A hypothesis test involving two means is most helpful when we are trying to detect a difference between two distributions that have a lot of overlap. For instance, an important societal issue is the pay difference between men and women. To investigate this difference, suppose we select a random sample of ten men and ten women. Suppose we observe: X1 = $49250 is the sample mean and s, = $8400 is the sample SD for men. X2 = $47640 is the sample mean and s2 = $8650 is the sample SD for women. Define: H1 = mean pay received by a man H, = mean pay received by a woman The difference in the sample means is X1- X, = $1610, so it is certainly true that the sample of men received higher pay as a group than the sample of women. The question for the statistician is, is this difference enough to tell us that men receive more pay than women on average for the entire population? In other words, does it tell us that uj > µ,? Find this QUESTION 12 (previous question continued...) To test the claim that men on average receive more pay than women on average, we calculated the test statistic t. For a right-tail test (like this one), the P-value is the probability of observing at statistic equal to or larger than the observed t value. Calculate the P-value, assuming that t follows a Student t-distribution with df = min(n, - 1,n, - 1) = min(9,9)=9. Round to four decimal places.

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A hypothesis test involving two means is most helpful when we are trying to detect a difference between two distributions that have a lot of overlap.
For instance, an important societal issue is the pay difference between men and women. To investigate this difference, suppose we select a random sample of ten
men and ten women.
Suppose we observe:
X1 = $49250 is the sample mean and s, = $8400 is the sample SD for men.
X2
= $47640 is the sample mean and s2 = $8650 is the sample SD for women.
Define:
H1 = mean pay received by a man
H, = mean pay received by a woman
The difference in the sample means is X1- X, = $1610, so it is certainly true that the sample of men received higher pay as a group than the sample of
women. The question for the statistician is, is this difference enough to tell us that men receive more pay than women on average for the entire population? In
other words, does it tell us that uj > µ,?
Find this
QUESTION 12
(previous question continued...) To test the claim that men on average receive more pay than women on average, we calculated the test statistic t.
For a right-tail test (like this one), the P-value is the probability of observing at statistic equal to or larger than the observed t value. Calculate the P-value, assuming that t follows a
Student t-distribution with df = min(n, - 1,n, - 1) = min(9,9)=9. Round to four decimal places.
Transcribed Image Text:A hypothesis test involving two means is most helpful when we are trying to detect a difference between two distributions that have a lot of overlap. For instance, an important societal issue is the pay difference between men and women. To investigate this difference, suppose we select a random sample of ten men and ten women. Suppose we observe: X1 = $49250 is the sample mean and s, = $8400 is the sample SD for men. X2 = $47640 is the sample mean and s2 = $8650 is the sample SD for women. Define: H1 = mean pay received by a man H, = mean pay received by a woman The difference in the sample means is X1- X, = $1610, so it is certainly true that the sample of men received higher pay as a group than the sample of women. The question for the statistician is, is this difference enough to tell us that men receive more pay than women on average for the entire population? In other words, does it tell us that uj > µ,? Find this QUESTION 12 (previous question continued...) To test the claim that men on average receive more pay than women on average, we calculated the test statistic t. For a right-tail test (like this one), the P-value is the probability of observing at statistic equal to or larger than the observed t value. Calculate the P-value, assuming that t follows a Student t-distribution with df = min(n, - 1,n, - 1) = min(9,9)=9. Round to four decimal places.
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