Question

One sample has n = 18 with SS = 5,471, and a second sample has n = 18 with SS = = 5,545.

Find the pooled variance for the two samples.

S2p = ________

Compute the estimated standard error for the sample mean difference.

S(m₁-m₂) = ___________

If the sample mean difference is 12.00 points, is this enough to reject the null hypothesis and conclude that there is a significant difference for a two-tailed test at the .05 level? (Use three decimal places for the critical value, two for the t statistic.)

t critical +- _________

t= __________

Conclusion:

__ Reject the null hypothesis or

__ Fail to reject the null hypothesis

If the sample mean difference is 18.00 points, is this enough to indicate a significant difference for a two-tailed test at the .05 level?

t = ___________

Conclusion:

___ Fail to reject the null hypothesis

___ Reject the null hypothesis

Calculate the percentage of variance accounted for (r²) to measure the effect size for a mean difference of 12.00 points and for a mean difference of 18.00 points. (Use three decimal places)

For M₁-M₂ = 12.00 r² = ________

For M₁-M₂ = 18.00 r² = ________

Trying to use the best formulas to calculate all of these and in layman terms, not graduate level understanding.

Answer 1

Find the pooled variance for the two samples.

S2p =   (SS1 + SS2)/(n1+ n2-2)
= (5471+ 5545)/(18 +18 -2)
= 324

Compute the estimated standard error for the sample mean difference.

S(m1-m2) = Sp *sqrt(1/n1 + 1/n2)
= sqrt(324 * 2/18)
= 6

If the sample mean difference is 12.00 points, is this enough to reject the null hypothesis and conclude that there is a significant difference for a two-tailed test at the .05 level? (Use three decimal places for the critical value, two for the t statistic.)
df = 34

t critical ==T.INV.2T(0.05,34)
= +- 2.0322

t= 12/6 = 2

Conclusion:
TS < critical value
Fail to reject the null hypothesis

If the sample mean difference is 18.00 points, is this enough to indicate a significant difference for a two-tailed test at the .05 level?

t = 3
Conclusion:
TS > critical value

Reject the null hypothesis