MyLab Statistics Answer Key: 90% and 95% Confidence Intervals

Answer

1. Find the sample proportion and standard error

Let $p$ be the population proportion of adults who made a New Year's resolution. The sample proportion is

$$\hat p=\frac{1302}{2173}=0.599171\ldots$$

and its estimated standard error is

$$SE=\sqrt{\frac{\hat p(1-\hat p)}{n}} =\sqrt{\frac{(0.599171\ldots)(0.400828\ldots)}{2173}} =0.0105129\ldots.$$

The large-count condition is satisfied because there are 1,302 successes and 871 failures.

2. Construct the 90% confidence interval

Using $z^*=1.644854$,

$$\hat p\pm z^*SE =0.599171\ldots\pm(1.644854)(0.0105129\ldots) =(0.581879\ldots,0.616464\ldots).$$

Rounded to three decimal places, the 90% confidence interval is

$$\boxed{(0.582,\ 0.616)}.$$

3. Construct the 95% confidence interval

Using $z^*=1.959964$,

$$\hat p\pm z^*SE =0.599171\ldots\pm(1.959964)(0.0105129\ldots) =(0.578567\ldots,0.619777\ldots).$$

Rounded to three decimal places, the 95% confidence interval is

$$\boxed{(0.579,\ 0.620)}.$$

4. Interpret and compare

Assuming the survey sample supports inference to its target population, we are 90% confident that the true proportion is between 0.582 and 0.616, and 95% confident that it is between 0.579 and 0.620.

The unrounded widths are approximately 0.0346 and 0.0412, respectively. The 95% interval is wider because a higher confidence level uses a larger critical value and therefore a larger margin of error.