Question

Random samples of students at 123 four-year colleges were interviewed several times since 1991. Of the students who reported drinking alcohol, the percentage who reported drinking daily was 32.2​% of 13,431 students in 1991 and 47.7% of 8507 students in 2003.

a. Estimate the difference between the proportions in 2003 and 1991​, and interpret.

b. Find the standard error for this difference.

c. Construct and interpret a 99​% confidence interval to estimate the true change.

d. State the assumptions for the confidence interval in​ (c) to be valid.

Answer 1

a. Estimate the difference between the proportions in 2003 and 1991​, and interpret.

Solution:

We are given P₁ = 0.477, P₂ = 0.322

Estimated difference in proportions = P₁ – P₂ = 0.477 - 0.322 = 0.155

b. Find the standard error for this difference.

Standard error = sqrt[(P1*(1 – P1)/N1) + (P2*(1 – P2)/N2)]

Standard error = sqrt[(0.477*(1 – 0.477)/8507) + (0.322*(1 – 0.322)/13431)]

Standard error = 0.0068

c. Construct and interpret a 99​% confidence interval to estimate the true change.

Confidence interval for difference between two population proportions:

Confidence interval = (P1 – P2) ± Z*sqrt[(P1*(1 – P1)/N1) + (P2*(1 – P2)/N2)]

Where, P1 and P2 are sample proportions for first and second groups respectively.

Confidence level = 99%

Critical Z value = 2.5758

(by using z-table)

Confidence interval = (P1 – P2) ± Z*sqrt[(P1*(1 – P1)/N1) + (P2*(1 – P2)/N2)]

Confidence interval = (.477 – .322) ± 2.5758* sqrt[(0.477*(1 – 0.477)/8507) + (0.322*(1 – 0.322)/13431)]

Confidence interval = (.477 – .322) ± 2.5758* 0.0068

Confidence interval = 0.155 ± 2.5758* 0.0068

Confidence interval = 0.155 ± 0.0174

Lower limit = 0.155 - 0.0174 = 0.1376

Upper limit = 0.155 + 0.0174 = 0.1724

Confidence interval = (0.1376, 0.1724)

d. State the assumptions for the confidence interval in​ (c) to be valid.

Both the sample sizes are greater and adequate for using normal distribution. Sample sizes are greater than 30 so that we can use normal approximation for the above confidence interval.