Answer:
(a) The data indicate a significant difference between the two treatments.
(b) The data do not indicate a significant difference between the two treatments.
(c) The data indicate a significant difference between the two treatments.
Step-by-step explanation:
Null hypothesis: There is no difference between the two treatments.
Alternate hypothesis: There is a significant difference between the two treatments.
Data given:
M1 = 55
M2 = 52
s1^2 = 8
s2^2 = 4
n1 = 12
n2 = 12
Pooled variance = [(n1-1)s1^2 + (n2-1)s2^2] ÷ (n1+n2-2) = [(12-1)8 + (12-1)4] ÷ (12+12-2) = 132 ÷ 22 = 6
Test statistic (t) = (M1 - M2) ÷ sqrt [pooled variance (1/n1 + 1/n2)] = (55 - 52) ÷ sqrt[6(1/6 + 1/6)] = 3 ÷ 1.414 = 2.122
Degree of freedom = n1+n2-2 = 12+12-2 = 22
(a) For a two-tailed test with a 0.05 (5%) significance level and 23 degrees of freedom, the critical values are -2.069 and 2.069.
Conclusion:
Reject the null hypothesis because the test statistic 2.122 falls outside the region bounded by the critical values.
(b) For a two-tailed test with a 0.01 (1%) significance level and 23 degrees of freedom, the critical values are -2.807 and 2.807.
Conclusion:
Fail to reject the null hypothesis because the test statistic 2.122 falls within the region bounded by the critical values.
(c) For a one-tailed test with 0.05 (5%) significance level and 23 degrees of freedom, the critical value is 1.714.
Conclusion:
Reject the null hypothesis because the test statistic 2.122 is greater than the critical value 1.714.
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