A 2018 poll of 3643 randomly selected users of a social media site found that 1759 get most of their news about world events on the site. Research done in 2013 found that only $46 %$ of all the site users reported getting their news about world events on this site.
a. Does this sample give evidence that the proportion of site users who get their world news on this site has changed since 2013 ? Carry out a hypothesis test and use a 0.05 significance level.
b. After conducting the hypothesis test, a further question one might ask is what proportion of all of the site users get most of their news about world events on the site in 2018. Use the sample data to construct a $95 %$ confidence interval for the population proportion. How does your confidence interval support your hypothesis test conclusion?
(Round to three decimal places as needed.)
Reject $H_{0}$ or do not reject $H_{0}$ and interpret the conclusion. Choose the correct answer below.
A. Reject $H_{0}$ . The percentage is significantly different from $46 %$ .
B. Do not reject $H_{0}$ . The percentage is significantly different from $46 %$ .
C. Reject $H_{0}$ . The percentage is not significantly different from $46 %$ .
D. Do not reject $H_{0}$ . The percentage is not significantly different from $46 %$ .
b. The $95 %$ confidence interval for the population proportion is $\frac{1}{◻}, ◻$ ). (Round to four decimal places as needed.)

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Final Answer: $b. The 95 % confidence interval for the population proportion is (0.467, 0.499) .$

Steps

Step 1 ：Define the sample size, n = 3643, and the number of successes, x = 1759.

Step 2 ：Calculate the sample proportion, $p_{h a t} = \frac{x}{n} = \frac{1759}{3643} = 0.483$ .

Step 3 ：Define the proportion in the null hypothesis, p0 = 0.46.

Step 4 ：Calculate the z-score using the formula $z = \frac{p_{h a t} - p 0}{\sqrt{\frac{p 0 * (1 - p 0)}{n}}} = \frac{0.483 - 0.46}{\sqrt{\frac{0.46 * (1 - 0.46)}{3643}}} = 2.77$ .

Step 5 ：Since the z-score is greater than the critical value of 1.96 for a 95% confidence level, we reject the null hypothesis and conclude that the proportion of site users who get their world news on this site has changed since 2013.

Step 6 ：Calculate the 95% confidence interval for the population proportion using the formula $p_{h a t} \pm z_{95} * \sqrt{\frac{p_{h a t} * (1 - p_{h a t})}{n}}$ , where $z_{95} = 1.96$ .

Step 7 ：The lower bound of the confidence interval is $0.483 - 1.96 * \sqrt{\frac{0.483 * (1 - 0.483)}{3643}} = 0.467$ .

Step 8 ：The upper bound of the confidence interval is $0.483 + 1.96 * \sqrt{\frac{0.483 * (1 - 0.483)}{3643}} = 0.499$ .

Step 9 ：Since the 95% confidence interval (0.467, 0.499) does not contain the value 0.46 (the proportion in 2013), this supports our conclusion from the hypothesis test that the proportion has changed.

Step 10 ：Final Answer: $a. Reject H_{0} . The percentage is significantly different from 46 % .$

Step 11 ：Final Answer: $b. The 95 % confidence interval for the population proportion is (0.467, 0.499) .$

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