Question

Suppose you want to test the claim the the paired sample data given below come from a population for which the mean difference is μd=0. xy7866678183687180529154928286 Use a 0.01 significance level to find the following: (a) The mean value of the differnces d for the paired sample data d¯= (b) The standard deviation of the differences d for the paired sample data sd= (c) The t test statistic t= (d) The positive critical value t= (e) The negative critical value t= (f) Does the test statistic fall in the critical region? A. No B. Yes (g) Construct a 99% conficence interval for the population mean of all differences x−y. <μd

Answer 1

Computational Table:

X	Y	d=x-y	d^2
78	66	12	144
67	81	-14	196
83	68	15	225
71	80	-9	81
52	91	-39	1521
54	92	-38	1444
82	86	-4	16
Total		-77	3627

a)

b)

c) Test statistic:

d)

Degrees of freedom = n-1 = 7-1 = 6

Critical value:

...............................From t table

Therefore,

Positive Critical value: 3.7074

Negative Critical value: -3.7074

Conclusion:

Test statistic (t) lies in the rejection region,i.e. -3.7074 < -1.35 < 3.7074, That is Fail to Reject Ho at 1% level of significance.   

f) Yes, Test statistic -1.35 is Falls in the critical region

g) Construct 99% Confidence interval:

We are 99% Confidence that population mean difference is lies in that interval.