Question

Please help calculate the p value and confidence interval. Thanks!

| Treatment Placebo A study was done using a treatment group and a placebo group. The results are shown in the table. Assume that the two samples are independent simple random samples selected from normally distributed populations, and do not assume that the population standard deviations are equal. Complete parts (a) and (b) below. Use a 0.10 significance level for both parts. XIF x s 27 2.31 0.92 32 2.61 0.51 a. Test the claim that the two samples are from populations with the same mean. What are the null and alternative hypotheses? CA. Ho 11 =12 Hp : Hq #2 | 0 C. Ho: H:1,212 OB. H. Hy = 42 H:14%H2 OD. Ho:44* H2 H:, <H2 The test statistic, t, is - 1.51. (Round to two decimal places as needed.) The P-value is (Round to three decimal places as needed.)

The confidence interval is:

Answer 1

Test Statistic :-
$t = (\bar{X_{1}} - \bar{X_{2}}) / \sqrt{ ( S_{1}^{2} / n1) &plus; (S_{2}^{2} / n2)}$
$t = ( 2.31 - 2.61 ) / \sqrt{ ( 0.92^{2} / 27) &plus; ( 0.51^{2} / 32)}$
t = -1.5099 ≈ -1.51

Decision based on P value
P - value = P ( t > 1.5099 ) = 0.1391

Looking for the value t = 1.51 in t table across 39 degree of freedom to find the P value

$DF = ( ( S_{1}^{2} / n1 &plus; S_{2}^{2} / n2))^{2} / ( (S_{1}^{2} / n1)^{2}/n1-1) &plus; (S_{2}^{2} / n2)^{2}/n2-1) )$
$DF = ( ( 0.92^{2} / 27 &plus; 0.51^{2} / 32 ))^{2} / ( ( 0.92^{2} / 27 )^{2}/ 27 -1) &plus; ( 0.51^{2} / 32 )^{2}/ 32 -1) )$
DF = 39

Reject null hypothesis if P value < α = 0.1 level of significance
P - value = 0.1391 > 0.1 ,hence we fail to reject null hypothesis
Conclusion :- Fail to Reject Null Hypothesis

Confidence interval :-
$\( \bar{X}_{1} - \bar{X}_{2}) \pm t_{\alpha /2 , DF}\sqrt{ (S_{1}^{2}/n1) &plus; (S_{2}^{2}/n2)}$
t(α/2, DF) = t(0.1 /2, 39 ) = 1.685
$( 2.31 - 2.61 ) \pm t_{ 0.1/2 , 39 }\sqrt{ ( 0.92^{2}/ 27 ) &plus; ( 0.51^{2}/ 32 )}$
Lower Limit = $( 2.31 - 2.61 ) - t_{ 0.1/2 , 39 }\sqrt{ ( 0.92^{2}/ 27 ) &plus; ( 0.51^{2}/ 32 )}$
Lower Limit = -0.6348
Upper Limit = $( 2.31 - 2.61 ) &plus; t_{ 0.1/2 , 39 }\sqrt{ ( 0.92^{2}/ 27 ) &plus; ( 0.51^{2}/ 32 )}$
Upper Limit = 0.0348
90% Confidence interval is ( -0.6348 , 0.0348 )

The confidence interval is:  ( -0.6348 , 0.0348 )