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in a survey of 2271 adults in a recent year. 1300 say they have made a New Year's resolution. Construct 90% and 95% confidence intervals for the population proportion. Interpret the results and compare the widths of the confidence intervals. The 90% confidence interval for the population proportion pis (Round to three decimal places as needed.)

User OrElse
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Final answer:

For a 90% confidence interval, the interval is approximately (0.549, 0.596). For a 95% confidence interval, the interval is approximately (0.543, 0.602). The 95% confidence interval is wider because it provides a higher level of confidence but also allows for a larger range of plausible values.

Step-by-step explanation:

To construct the confidence intervals for the population proportion, we can use the formula:

Sample proportion ± Z * sqrt((sample proportion * (1 - sample proportion))/sample size)

For the 90% confidence interval, with a sample size of 2271 and 1300 people saying they made a New Year's resolution, the point estimate is 1300/2271 = 0.5724.

The Z-value for a 90% confidence level is 1.645.

Plugging in these values, we get:

0.5724 ± 1.645 * sqrt((0.5724 * (1 - 0.5724))/2271)

Calculating the values, the 90% confidence interval is approximately (0.549, 0.596).

Similarly, for the 95% confidence interval, the Z-value is 1.96. Plugging in the values, we get:

0.5724 ± 1.96 * sqrt((0.5724 * (1 - 0.5724))/2271)

Calculating the values, the 95% confidence interval is approximately (0.543, 0.602).

The 95% confidence interval is wider than the 90% confidence interval because a higher confidence level requires a larger margin of error, making the interval wider.

This means that we can be more confident that the true population proportion falls within the 95% confidence interval, but it also means that the range of plausible values is larger.

User HokaHelal
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