The following data, recorded in days, represents the length of time to recovery for patients randomly treated with one of two medications to clear up severe bladder infections. Assume that the recovery times are normally distributed. Medication 1 = n = 13, x̄ = 17, s2 = 1.5, Medication 2 = n = 10, x̄ = 19, s2 = 1.8. (a) Is there a difference in the mean recovery times for the two medications? Test at the 5% level of significance, assuming equal variances. (b) Is the assumption of equal variances made in (a) valid? Test at the 5% level of significance.

You will need to use an independent t-test to see if there is a difference between the means. To perform an independent t-test, you need to find the t-statistic and compare it with the significance value (in this case, 5%).
The t-statistic is given by the following formula:
T = (x̄1 - x̄2) / sqrt[s^2 * (1/n1 + 1/n2)]
The only value we don't have from that formula is s^2, or the pooled sample variance. To find that value, we need to sum the squares of the s values, 1.5 and 1.8—so (1.5^2 + 1.8^2) = 5.49.
T = (17 - 19) / sqrt[5.49 * (1/13 + 1/10)] = (-2)/0.9855 = -2.029.
For the given degrees of freedom (21), the 5% confidence interval is at a t-value of +/- 1.720743. This value is outside of that range, and therefore it is statistically significant.