Question

In Exercises 6.139 to \(6.142,\) use the normal distribution to find a confidence interval for a difference in proportions \(p_{1}-p_{2}\) given the relevant sample results. Give the best estimate for \(p_{1}-p_{2},\) the margin of error, and the confidence interval. Assume the results come from random samples. A 99\% confidence interval for \(p_{1}-p_{2}\) given counts of 114 yes out of 150 sampled for Group 1 and 135 yes out of 150 sampled for Group 2 .

Short answer

The best estimate for \(p_{1}-p_{2}\) is -0.14. The margin of error is about 0.137. Therefore, the 99% confidence interval is \(-0.277\) to \(-0.003\)

Step-by-step solution

Calculate the proportions

The proportions were calculated as follows: \(p_1 = \frac{114}{150} = 0.76\) and \(p_2 = \frac{135}{150} = 0.9\). Hence the proportions for Group 1 and Group 2 are 0.76 and 0.9 respectively.

Calculate the best estimate for \(p_{1}-p_{2}\)

The best estimate is simply the difference in the sample proportions: \(p_{1} - p_{2} = 0.76 - 0.9 = -0.14 \). Hence the best estimate for the difference in proportions is -0.14.

Compute the Standard Error (SE)

The SE of the difference in proportions is given by the square root of the sum of the variances of \(p_{1}\) and \(p_{2}\). The SE can be calculated as: SE = sqrt[(\(p_{1}(1-p_{1})/n_{1}\) + \(p_{2}(1-p_{2})/n_{2}\)] where \(n_{1}\) and \(n_{2}\) are the sizes of group 1 and 2 respectively. In this case, SE = sqrt[((0.76)(1-0.76)/150) + ((0.9)(1-0.9)/150)] = 0.053.

Compute the margin of error (ME)

The ME is the product of the z-score and the standard error (SE). A 99\% confidence level corresponds to a z-score of 2.58. Therefore, ME = z * SE = 2.58 * 0.053 = 0.137

Compute Confidence Interval

The confidence interval is computed by subtracting ME from the best estimate (lower limit) and adding ME to the best estimate (upper limit). Therefore, the 99\% confidence interval for \(p_{1} - p_{2}\) is \(-0.14 - 0.137\) to \(-0.14 + 0.137\) or \(-0.277\) to \(-0.003\)