In: Statistics and Probability
Following complaints about the working conditions in some apparel factories both in the United States and abroad, a join government and industry commission recommended in 1998 that companies that monitor and enforce proper standards be allowed to display a “No Sweat” label on their products. Does the presence of these labels influence consumer behavior? A survey of U.S. residents aged 18 or older asked a series of questions about how likel they would be to purchase a gament under various conditions. For some conditions, it was stated that the garment had a “No Sweat” label; for others, there was no mention of such a label. On the basis of the responses, each person was classified as a “label use” or a “label nonuser” there were 296 women surveyed. Of these, 63 were label users. On the other hand, 27 of 251 men were classified as users.
(a) Give a 95% confidence interval for the difference in the proportions.
(b) You would like to compare the women with the men. Set up appropriate hypotheses, and find the test staitistic and the P-value. What do you conclude?
a) = 63/296 = 0.2128
= 27/251 = 0.1076
The pooled sample proportion(P) = * n1 + * n2)/(n1 + n2)
= (0.2128 * 296 + 0.1076 * 251)/(296 + 251)
= 0.1645
SE = sqrt(P(1 - P)(1/n1 + 1/n2))
= sqrt(0.1645 * (1 - 0.1645) * (1/296 + 1/251))
= 0.0318
At 95% confidence interval the critical value is z* = 1.96
The 95% confidence interval for difference in the population proportion is
() +/- z* * SE
= (0.2128 - 0.1076) +/- 1.96 * 0.0318
= 0.1052 +/- 0.0623
= 0.0429, 0.1675
b) H0: P1 = P2
H1: P1 P2
The test statistic z = ()/SE
= (0.2128 - 0.1076)/0.0318
= 3.31
P-value = 2 * P(Z > 3.31)
= 2 * (1 - P(Z < 3.31))
= 2 * (1 - 0.9995)
= 0.001
Since the P-value is less than the significance level(0.001 < 0.05), so we should reject the null hypothesis.
At 5% significance level, we can conclude that there is a significant difference in the proportion of women and men label users.