Question

An advertising executive claims that there is a difference in the mean household income for Visa Gold and MasterCard Gold credit card holders. A random survey of these customers was conducted and the results are reported in the table below. Is there enough evidence to support the executive's claim? Let  be the true mean household income for Visa Gold cardholders and be the true mean household income for MasterCard Gold cardholders. Use a significance level of 0.001 for the test. Assume that the population variances are equal and that the two populations are normally distributed.

	Visa Gold	MasterCard Gold
Mean	$55,290	$49,241
Standard deviation	$6,123	$8,752
Observations	18	15

What can we say about the value of the range of the P-value?

		P-value < 0.01
		0.02 < P-value < 0.05
		0.05 < P-value < 0.10
		P-value > 0.10
		0.20 < P-value < 0.50

Answer 1

V_is Gold Master Card Gold vector x_1 = $ 55290 vector x_2 = $ 49241 Delta_1 = $ 6123 Delta_2 = $ 8752 n_1 = 18 n_2 = 15 Under assumption of equal variances of the two population, The pooled standard deviation is r = squareroot Delta^2_1 (n_1 - 1) + Delta^2_2 (n_2 - 1)/n_1 + n_2 - 2 = squareroot 6123^2 (18 - 1) + 8752^2 (15 - 1)/18 + 15 - 2 = 7426.44 Hypothesis H_0: M_1 = M_2 H_1: M_1 notequalto M_2 test statistic under H_0 + = vector x_1 - vector x_2/Delta squareroot 1/n_1 + 1/n_2 ~ R_n_1 + n_2 - 2 = 55290 - 49241/7426.94 squareroot 1/18 + 1/15 = 2.3299 P - value = 2 P ( > 2.3299) at 18 + 15 - 2 df = 0.0275 Comment: Retain the null hypothesis and conclude that the claim of the difference in the mean household income for visa Gold and master Card credit card holder by advertising executive is rejected at 0.01 level of significance

The value of the range of the P-value is 0.02 < P-value < 0.05.