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Suppose that a simple random sample of 29 college-age men and women was randomly divided into two groups. The first group. n1= 15 was given

Suppose that a simple random sample of 29 college-age men and women was randomly divided into two groups. The first group. n1= 15 was given liter of red wine before going to sleep. The second group of n2= 14 bull. Was given no alcohol before going to sleep. Everyone in both groups went to sleep at 11:00 PM. The average brain wave activity. From 4:00 to 6:00 AM was determined for each individual in the groups. Their results follow. Group 1: 16.0 ,18.8 ,19.6 ,19.1 ,19.9 ,17.4 20.9, 21.1, 20.3 22.1, 20.1, 16.4 ,20.6, 20.1, 22.3

Group 2: 8.2 ,5.4, 6.8 ,6.5, 4.7, 5.9, 2.9, 7.6, 10.2, 6.4, 8.8,5.4,8.3,5.1

The mean for group one is a little over 13 points higher than for group 2, but that might not be the case for the population at large. Assuming the two groups are independent and that both groups are normally distributed, use a hypothesis test with significance level a = 0.10 to test the claim that the average brain wave for people of college age who are not given liter of red wine before going to sleep is at least 13 points higher those who are not. [Hint: In your hypothesis and claim, instead of comparing 1and 2 directly, you should compare(1-13) with 2 .You can do compare ( x1-13) with x2 , by putting(x1 -13) values into the entry in your calculator.]

[ZTest] OR [TTest] OR [2-SampZTest] OR [2-SampTTest] ? (Circle one)

a: or Sx1: 1: n1:

orSx1: 2:n2:

(Circle the original claim)-H0:Ha:

One- sided leftOne- sided rightTwo-sided(Circle One)?

Test Statistic:z or t

Critical Value(s) (based on a or ) :

Graph( Sketch):

P-value:

Conclusions? (Circle One): Reject H0Fail to Reject H0

Interpretation?

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