Question

3) A supplier finds that after surveying 1200 parts managers only 90 of them need his product. Construct a 99% confidence interval estimate for the true proportion for all parts managers who use his product.

4) A manufacturer wants to do a study to determine the proportion of people with cell phones who take selfies. How many people with cell phones must be sampled if the manufacturer wants to be 92% confident that the sample proportion is in error by no more than 0.035?

Answer 1

3)

Solution :

Given that,

Point estimate = sample proportion = = x / n = 90 / 1200 = 0.075

Z_/2 = 2.576

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

= 2.576 * (((0.075 * 0.925) / 1200)

= 0.020

A 99% confidence interval for population proportion p is ,

- E < p < + E

0.075 - 0.020 < p < 0.075 + 0.020

0.055 < p < 0.095

(0.055 , 0.095)

4)

= 0.5

1 - = 0.5

margin of error = E = 0.035

Z_/2 = 1.75

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

= (1.75 / 0.035)² * 0.5 * 0.5

= 625

sample size = n = 625