Question

1. A recent issue of the AARP Bulletin reported that the average weekly pay for a woman with a high school degree is $520. Suppose you would like to determine if the average weekly pay for all working women is significantly greater than that for women with a high school degree. Data providing the weekly pay for a sample of 50 working women are available in the file named WeeklyPay. These data are consistent with the findings reported in the AARP article.

1. State the hypotheses that should be used to test whether the mean weekly pay for all women is significantly greater than the mean weekly pay for women with a high school degree.
2. Use the data in the file named WeeklyPay to compute the mean, the test statistic, and the p-value.
3. Use   α = 0.05, what is your conclusion?
4. Repeat the hypothesis test using the critical value approach
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Answer 1

Date: 14/05/2019 Answer To test the hypothesis is that the mean weekly pay for all women is significantly greater than $520 at 5% significance level. a) The null and alternative hypothesis is, Ho: μ-3520 Ha: μ> $520 b) The t-test statistics is First, compute mean and standard deviation then find the -test statistics. The sample mean is

582 333 +...772 691 50 -637.94 The sample standard deviation is, (x-x)2 (582 - 637.94)2(333 - 637.94)2 +.(772 637.94)2 (691 637.94)2 50-1 - V22053.77 -148.51 The t-test statistics is,

x-μ (s/Vn) 637.94-520 1 148.51/ V50 5.616 The t-test statistics is 5.616. The p-value is, By using Imathas hypothesis test graph generator, find p-value with the help of following steps 1) Select t distribution and give the Sample size. 2) Select the Right tailed type 3) Give Test Statistics Value. 4) Select Shade P-value region. 5) Click Draw button

P-Value =0.000 .2 .3 0 test t From the Imathas hypothesis test graph generator, thep-value is 0.0001 c) Decision 0.0001, so the null hypothesis is rejected at 5% level of significance. There is sufficient evidence to indicate that that the mean weekly pay for all women is significantly greater than S520. The result is statistically significant.

d) Critical value approach To test the hypothesis is that the mean weekly pay for all women is significantly greater than $520 at 5% significance level. The null and alternative hypothesis is, Ha : μ > $520 The t-test statistics is same as part (b) which is 5.616. The t critical value is, The sample size is small and right-tailed test. Look in the column headed α-0.05 and the row headed in the t distribution table by using degree of freedom is, d.f.= n-1 50-1 49 The t critical value is 1.677

Decision The conclusion is that the t value corresponds to sample statistics is falls in the critical region, so the null hypothesis is rejected at 5% level of significance. There is sufficient evidence to indicate that the mean weekly pay for all women is significantly greater than S520. The result is statistically significant.