How to calculate a correct 95% confidence interval based on provided sample data?

Question

A researcher interested in security events. Researcher suspected that people in the 20-29 age group were more likely to say they had experienced security events than people in the 30-39 age group. Researcher obtained separate random samples of people from each age group. Here are the results: 

Impacted?	20-29	30-39
Yes	24	24
No	56	96
Total	80	120

Researcher wants to use above results to construct a 95% confidence interval to estimate the difference between the proportion of people in each age group who would say they have been impacted (x=p20s-p30s). Assume that all of the conditions for inference have been met.

What is the correct 95% confidence interval (x) based on researcher's samples? 

Answer 1

We are dealing with proportions so the proportion of 20s saying yes is 24/80=0.3, while that for 30s is 24/120=0.2. So the difference is 0.1. It's this difference we need to examine more closely later. We also need to note the sample sizes.

Let p₂₀=0.3 and p₃₀=0.2 and let μ=p₂₀-p₃₀=0.1. Statistically we can equate the proportions to the two means in a probability distribution. We can determine the variance because we have a binary situation－yes and no are the only two states. So for the individual groups: 

μ₂₀=p₂₀=0.3 and μ₃₀=p₃₀=0.2, variance σ₂₀²=p₂₀(1-p₂₀)/n₂₀=0.21/80=0.002625; variance σ₃₀²=p₃₀(1-p₃₀)/n₃₀=0.16/120=0.001333. (We can deduce the standard deviations from these if we need them.)

What we called μ (p₂₀-p₃₀) is the mean for the probability distribution for the difference in proportions. The variance is simply the sum of the individual variances:

σ²=σ₂₀²+σ₃₀²=0.002625+0.001333=0.003958, so standard deviation σ=√0.003958=0.06292 approx.

The sample sizes n₂₀=80 and n₃₀=120 are reasonably large (bigger than 30) so we can assume a normal distribution.

A 95% confidence interval corresponds to a Z-score of 1.96, a two-tailed value because we are considering a data range [low,high] on either side of the mean, so the 100-95=5% is distributed equally between the left and right tails of the distribution. We are actually looking at the Z-score corresponding to 100-5/2=97.5%=0.975 (Z=1.96) when we inspect the body of the distribution table. Z=(X-μ)/σ for some value X in the dataset. We are looking for a low X value and a high X value to give us a 95% confidence interval. So:

|X-μ|/σ=1.96, from which X=μ±1.96σ=0.1±1.96×0.06292. The confidence interval is about [-0.02,0.22]. We can be 95% confident that the proportion difference lies between these two extremes. This means that in a population -0.02≤p₂₀-p₃₀≤0.22. (This appears to allow for a small minority of 20-29 year olds to actually be a little less impacted than 30-39 year olds; but the majority of 20-29 year olds will be impacted at least as much as 30-39 year olds.)

Comments

Hello Rod, thank you for your help as always.