Question

The Census Bureau stated that for Americans in the age group 18-24 years, 64.8 percent had private health insurance.. In the age group 25-34 years, the percentage was 72.1. Assume that these percentages are the population parameters in those age groups for the United States. Suppose we select random sample of 250 Americans from the 18-24 age group (Group A) and an independent random sample of 200 Americans from the age group 25-34 (Group B); find the probability that the difference in proportions is less than 6 percent.

Answer 1

To find the probability that the difference in proportions is less than 6 percent, calculate the standard deviation of the difference in proportions and use the standard normal distribution.

Step-by-step explanation:

To find the probability that the difference in proportions is less than 6 percent, we need to calculate the standard deviation of the difference in proportions.

First, let's find the standard deviation of the proportion in Group A:

Standard Deviation (Group A) = sqrt((P_a * (1 - P_a)) / n_a) = sqrt((0.648 * (1 - 0.648)) / 250) ≈ 0.0256

Next, let's find the standard deviation of the proportion in Group B:

Standard Deviation (Group B) = sqrt((P_b * (1 - P_b)) / n_b) = sqrt((0.721 * (1 - 0.721)) / 200) ≈ 0.0246

Now, let's find the standard deviation of the difference in proportions:

Standard Deviation (Difference) = sqrt((σ_a^2 / n_a) + (σ_b^2 / n_b)) = sqrt((0.0256^2 / 250) + (0.0246^2 / 200)) ≈ 0.0035

Finally, to find the probability that the difference in proportions is less than 6 percent, we calculate the z-score of 6 percent and use the standard normal distribution:

z = (0.06 - 0) / 0.0035 ≈ 17.14

The probability that the difference in proportions is less than 6 percent is the area to the left of 17.14 on the standard normal distribution curve, which is practically 1. Therefore, the probability is approximately 1.