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 4 28.00 3. 00 1 . 5 Sample 2 27.87 3. 69 1 . 3 Difference = mu (Sample 1) - mu (Sample 2) Estimate for difference: 0.13 954 CI for difference: (-4.6, 4.9) T-Test of difference = 0 (vs not =) : T-Value = 0.06 P-Value = 0.95 DF = 10 Both use Pooled StDev = 3.50 (a) Do the two sample standard deviations indicate that the assumption of a common population variance is reasonable? O Yes, the ratio of the two variances is less than three. O Yes, the ratio of the two variances is more than three. O No, the ratio of the two variances is less than three. O No, the ratio of the two variances is more than three. O 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 s of the population variance? (Round your answer to two decimal places.) $2 = (d) Use the answers to part (b) to draw conclusions about the difference in the two population means. (Use a = 0.10.) O 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. O 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. O 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. O 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. to Does this interval confirm your conclusions in part (d)? O Yes, since 0 falls in the confidence interval, there is sufficient evidence to indicate a difference in the two population means. ndiratA study was conducted on the effect of an oral antiplaque rinse on plaque buildup on teeth. Fourteen people whose teeth were thoroughly cleaned and polished were randomly assigned to two groups of seven subjects each. Both groups were assigned to use oral rinses (no brushing) for a two-week period. Group 1, the control group, received a rinse that, unknown to the subjects, contained no antiplaque agent. Group 2 used a rinse that contained an antiplaque agent. A plaque index x, a measure of plaque buildup, was recorded at 4, 7, and 14 days. The mean and standard deviation for the 14-day plaque measurements are shown in the table for the two groups. Control Group Antiplaque Group Sample size 7 7 Mean 1.29 0.78 Standard deviation 0.33 0.33 (a) State the null and alternative hypotheses that should be used to test the effectiveness of the antiplaque oral rinse. O Ho: (H1 - 12) = 0 versus Ha: (1 1 - 12) $ 0 O Ho: ( M 1 - 12) = 0 versus Ha: (1 1 - 12 ) > 0 O Ho: ( 1 1 - 12 ) 0 O Ho: ( H 1 - 12 ) = 0 versus Ha: (1 1 - 12 ) = 0 O Ho: (M 1 - 12 ) = 0 versus Ha: (1 1 - 12) t<