Question

An independent-measures research study was used to compare two treatment conditions with n= 12 participants in each treatment. The first treatment had a mean of M= 55 with variance s2 = 8, and the second treatment had M= 52 and s2= 4.

a) Use a two-tailed test with Alpha= 0.05 to determine whether these date indicate a significant difference between the two treatments.

b) Use a two-tailed test with Alpha= 0.01 to determine whether these date indicate a significant difference between the two treatments.

c) Use a one-tailed test with Alpha= 0.05 to determine whether these date indicate a significant difference between the two treatments.

Answer 1

(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.

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.