Question

An article presents a study of health outcomes in women with symptoms of heart disease. In a sample of 110 women whose test results suggested the presence of coronary artery disease, the mean peak systolic blood pressure was 169.9 mmHg, with a standard deviation of 24.8 mmHg. In a sample of 225 women whose test results suggested an absence of coronary artery disease, the mean peak systolic blood pressure was 163.3 mmHg, with a standard deviation of 25.8 mmHg. Find a 95% confidence interval for the difference in mean systolic blood pressure between these two groups of women. (Round the final answers to three decimal places.)

Answer 1

(0.869, 12.33104)

Explanation:

N1 = 110

N2 = 225

S1 = 24.8

S2 = 25.8

Bar x1 = 169.9

Bar x2 = 163.3

Difference in mean = 169.9 - 163.3 = 6.6

We use the z score

Alpha = 0.05

Z critical value at this alpha level = 1.96

We find standard error

√s1²/n1+s2²/n2

= √(24.8²/110 + 25.8²/225)

= 2.9239

Confidence interval

6.6 +-1.96(2.924)

(6.6-5.73104, 6.6+5.73104)

= (0.869, 12.331)