Question

a random sample of 5002 adults in a country includes 757 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 46%, 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. % < p < % (round to one decimal place as needed.)

Explanation:

Step1: Calculate the sample proportion $\hat{p}$

The sample proportion $\hat{p}=\frac{x}{n}$, where $x = 757$ (number of successes, i.e., adults who do not use the Internet) and $n=5002$ (sample size). So, $\hat{p}=\frac{757}{5002}\approx0.151$.

Step2: Calculate $q$

Since $q = 1-\hat{p}$, then $q=1 - 0.151=0.849$.

Step3: Find the critical value $z_{\alpha/2}$

For a 95% confidence interval, $\alpha=1 - 0.95 = 0.05$, and $\alpha/2=0.025$. Looking up in the standard normal distribution table, $z_{\alpha/2}=z_{0.025}=1.96$.

Step4: Calculate the margin of error $E$

The formula for the margin of error for a proportion is $E = z_{\alpha/2}\sqrt{\frac{\hat{p}q}{n}}$.
Substitute the values: $E=1.96\sqrt{\frac{0.151\times0.849}{5002}}$.
First, calculate $\frac{0.151\times0.849}{5002}=\frac{0.128199}{5002}\approx2.563\times10^{-5}$.
Then, $\sqrt{2.563\times 10^{-5}}\approx0.00506$.
Multiply by $1.96$: $E=1.96\times0.00506\approx0.01$.

Step5: Calculate the confidence interval

The confidence interval for the proportion $p$ is $\hat{p}-ESubstitute $\hat{p}=0.151$ and $E = 0.01$: $0.151-0.01

Answer:

$14.1\%