Answer:
There is a difference between the population means at 0.10 significance level.
(a) The decision rule is to reject H0 if the test statistic (t) falls outside the region bounded by the critical values.
(b) Pooled estimate of the population variance is 17.525.
(c) Test statistic is -2.027
(d) Reject H0
Step-by-step explanation:
At 0.10 significance level, there is a difference between the population means because the null hypothesis is rejected.
(a) The decision rule is to reject H0 if the test statistic (t) falls outside the region bounded by the critical values.
(b) n1 = 9
s1 = 4.7
s1^2 = 4.7^2 = 22.09
n2 = 9
s2 = 3.6
s2^2 = 3.6^2 = 12.96
Pooled variance = [(n1-1)s1^2 + (n2-1)s2^2] ÷ (n1+n2-2) = [(9-1)22.09 + (9-1)12.96] ÷ (9+9-2) = 280.4 ÷ 16 = 17.525
(c) Test statistic (t) = (mean1 - mean2) ÷ sqrt[pooled variance (1/n1 + 1/n2)] = (22 - 26) ÷ sqrt[17.525(1/9 + 1/9)] = -4 ÷ 1.973 = -2.027
(d) degree of freedom = n1+n2-2 = 9+9-2 = 16
significance level = 0.10 = 10%
The test is a two-tailed test because the alternate hypothesis is expressed using not equal to.
critical values corresponding to 16 degrees of freedom and 10% significance level are -1.746 and 1.746
My decision rule:
Reject H0 because the test statistic -2.027 falls outside the region bounded by the critical values -1.746 and 1.746.
