Question

A study was done to determine whether there is a relationship between snoring and the risk of heart disease. Among 1,105 snorers in the study, 87 had heart disease, while only 27 of 1,379 nonsnorers had heart disease.
1) Determine a 95% confidence interval that estimates p1 − p2 = difference in proportions of snorers and nonsnorers who have heart disease. (Let p1 be the proportion of snorers and p2 be the proportion of nonsnorers. Round your answers to three decimal places.)

Answer 1

The 95% confidence interval for the difference in proportions of snorers and non snorers who have heart disease is 0.0591 ± 0.0484 = (0.0107, 0.1075).

To calculate the 95% confidence interval for the difference in proportions of snorers and non snorers who have heart disease, we need to use the following formula:

CI95% (p1 − p2) = pˉ1 − pˉ2 ± 1.96 × √(pˉ1 × (1 − pˉ1) / n1 + pˉ2 × (1 − pˉ2) / n2)

Where pˉ1 is the proportion of snorers with heart disease, pˉ2 is the proportion of non snorers with heart disease, and n1 and n2 are the sample sizes of snorers and nonsnorers respectively.

In our case, pˉ1 = 87/1105 = 0.0786, pˉ2 = 27/1379 = 0.0195, n1 = 1105, and n2 = 1379.

Therefore, the 95% confidence interval for the difference in proportions of snorers and non snorers who have heart disease is 0.0591 ± 1.96 × √(0.0786 × (1 − 0.0786) / 1105 + 0.0195 × (1 − 0.0195) / 1379) = 0.0591 ± 0.0484 = (0.0107, 0.1075).

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