Question

9. We survey the systolic blood pressure of 43 married couples in which the ages of the spouses are within 5 years. The men have an average systolic pressure of 135.2 (millibars), and the women an average of 118.7. If the standard deviation of the difference is 9.6, find a 95% confidence interval for the difference.

Answer 1

Solution:

We are given n = 43

df = n – 1 = 42

Dbar = 135.2 - 118.7 = 16.5

S_d = 9.6

Confidence level = 95%

Critical t value = 2.0181

(by using t-table)

Confidence interval is given as below:

Confidence interval = Dbar ± t*S_D/sqrt(n)

Confidence interval = 16.5 ± 2.0181*9.6/sqrt(43)

Confidence interval = 16.5 ± 2.0181*1.463986275

Confidence interval = 16.5 ± 2.9544

Lower limit = 16.5 - 2.9544 = 13.5456

Upper limit = 16.5 + 2.9544 = 19.4544

Confidence interval = (13.5456, 19.4544)