Questions: A random sample of 5003 adults in a country includes 760 who do not use the internet. Construct a 95% confidence interval estimate of the percentage of adults in the country who do not use the Internet. Based on the result, does it appear that the percentage of adults in the country who do not use the internet is different from 47%, which was the percentage in the year 2000 ? Construct a 95% confidence interval estimate of the percentage of adults in the country who do not use the internet. 142% < p < 16.2% (Round to one decimal place as needed.) Does it appear that the percentage of adults in the country who do not use the internet is different from 47%? A. Because 47% is not contained within the confidence interval, it appears that the percentage of adults in the country who do not use the internet is different from 47%. B. Because 47% is not contained within the confidence interval, it does not appear that the percentage of adults in the country who do not use the internet is different from 47%. C. Because 47% is contained within the confidence interval, it does not appear that the percentage of adults in the country who do not use the internet is different from 47%. D. Because 47% is contained within the confidence interval, it appears that the percentage of adults in the country who do not use the internet is different from 47%.

Transcript text: A random sample of 5003 adults in a country includes 760 who do not use the internet. Construct a $95 \%$ confidence interval estimate of the percentage of adults in the country who do not use the Internet. Based on the result, does it appear that the percentage of adults in the country who do not use the internet is different from $47 \%$, which was the percentage in the year 2000 ? Construct a $95 \%$ confidence interval estimate of the percentage of adults in the country who do not use the internet. \[ 142 \%

Solution

Solution Steps

Step 1: Calculate the Sample Proportion

The sample proportion of adults who do not use the internet is calculated as follows:

\[ \hat{p} = \frac{x}{n} = \frac{760}{5003} \approx 0.1519 \]

Step 2: Calculate the Confidence Interval

To construct a \(95\%\) confidence interval for the proportion, we use the formula:

\[ \hat{p} \pm z \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} \]

Where:

• \(\hat{p} \approx 0.1519\)
• \(n = 5003\)
• The critical value \(z\) for a \(95\%\) confidence level is approximately \(1.96\).

Calculating the margin of error:

\[ \text{Margin of Error} = z \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} = 1.96 \cdot \sqrt{\frac{0.1519(1 - 0.1519)}{5003}} \approx 0.0200 \]

Thus, the confidence interval is:

\[ (0.1519 - 0.0200, 0.1519 + 0.0200) = (0.1319, 0.1719) \]

Step 3: Interpret the Confidence Interval

The \(95\%\) confidence interval for the proportion of adults who do not use the internet is:

\[ (0.1319, 0.1719) \]

Step 4: Compare with Historical Percentage

We need to determine if the historical percentage of \(47\%\) (or \(0.47\)) is contained within the confidence interval:

\[ 0.1319 < 0.47 < 0.1719 \]

Since \(0.47\) is not contained within the confidence interval, we conclude:

Final Answer

The answer is A. Because \(47\%\) is not contained within the confidence interval, it appears that the percentage of adults in the country who do not use the internet is different from \(47\%\).

\(\boxed{A}\)

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