Question 331543: C. A researcher wishes to determine whether people with high blood pressure can reduce their blood pressure by following a particular diet. Use the sample data below to test the claim that the treatment population mean 1 is smaller than the control population mean 2. Test the claim using a significance level of 0.01. You may assume the samples are independent and were randomly selected.
Treatment Group Control Group
n1 = 85 n2 = 75
x-bar1 = 189.1 x-bar2 = 203.7
s1 = 38.7 s2 = 89.2
You will be asked to identify each of the following.
11. The justification for normality in this situation.
12. The null and alternative hypotheses appropriate for this test.
13. The calculated value of the test statistic. (To match my answer you will need to take Treatment Group – Control Group.)
14. The prob-value calculated for this test.
15. The decision reached by this test.
Answer by jrfrunner(365)   (Show Source):
You can put this solution on YOUR website! Since analysis is on averages and the samples are considerably larger than 30, by the Central Limit theorem this makes the assumption of normally distributed averages reasonable
Ho: Mu1=Mu2
Ha: Mu1 not equal Mu2
testing at alpha=0.01
The test statistic is a t distribution since population variances are unknown, but there at two t distributions one where the pop variances are equal and one where the pop variance are unequal
You should test for equality of population variances to determine the proper t test.
Ho: sigma1 = sigma2
Ha: sigma1 not equal sigma2
Test statistic 
Critical value F(alpha, dof num, dof dem)=F(.01,74,84)=1.691
Since the test statistic exceeds the critical value, reject Ho and conclude that the population variances are not equal (ie cannot pool sample variances)
therefore, the test statistic for the means is
the degrees of freedom for this statistic is kind of messy
 or 99 dof
pvalue = P(t<-1.313)+P(t>1.313) = 0.192 (this a two tail test)
since pvalue > alpha ie 0.192 > 0.01 this means the a t value as extreme as -1.313 could have come from distributions whose means are equal by pure chance ( 19.2%)
In other words, this is possible under the null, so cannot reject Ho and conclude no evidence agains the means being equal
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