Question

The MINITAB printout shows a test for the difference in two population means.

Two-Sample T-Test and CI: Sample 1, Sample 2
Two-sample T for Sample 1 vs Sample 2
	N	Mean	StDev	SE Mean
Sample 1	5	29.00	4.00	1.8
Sample 2	9	28.86	4.61	1.5
Difference = mu (Sample 1) - mu (Sample 2)
Estimate for difference: 0.14
95% CI for difference: (-5.2, 5.5)
T-Test of difference = 0 (vs not =):
T-Value = 0.06 P-Value = 0.96 DF = 12
Both use Pooled StDev = 4.42

(a) Do the two sample standard deviations indicate that the assumption of a common population variance is reasonable?

Yes, the ratio of the two variances is less than three.Yes, the ratio of the two variances is more than three.    No, the ratio of the two variances is less than three.No, the ratio of the two variances is more than three.It is not possible to check that assumption with the given information.

(b) What is the observed value of the test statistic?
t =  

What is the p-value associated with this test?
p-value =  

(c) What is the pooled estimate s² of the population variance? (Round your answer to two decimal places.)
s² =  

(d) Use the answers to part (b) to draw conclusions about the difference in the two population means. (Use ? = 0.10.)

Since the p-value is less than 0.10, the results are significant. There is sufficient evidence to indicate a difference in the two population means.Since the p-value is greater than 0.10, the results are not significant. There is insufficient evidence to indicate a difference in the two population means.    Since the p-value is less than 0.10, the results are not significant. There is insufficient evidence to indicate a difference in the two population means.Since the p-value is greater than 0.10, the results are significant. There is sufficient evidence to indicate a difference in the two population means.

(e) Find the 95% confidence interval for the difference in the population means.
to  

Does this interval confirm your conclusions in part (d)?

Yes, since 0 falls in the confidence interval, there is sufficient evidence to indicate a difference in the two population means.Yes, since 0 falls in the confidence interval, there is insufficient evidence to indicate a difference in the two population means.    Yes, since 0 falls outside the confidence interval, there is sufficient evidence to indicate a difference in the two population means.No, since 0 falls in the confidence interval, there is sufficient evidence to indicate a difference in the two population means.No, since 0 falls outside the confidence interval, there is insufficient evidence to indicate a difference in the two population means.

Answer 1

(a) Do the two sample standard deviations indicate that the assumption of a common population variance is reasonable?

Yes, the ratio of the two variances is less than three.

( 4.61 / 4 = 1.1525 which is less than 3)

(b) What is the observed value of the test statistic?

t = t - Value = 0.06

What is the p-value associated with this test?
p-value = 0.96

(c) What is the pooled estimate s2 of the population variance? (Round your answer to two decimal places.)

s² = 4.42^2 = 19.5364

(d) Use the answers to part (b) to draw conclusions about the difference in the two population means. (Use ? = 0.10.)

Decision rule:

1) If p-value < level of significance (alpha) then we reject null hypothesis

2) If p-value > level of significance (alpha) then we fail to reject null hypothesis.

Here p value = 0.96 > 0.10 so we used 2nd rule.

That is we fail to reject null hypothesis

Since the p-value is greater than 0.10, the results are not significant. There is insufficient evidence to indicate a difference in the two population means.

(e) Find the 95% confidence interval for the difference in the population means.

95% CI for difference: (-5.2, 5.5)

Yes because zero include in the above confidence interval.

Yes, since 0 falls in the confidence interval, there is insufficient evidence to indicate a difference in the two population means.