Question

Pair Population 1 60 Population 2 63 1 2. 24 25 3 53 55 The data for a random sample of 8 paired observations are shown in the table to the right. a. What are the appropriate null and alternative hypotheses to test whether the mean for population 2 is larger than that for population 1? b. Conduct the test identified in part a using a = 0.10. c. Find a 90% confidence interval for Hd Interpret this result. d. What assumptions are necessary to ensure the validity of this analysis? 4 34 34 19 5 21 6 43 19 45 25 7 8 31 35

b.

A. t > __

B. |t| > ___

C. t < __

d:

A) The variances of the two populations are approximately the same.

B) The population of differences is approximately normal

C) The populations being sampled are approximately normal

D) No assumptions are required

Answer 1

Solution: Paired t-test d² 1 NA Pair Population 1 population 2 | d={population 1 – population) 60 63 2 24 25 3 53 55 -2 4 34 34 5 19. 21 -2 43 -2 7 19 25 -6 8 31 35 -4 Ed=-20 6 45 4 0 4 4 36 16 Ed²=74 a = Ed h -20 -2.5 -2.5 Ho! Ma=0 LH: Md co mean of Population 2 is larger than that of population 1 This is one tailed test C left tailed) Under Ho, test statistic is given by with degree of freedom = N-1 sun t= d where , s= mi [Ed - Ledley

S = 1-1 [74 6-2027 8-12 8 Is = 1 1/2 [74 400 ] = [74-50] IS Ix24 7 13.4286 is 1.852 Now, -2.5 0.6548 t = da -2.5 s/n 1.852718 It= -3.818 Significance level D = 0.10 deguce of freedom = nel df=8-1 [df=7 a using Now critical value of t at 0.10 level of Significance for 7 degree of freedom 't-tables One tailed teritical 1.415 [from t- tables] Our test is left -tailed So Ecritical becomes teritical - 1.415

Our test statistic t= -3.818 is less than Critical value =-1-415 So, test statistic is in the Rejection Region So, we reject the null hypothesis Ho at Dolo level of significance. Conclusion: It is concluded that null Hypothesis Ho is rejected. Therefore, there is enough evidence to claim that Mean of population 2 is larger than mean of population 1 at 0.10 significance level.

Ma 90% confidence interval for confidence level Significance level K = 0.10 90% = 0.90 1-0.90 formula to calculate confidence Interval : Mi ū i texitial s vñ where tcritical is the critical value of t at Outo level of significance for (B-1)= 7 degree of freedom tanto = 1.895 [ using t-tables ztailed] Critical value can be determined by I tailed tebles aleo, Just divide toos = 1.895 [uring t-tables 1 tailed] Now, we have J = -2.5 tcritical 1.895 Oil0 =0.05 2 n=8 S= 1.852 put these in the formula,

Md = (-2,5 I 1.895 x 1.852 V8 - My 7-2.5 † 1.895% 0.6548 Ma (-2.5 – 1.2408) Me (-2.5 - 1.2408 -2.5+ 1.24082 (-3.74 2-1.26) = (-3, 74, -1:26) I Md It means, we are go% confident that the Mean difference (Ud) is lies within -3.74 and -1-26 Both lower limit and upper limit are negative It shows that (population I mean is larger than (population 1) mean. Assump How of paised t-test. The population of differences is approximately normal.