Question

In Exercises 6.150 and \(6.151,\) use StatKey or other technology to generate a bootstrap distribution of sample differences in proportions and find the standard error for that distribution. Compare the result to the value obtained using the formula for the standard error of a difference in proportions from this section. Sample A has a count of 90 successes with \(n=120\) and Sample \(\mathrm{B}\) has a count of 180 successes with \(n=300\).

Short answer

The difference in proportions for the samples A and B is 0.15. The standard error of this difference computed using the formula is approximately 0.061. The standard error computed using a bootstrap distribution should be approximately the same, reaffirming the accuracy of the formula.

Step-by-step solution

Calculate the proportions

First, calculate the proportions for both samples A and B. The proportion is given by the number of successes divided by the total number of trials. For sample A, it's \(90/120=0.75\). For sample B, it's \(180/300 = 0.60\). The difference in proportions is \(0.75 - 0.60 = 0.15\).

Compute the standard error using the formula

The formula for the standard error of a difference in proportions is \(\sqrt{((p_1(1-p_1))/n_1) + ((p_2(1-p_2))/n_2)}\), where \(p_1\) and \(p_2\) are the proportions of the two samples, and \(n_1\) and \(n_2\) are the sizes of the two samples. Plugging in the calculated proportions and sample sizes gives: \(\sqrt{((0.75)(1-0.75))/120 + ((0.60)(1-0.60))/300} \approx 0.061\)

Generate a bootstrap distribution

Using software like StatKey, generate a bootstrap distribution from the observed samples A and B, calculate the difference in proportions for each resample, and compute the standard error of these differences. This simulation-based standard error is likely to be quite close to the standard error computed in step 2, reaffirming the accuracy of the standard error formula.