Question

A researcher is interested in the relationship between oral contraceptive (OC) use and blood pressure in women. Suppose a sample of ten 35-to 39 year old nonpregnant, premenopausal OC users who work in a company and have a mean systolic blood pressure (SBP) of 132.86 mm Hg and sample standard deviation of 15.34 mm Hg are identified. A sample of 21 nonpregnant, premenopausal, non-OC users in the same age group are similarly identified who have mean SBP of 127. 44 mm Hg and a sample standard deviation of 18.23 mm Hg. A researcher assumes that the two groups are independent and normal will equal variances. 1. Using the p-value, does the underlying mean difference in blood pressure between the two groups differ from zero? Use the significance level α = 0.05. 2. Compute a 95% confidence interval for the true mean difference in systolic blood pressure (SBP) between 35 to 39 year old OC users and non-OC users. Interpret this result.

Answer 1

The provided sample means are shown below: X1 = 132.86 X2 = 127.44 Also, the provided sample standard deviations are: Si = 15.34 82 = 18.23 and the sample sizes are ni = 10 and n2 = 21. (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: Mi = M2 Ha: Mi 7 M2

This corresponds to a two-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used. (2) Rejection Region Based on the information provided, the significance level is a = 0.05, and the degrees of freedom are df = 29. In fact, the degrees of freedom are computed as follows, assuming that the population variances are equal: Hence, it is found that the critical value for this two-tailed test is to = 2.045, for a = 0.05 and df = 29. The rejection region for this two-tailed test is R = {t: \t| > 2.045}. (3) Test Statistics Since it is assumed that the population variances are equal, the t-statistic is computed as follows:

Final conclusion: Mean difference in blood pressures between the two groups not different from zero

t= Ã1 - X2 (ni-1)s+(n2-1) să (17+) ni+n2 - 2 0.811 132.86 – 127.44 (10-1)15.342+(21-1)18.232 (o + 11) 10+21-2 (4) Decision about the null hypothesis Since it is observed that t = 0.811 <tc = 2.045, it is then concluded that the null hypothesis is not rejected. Using the P-value approach: The p-value is p = 0.4237, and since p= 0.4237 > 0.05, it is concluded that the null hypothesis is not rejected. (5) Conclusion It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean Mi is different than 42, at the 0.05 significance level.

We need to construct the 95% confidence interval for the difference between the population means Mi - M2, for the case that the population standard deviations are not known. The following information has been provided about each of the samples: Sample Mean 1 (X1) = 132.86 Sample Standard Deviation 1 (81) = 15.34 10 Sample Size 1 (N1) = Sample Mean 2 (X2) = 127.44 Sample Standard Deviation 2 (S2) = 18.23 Sample Size 2 (N2) = 21 Based on the information provided, we assume that the population variances are equal, so then the number of degrees of freedom are df = ni + n2 – 2 = 10 + 21 – 2 = 29.

The critical value for a = 0.05 and df = 29 degrees of freedom is te = t1-a/2;n-1 = 2.045. The corresponding confidence interval is computed as shown below: Since the population variances are assumed to be equal, we need to compute the pooled standard deviation, as follows: 1)s = (n1 – 1)s +(n2 ni + n2 2 Sp = (10 – 1) × 15.342 + (21 – 1) × 18.232 10 + 21 – 2 17.385 Since we assume that the population variances are equal, the standard error is computed as follows:

1 1 se = Sp + ni n2 1 1 = 17.385 x + 10 21 = 6.679 Now, we finally compute the confidence interval: CI = (X1 – X2 – te X se, Ã1 - 12 + te x se) = (132.86 – 127.44 – 2.045 x 6.679, 132.86 – 127.44 + 2.045 x 6.679) (-8.241, 19.081)

Final conclusion: Here we can see that within the confidence interval include zero that means, Mean difference in blood pressures between the two groups not different from zero