Question

Psychiatrists wish to determine if there is a higher incidence of divorce among couples who do not drink much and couples where one or more of the spouses drink heavily. A simple random sample of couples was taken, assuming the samples were taken independently. Construct a 90% confidence interval for the difference in the proportion of divorced couples where one or more of the spouses drink heavily and the proportion of couples who do not drink. Round proportions to 3 decimal places. (11 points) Heavy alcohol: 16 married couples, 34 divorced couples; No alcohol: 43 married couples, 12 divorced couples.

A) (0.077, 0.386)
B) (-0.386, -0.077)
C) (0.115, 0.346)
D) (-0.346, -0.115)

Answer 1

After calculating the confidence interval for the difference in divorce proportions between heavy drinking couples and non-drinking couples, the result is (0.3201, 0.6039), which does not match the provided options.

Step-by-step explanation:

To answer the student's question, we need to construct a 90% confidence interval (CI) for the difference in the proportion of divorced couples where one or more of the spouses drink heavily and the proportion of couples who do not drink heavily.

Let's denote the proportion of divorced couples among heavy drinkers as p1 and the proportion of divorced couples among non-drinkers as p2.

Based on provided data:
p1 = number of divorced heavy drinking couples / total heavy drinking couples = 34 / (16+34) = 34 / 50 = 0.680
p2 = number of divorced non-drinking couples / total non-drinking couples = 12 / (43+12) = 12 / 55 = 0.218

The difference in proportions is p1 - p2 = 0.680 - 0.218 = 0.462.

Using a standard formula for the confidence interval of the difference between two independent proportions:
CI = (p1 - p2) ± Z*sqrt((p1*(1-p1)/n1) + (p2*(1-p2)/n2))

Where Z is the Z-value corresponding to the 90% confidence level which is approximately 1.645. Plugging in the numbers:

CI = 0.462 ± 1.645*sqrt((0.680*(1-0.680)/50) + (0.218*(1-0.218)/55))
CI = 0.462 ± 1.645*sqrt((0.2176/50) + (0.1705/55))
CI = 0.462 ± 1.645*sqrt((0.004352) + (0.0031))
CI = 0.462 ± 1.645*sqrt(0.007452)
CI = 0.462 ± 1.645*0.0863
CI = 0.462 ± 0.1419

Hence, the confidence interval is (0.462-0.1419, 0.462+0.1419) = (0.3201, 0.6039).

This result does not match any of the options provided by the student (A, B, C, or D). It's important to use rounded proportions to three decimal places, as specified in the question.