For hypothesis tests be sure to show all the information req

For hypothesis tests, be sure to show all the information requested: hypotheses, definitions of population parameters used in the hypotheses, the distribution of the test statistic under the null and justification for its use, the test statistic formula and calculation from SAS, pvalue from SAS, the decision (reject or fail to reject H₀) and the final conclusion (stated in the context of the problem).

To study the effects of a drug on blood clotting time, a random sample of six people were treated with the drug and an independent random sample of seven people were treated with a placebo. Each subject’s blood clotting time (in minutes) was recorded. The sample results are given below:

Drug Group     Placebo group

8.8                   9.9

8.4                   9.0

7.9                   11.1

8.7                   9.6

9.1                   8.7

9.61                 10.4

                        9.5

Test whether the true average clotting time is lower for drug treated people. Assume that the clotting times are close to normal for both groups, with equal population variances.  

Solution

Set Up Hypothesis
Null, Ho: u1 > u2
Alternative, true average clotting time is lower for drug treated people H1: u1 < u2
Test Statistic
X (Mean)=8.7517; Standard Deviation (s.d1)=0.5852
Number(n1)=6
Y(Mean)=9.7429; Standard Deviation(s.d2)=0.8182
Number(n2)=7
Value Pooled variance S^2= (n1-1*s1^2 + n2-1*s2^2 )/(n1+n2-2)
S^2 = (5*0.3425 + 6*0.6695) / (13- 2 )
S^2 = 0.5208
we use Test Statistic (t) = (X-Y)/Sqrt(S^2(1/n1+1/n2))
to=8.7517-9.7429/Sqrt((0.5208( 1 /6+ 1/7 ))
to=-0.9912/0.4015
to=-2.4688
| to | =2.4688
Critical Value
The Value of |t | with (n1+n2-2) i.e 11 d.f is 1.796
We got |to| = 2.4688 & | t | = 1.796
Make Decision
Hence Value of | to | > | t | and Here we Reject Ho
P-Value: Left Tail - Ha : ( P < -2.4688 ) = 0.0156
Hence Value of P0.05 > 0.0156,Here we Reject Ho

We have evidence that true average clotting time is lower for drug treated people