Question

The distribution of diastolic blood pressures for the population of female diabetics between the ages of 30 and 34 has an unknown mean and standard deviation. A sample of 10 diabetic women is selected; their mean diastolic blood pressure is 84 mm Hg. We want to determine whether the diastolic blood pressure of female diabetics are different from the general population of females in this age group, where the mean u = 74.4 mmHg and standard deviation o = 9.1 mm Hg. Diastolic blood pressure is normally distributed.

a) Create a two-sided 95% confidence interval to determine whether diabetic women have a different mean diastolic blood pressure compared to the general population.
b) Now, conduct a two-sided hypothesis test at the a = 0.05 level of significance to determine whether diabetic women have a different mean diastolic blood pressure compared to the general population. Use both critical value and p-value methods. b) For either method, would your conclusion have been different if you had chosen a = 0.01 instead of a = 0.05?

Answer 1

To analyze diastolic blood pressure differences in female diabetics compared to the general population, a two-sided 95% confidence interval and a two-sided hypothesis test are conducted using a z-test approach with the known standard deviation of the general population. The choice of alpha level, 0.05 vs. 0.01, impacts the critical values and the robustness needed to reject the null hypothesis in the test.

Step-by-step explanation:

The goal is to create a two-sided 95% confidence interval for the mean diastolic blood pressure of female diabetics and to conduct a two-sided hypothesis test at the alpha (a) = 0.05 level of significance to determine if there is a significant difference from the general population. Since the population standard deviation is not known and the sample size is small, we would normally use the t-distribution. However, since the question specifies that we are to use the known standard deviation of the general population, we will proceed with a z-test, which is unusual in such contextual problems.

1. To compute the 95% confidence interval, we use the formula: \(\bar{x} \pm z\frac{\sigma}{\sqrt{n}}\), where \(\bar{x}\) is the sample mean, \(z\) is the z-score corresponding to the 95% confidence level (1.96 for two-tailed), \(\sigma\) is the population standard deviation, and \(n\) is the sample size.
2. For hypothesis testing, we state the null hypothesis \(H_0: \mu = 74.4\) mm Hg (no difference) and alternative hypothesis \(H_1: \mu \\eq 74.4\) mm Hg (there is a difference). We calculate the test statistic \(z = \frac{\bar{x} - \mu_{0}}{\sigma/\sqrt{n}}\) and compare it to the critical z-value or check the p-value against \(\alpha\).
3. If \(\alpha\) was 0.01 instead of 0.05, we would need more extreme results for the hypothesis test to reject the null hypothesis, as the critical z-value will be larger (about \(\pm2.576\) for a two-tailed test).