Question

Q4. To determine the effectiveness of the advertising campaign for a new digital video recorder, management would like to know what proportion of the households is aware of the brand. The advertising agency thinks that this figure is close to .55. The management would like to have a margin of error of ±.025 at the 99% confidence level.

a) What sample size should be used?

b) A sample of the size calculated in a) has been taken. The management found the sample proportion to be .575. Construct a 99% CI for the true proportion.

c) If someone insists that the true proportion is .59. Based your answer to b), would you agree or disagree with this person? Why agree or why not agree?

Answer 1

a)

The following information is provided,
Significance Level, α = 0.01, Margin of Error, E = 0.025

The provided estimate of proportion p is, p = 0.55
The critical value for significance level, α = 0.01 is 2.58.

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 = 0.55*(1 - 0.55)*(2.58/0.025)^2
n = 2635.93

Therefore, the sample size needed to satisfy the condition n >= 2635.93 and it must be an integer number, we conclude that the minimum required sample size is n = 2636
Ans : Sample size, n = 2636

b)

sample proportion, = 0.575
sample size, n = 2636
Standard error, SE = sqrt(pcap * (1 - pcap)/n)
SE = sqrt(0.575 * (1 - 0.575)/2636) = 0.0096

Given CI level is 99%, hence α = 1 - 0.99 = 0.01
α/2 = 0.01/2 = 0.005, Zc = Z(α/2) = 2.58

Margin of Error, ME = zc * SE
ME = 2.58 * 0.0096
ME = 0.0248

CI = (pcap - z*SE, pcap + z*SE)
CI = (0.575 - 2.58 * 0.0096 , 0.575 + 2.58 * 0.0096)
CI = (0.5502 , 0.5998)

c)

Disagree because confidence interval contains 0.59