Question

In: Statistics and Probability

Following complaints about the working conditions in some apparel factories both in the United States and...

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?

Solutions

Expert Solution

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.


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