Question

According to a national representative survey done by Consumer Reports, you should always try to negotiate for a better deal when shopping or paying for services.† Tips include researching prices at other stores and on the Internet, timing your visit late in the month when salespeople are trying to meet quotas, and talking to a manager rather than a salesperson. Suppose that random samples of 200 men and 200 women are taken, and that the men were more likely than the women to say they "always or often" bargained (25% compared with 20%). (Use p₁ and p₂ for the proportions of men and women, respectively, who say they "always or often" negotiate for a better deal.)

(a) Construct a 95% confidence interval for the difference in the proportion of men and women who say they "always or often" negotiate for a better deal. (Round your answers to three decimal places.)
______ to _____

Answer 1

Solution:

Confidence interval for difference between two population proportions:

Confidence interval = (P1 – P2) ± Z*sqrt[(P1*(1 – P1)/N1) + (P2*(1 – P2)/N2)]

P1 is sample proportion of first group (Male)

P2 is sample proportion of second group (Female)

N1 = sample size for male group

N2 = sample size for female group

Z is critical value at 95% confidence level

We are given P1 = 0.25, P2 = 0.20, N1 = 200, N2 = 200, c=0.95, so Z = 1.96 (by using z-table)

Confidence interval = (0.25 – 0.20) ± 1.96*sqrt[(0.25*(1 – 0.25)/200) + (0.20*(1 – 0.20)/200)]

Confidence interval = (0.25 – 0.20) ± 1.96* 0.0417

Confidence interval = (0.25 – 0.20) ± 0.0817

Confidence interval = 0.05 ± 0.0817

Lower limit = 0.05 - 0.0817 = -0.0317

Upper limit = 0.05 + 0.0817 = 0.1317

Lower limit = -0.032

Upper limit = 0.132

Confidence interval for difference in the proportion of men and women who say they “always or often” is given as

-0.032 to 0.132