A 2018 poll of 3634 randomly selected users of a social media site found that 1799 get most of their news about world events on the site. Research done in 2013 found that only 49% 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?

a) The calculated p-value is more than the significance level, hence, we fail to reject the null hypothesis & conclude that the proportion of all the site users that get their world news from their site hasn't changed from 49% since 2013

b) 95% Confidence interval for the proportion = (0.4787, 0.5113)

c) The result of the 95% confidence interval agrees with the result of the hypothesis testing performed in (a) because the value of p₀ lies within this confidence interval obtained.

Step-by-step explanation:

For hypothesis testing, we first clearly state our null and alternative hypothesis.

The null hypothesis is that the proportion of all the site users that get their world news from their site hasn't changed from 49% since 2013

And the alternative hypothesis is that the proportion of site users who get their world news on this site has changed from 49% since 2013.

Mathematically, the null hypothesis is

H₀: p₀ = 0.49

The alternative hypothesis is

Hₐ: p₀ ≠ 0.49

To do this test, we will use the z-distribution because, the degree of freedom is so large, it checks out.

So, we compute the z-test statistic

z = (x - μ)/σₓ

x = the proportion from the 2018 poll = p =(1799/3634) = 0.495

μ = p₀ = the proportion from 2013, which we want to check if it has changed = 0.49

σₓ = standard error of the poll proportion = √[p(1-p)/n]

where n = Sample size = 3634

σₓ = √[0.495×0.505/3634] = 0.0082938433 = 0.008294

z = (0.495 - 0.49) ÷ 0.008294

z = 0.603 = 0.60

checking the tables for the p-value of this z-statistic

p-value (for z = 0.60, at 0.05 significance level, with a two tailed condition) = 0.5485

The interpretation of p-values is that

When the (p-value > significance level), we fail to reject the null hypothesis and when the (p-value < significance level), we reject the null hypothesis and accept the alternative hypothesis.

So, for this question, significance level = 5% = 0.05

p-value = 0.5485

0.549 > 0.05

Hence,

p-value > significance level

This means that we fail to reject the null hypothesis & conclude that the proportion of all the site users that get their world news from their site hasn't changed from 49% since 2013

b) For confidence interval,

Confidence Interval for the population proportion is basically an interval of range of values where the true population proportion can be found with a certain level of confidence.

Mathematically,

Confidence Interval = (Sample proportion) ± (Margin of error)

Sample proportion = 0.495

Margin of Error is the width of the confidence interval about the mean.

It is given mathematically as,

Margin of Error = (Critical value) × (standard Error)

Critical value at 95% confidence interval for sample size of 3634 is obtained from the z-tables.

Critical value = 1.960

standard Error has already been calculated in (a),

σₓ = 0.008294

95% Confidence Interval = (Sample proportion) ± [(Critical value) × (standard Error)]

CI = 0.495 ± (1.96 × 0.008294)

CI = 0.495 ± 0.01625624

95% CI = (0.47874376, 0.51125624)

95% Confidence interval = (0.4787, 0.5113)

c) The result of the 95% confidence interval agrees with the result of the hypothesis testing performed in (a) because the value of p₀ lies within this confidence interval obtained.

Hope this Helps

r and t.

both are 64 degrees.