Question

1. A retail store chain with 2 locations is worried that Store #2 is not being patroned by enough male customers. They want to run a hypothesis test that the percent of customers that are male is the same at both locations. They have collected the following data. Refer to Store #1 as Population 1 and Store #2 as Population 2, and answer the following questions leading up to the hypothesis test.

Store #1(population) 1 Customers surveyed = 400 ,Number that were male = 192

Store 2- (population -2)  Customers surveyed = 300 ,  Number that were male = 138

(a) What is the estimate of the proportion in population 1? (b) What is the estimate of the proportion in population 2?

(c) What is the estimate of the difference between the two population proportions?

(d) What is the standard error of the difference between the two population proportions?

(e) Develop a 90% confidence interval for the difference between the two population proportions.

(f) Develop a 95% confidence interval for the difference between the two population proportions.

(g) Develop a 99% confidence interval for the difference between the two population proportions.

(h) What is the pooled estimate of p?

(i) What is the test statistic (z -value)?

(j) What is the p-value? (k) Given α = 0.10, what is your conclusion? (l) Given α = 0.05, what is your conclusion? (m) Given α = 0.01, what is your conclusion?

Answer 1

(a)

Estimate of the proportion in population 1, p1 = 192 / 400 = 0.48

(b)

Estimate of the proportion in population 2, p2 = 138 / 300 = 0.46

(c)

Estimate of the difference between the two population proportions = p1 - p2 = 0.48 - 0.46 = 0.02

(d)

Pooled estimate of p, p = (p1 n1 + p2 n2) / (n1 + n2)

= (0.48 * 400 + 0.46 * 300) / (400 + 300)

= 0.4714

Standard error of the difference between the two population proportions =

= 0.0381

(e)

z value for 90% confidence interval is 1.645

Margin of error = z * standard error = 1.645 * 0.0381 = 0.0627

90% confidence interval for the difference between the two population proportions is,

(0.02 - 0.0627, 0.02 + 0.0627)

= (-0.0427,  0.0827)

(f)

z value for 95% confidence interval is 1.96

Margin of error = z * standard error = 1.96 * 0.0381 = 0.0747

90% confidence interval for the difference between the two population proportions is,

(0.02 - 0.0747, 0.02 + 0.0747)

= (-0.0547,  0.0947)

(g)

z value for 99% confidence interval is 2.576

Margin of error = z * standard error = 2.576 * 0.0381 = 0.0981

90% confidence interval for the difference between the two population proportions is,

(0.02 - 0.0981, 0.02 + 0.0981)

= (-0.0781,  0.1181)

(h)

Pooled estimate of p, p = (p1 n1 + p2 n2) / (n1 + n2)

= (0.48 * 400 + 0.46 * 300) / (400 + 300)

= 0.4714

(i)

z -value = (Observed differences - Hypothesized differences) / Standard error

= (0.02 - 0) / 0.0381

= 0.5249

(j)

For two tail test, p-value = 2 * P(z > 0.5249) = 0.5997

(k)

Since, p-value is greater than 0.10 significance level, we fail to reject null hypothesis H0 and conclude that there is no strong evidence that the percent of customers that are male is different at both locations.

(l)

Since, p-value is greater than 0.05 significance level, we fail to reject null hypothesis H0 and conclude that there is no strong evidence that the percent of customers that are male is different at both locations.

(m)

Since, p-value is greater than 0.01 significance level, we fail to reject null hypothesis H0 and conclude that there is no strong evidence that the percent of customers that are male is different at both locations.