Question

As the newly hired manager of a company that provides cell-phone service, you want to determine the proportion of adults in your state that live in a home with a cell phone but no land line service.

1.Assuming you have no prior information, how many people must you survey to ensure that your estimate has no more than a 4% margin of error, with 95% confidence?

(Hint: use p=50% when you haven't done a pilot study)

2.Suppose you do a small pilot study of 25 adults in your state, and find that 5 of those people live in a home with a cell phone but do not have land line service. Using that information, how many people must you survey to ensure that your estimate has no more than a 4% margin of error, with 95% confidence?

Answer 1

1)

The following information is provided,
Significance Level, α = 0.05, Margin of Error, E = 0.04

The provided estimate of proportion p is, p = 0.5
The critical value for significance level, α = 0.05 is 1.96.

The following formula is used to compute the minimum sample size required to estimate the population proportion p within the required margin of error:
n >= p*(1-p)*(zc/E)^2
n = 0.5*(1 - 0.5)*(1.96/0.04)^2
n = 600.25

Therefore, the sample size needed to satisfy the condition n >= 600.25 and it must be an integer number, we conclude that the minimum required sample size is n = 601
Ans : Sample size, n = 601 or 600

2)
The following information is provided,
Significance Level, α = 0.05, Margin of Error, E = 0.04

The provided estimate of proportion p is, p = 0.2
The critical value for significance level, α = 0.05 is 1.96.

The following formula is used to compute the minimum sample size required to estimate the population proportion p within the required margin of error:
n >= p*(1-p)*(zc/E)^2
n = 0.2*(1 - 0.2)*(1.96/0.04)^2
n = 384.16

Therefore, the sample size needed to satisfy the condition n >= 384.16 and it must be an integer number, we conclude that the minimum required sample size is n = 385
Ans : Sample size, n = 385 or 384