1 - )) / n)

= 1.645 (((0.60 * 0.40) / 160)

= 0.064

A 90% confidence interval for population proportion p is ,

± E  

= 0.60  ± 0.064

=( 0.536, 0.664 )

4.11) Margin of error = E = 0.01

sample size = n = (Z / 2 / E )2 * * (1 - )

= (1.645 / 0.01)2 * 0.60 * 0.40

= 6494.46

sample size = n = 6495

4.12) =  1 - =   0.5

sample size = n = (Z / 2 / E )2 * * (1 - )

= (1.645 / 0.01)2 * 0.5 * 0.5

= 6765.06

sample size = n = 6766