Question

A survey was nun by a high school student in order to determine what proportion of mortgage-holders in his town expect to own their house within 10 years. He purveyed 179 mortgage holders and found that the proportion of these that did expect to own their house within 10 years is 0.69. The student decides to construct a 95% canfidence interval for the population proportion. a) Calculate the margin of error that the high school student will have. Give your answer as a decimal to 3 decimal places. Margin of error = A different high school student sees this survey and decides to try and repeat it. This second high school student also surveys 179 mortgage holders, and this student also finds that the proportion of these that do expect to own their house within 10 years is 0.69. However, this student wants to construct a 99% confidence interval for the population proportion. b) Calculate the margin of errar that the second student will have. Give your answer as a decimal to 3 decimal places. Margin of error =

Answer 1

To calculate the margin of error for a confidence interval, you need to use the formula:

Margin of Error = Critical Value * Standard Error

a) For a 95% confidence interval:
The critical value for a 95% confidence interval can be found using a standard normal distribution table. The critical value for a 95% confidence interval is approximately 1.96.

Given that the proportion of mortgage holders who expect to own their house within 10 years is 0.69, and the sample size is 179, we can calculate the standard error:

Standard Error = sqrt((p * (1 - p)) / n)

Standard Error = sqrt((0.69 * (1 - 0.69)) / 179)

Standard Error ≈ 0.0403

Now we can calculate the margin of error:

Margin of Error = 1.96 * 0.0403

Margin of Error ≈ 0.079 (rounded to 3 decimal places)

Therefore, the margin of error for the first student's 95% confidence interval is approximately 0.079.

b) For a 99% confidence interval:
The critical value for a 99% confidence interval is approximately 2.576.

Using the same proportion and sample size as before, we can calculate the standard error:

Standard Error = sqrt((0.69 * (1 - 0.69)) / 179)

Standard Error ≈ 0.0403

Now we can calculate the margin of error:

Margin of Error = 2.576 * 0.0403

Margin of Error ≈ 0.104 (rounded to 3 decimal places)

Therefore, the margin of error for the second student's 99% confidence interval is approximately 0.104.