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a random sample of 5002 adults in a country includes 757 who do not use…

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?
(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 46%?
○ a. because 46% 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 46%.
○ b. because 46% 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 46%.
○ c. because 46% 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 46%.
○ d. because 46% 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 46%.

Explanation:

Step1: Calculate the sample proportion

The sample proportion \( \hat{p}=\frac{757}{5002}\approx0.151 \)

Step2: Calculate the standard error

The standard error \( SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}=\sqrt{\frac{0.151\times(1 - 0.151)}{5002}}\approx\sqrt{\frac{0.151\times0.849}{5002}}\approx\sqrt{\frac{0.128}{5002}}\approx0.005 \)

Step3: Find the critical value

For a 95% confidence interval, the critical value \( z = 1.96 \)

Step4: Calculate the margin of error

The margin of error \( ME=z\times SE = 1.96\times0.005 = 0.010 \)

Step5: Calculate the confidence interval

The confidence interval is \( \hat{p}-ME\( 0.151- 0.010 < p < 0.151+0.010 \)
\( 0.141 < p < 0.161 \) or \( 14.1\%

Since \( 46\% \) is not in the interval \( (14.1\%,16.1\%) \)

Answer:

D. Because 46% 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 46%.