Question

In a survey of 2003 adults in a recent​ year, 743 made a New​ Year's resolution to eat healthier.

Construct​ 90% and​ 95% confidence intervals for the population proportion

Answer 1

The 90% confidence interval is approximately (0.3542, 0.3876) and the 95% confidence interval is approximately (0.3513, 0.3905).

How to construct confidence intervals

To construct confidence intervals for the population proportion based on the given data, use the formula below:

Confidence Interval = Sample Proportion ± Margin of Error

where the margin of error depends on the desired confidence level and the sample size.

Let's calculate the confidence intervals for the population proportion using the given information:

Sample size (n) = 2003

Number of adults who made a New Year's resolution to eat healthier (x) = 743

First, calculate the sample proportion (
`\hat{p}`):

Sample Proportion (
`\hat{p}`) = x / n

For the given data:

`\hat{p}` = 743 / 2003 ≈ 0.3709

Now, calculate the margin of error for the 90% confidence interval:

For a 90% confidence interval, the critical z-value (Zα/2) is approximately 1.645.

Margin of Error (ME) = Zα/2 * sqrt((
`\hat{p}` * (1 -
`\hat{p}`)) / n)

ME = 1.645 *
`\sqrt`((0.3709 * (1 - 0.3709)) / 2003)

ME ≈ 0.0167

Therefore, the 90% confidence interval for the population proportion is:

0.3709 ± 0.0167

Next, calculate the margin of error for the 95% confidence interval:

For a 95% confidence interval, the critical z-value (Zα/2) is approximately 1.96.

Margin of Error (ME) = Zα/2 *
`\sqrt`((
`\hat{p}` * (1 -
`\hat{p}`)) / n)

ME = 1.96 *
`\sqrt`((0.3709 * (1 - 0.3709)) / 2003)

ME ≈ 0.0196

Therefore, the 95% confidence interval for the population proportion is:

0.3709 ± 0.0196

The 90% confidence interval is approximately (0.3542, 0.3876) and the 95% confidence interval is approximately (0.3513, 0.3905).