a. To test whether the proportion of site users who get their world news on this site has changed since 2013, we can perform a hypothesis test. Let p be the true population proportion of site users who get their world news on this site in 2018. Then the null hypothesis is H0: p = 0.46, and the alternative hypothesis is Ha: p ≠ 0.46. We will use a two-tailed z-test with a 0.05 significance level.
Using the sample data, the sample proportion of site users who get their world news on this site is 1704/3612 = 0.471. The standard error of the sample proportion is sqrt(0.46*0.54/3612) = 0.013. The z-score for the sample proportion is (0.471 - 0.46) / 0.013 = 0.846. The corresponding p-value for this test is P(|Z| > 0.846) = 0.397 > 0.05.
Since the p-value is greater than 0.05, we fail to reject the null hypothesis and conclude that there is not enough evidence to suggest that the proportion of site users who get their world news on this site has changed since 2013.
b. To construct a 95% confidence interval for the population proportion of site users who get their world news on this site in 2018, we can use the sample proportion and the margin of error formula:
Margin of error = z*sqrt(p*(1-p)/n)
where z = 1.96 for a 95% confidence level, p = 0.471, and n = 3612.
Plugging in these values, we get:
Margin of error = 1.96*sqrt(0.471*(1-0.471)/3612) = 0.014
Therefore, the 95% confidence interval for the population proportion is:
0.471 ± 0.014, or (0.457, 0.485)
Since the null hypothesis in part (a) was that the proportion of site users who get their world news on this site is 0.46, and the confidence interval does not contain 0.46, this supports the conclusion of the hypothesis test that there is not enough evidence to suggest that the proportion has changed since 2013.
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