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Problem 51 Hard Difficulty

A recent issue of the AARP Bulletin reported that the average weekly pay for a woman with a high school diploma was $\$ 520$ (AARP Bulletin, January, 2010 ). 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 diploma. 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 article mentioned above.
a. 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 diploma.
b. Use the data in the file named WeeklyPay to compute the sample mean, the test statistic, and the $p$ -value.
c. Use $\alpha=.05 .$ What is your conclusion?
d. Repeat the hypothesis test using the critical value approach.

Answer

a. $H_{0} : \mu \leq 520, H_{a} : \mu>520$
b. $t=5.62, P<0.005$
c. There is sufficient evidence to support that the claim of a significant
d. There is sufficient evidence to support that the claim of a significant
increase.

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Video Transcript

in question 51 were given that the average weekly pay for a woman with high school education was $520 and were asked to determine if the average is significantly higher for all women compared to women with only the high school diploma. Were also given a sample of 50 working women in the file accompanying the textbook. And for part, they were asked to state or hypotheses to test our claim. And so we'll start with the alternative hypothesis. And that would be that the average weekly earnings that women is of all women in general is greater than $520 and therefore, the no hypothesis is that the mean is less than or equal to 520 for Part B. We're supposed to use the sample to find the sample average, the sample standard deviation, the test statistic and the P value. So I've looked at the file when I've calculated the sample average and that you know to 637 0.94 a sample. Standard deviation was 148.47 and the next step is to calculate the test statistic. Since we're not given the populations standard deviation. Our sample average will follow the T distribution. So tea is there test statistic and recall that it is equal to the sample average minus the population average over the sample standard deviation divided by the square root of the sample size. And that comes out to you 5.62 So before we go to the tea table, we can note that our degrees of freedom is 49. We're gonna go to the tea table to look up a p value. So 49 degrees of freedom T score of 5.62 is off the charts, which means that RPI value is actually smaller than 0.5 So we can scratch that and say p value is lesson 0.5 part C. We're told at the significance level of Alfa equals 0.5 What is their conclusion regarding their test? And we can say that a P value, which is less than 0.5 is less than 0.5 which is Alva. So therefore, the P value is a lesson Alfa and therefore we reject the no hypothesis for Part D. We're told to repeat the hypothesis test using the critical value approach so the critical values were looking for the test statistic. Or that we're looking for the T score that corresponds to an area in the tail of 0.5 And so again, we go to our table for this. So an area that retail 0.5 corresponds to 1.677 so we have critical value of one point 667 So that's 1.677 And we already have our test statistic from the previous part of the question that he was 5.62 And since that is greater than the critical value, we can reject the null hypothesis.