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.

(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.

true? is that what you're asking or do you mean something else?

125 yards

step-by-step explanation:

i take 75 plus 50.