Math  /  Data & Statistics

QuestionA 2018 poll of 2236 randomly selected U.S. adults found that 36.58%36.58 \% planned to watch at least a "fair amount" of a particular sporting event in 2018. In 2014, 46\% of U.S. adults reported planning to watch at least a "fair amount." a. Does this sample give evidence that the proportion of U.S. adults who planned to watch the 2018 sporting event was less than the proportion who planned to do so in 2014? Use a 0.05 significance level. b. After conducting the hypothesis test, a further question one might ask is what proportion of all U.S. adults planned to watch at least a "fair amount" of the sporting event in 2018. Use the sample data to construct a 90%90 \% confidence interval for the population proportion. How does your confidence interval support your hypothesis test conclusion? a. State the null and alternative hypotheses. Let p be the proportion of U.S. adults that planned to watch at least a "fair amount" of the sporting event. H0:p=0.46Ha:p<0.46\begin{array}{l} H_{0}: p=0.46 \\ H_{a}: p<0.46 \end{array} (Type integers or decimals. Do not round.) Compute the z-test statistic. z=8.97z=-8.97 (Round to two decimal places as needed.) Compute the p-value. p-value =\mathrm{p} \text {-value }= \square (Round to three decimal places as needed.)

Studdy Solution
a. p-value = 0.000.
We reject the null hypothesis.
There is sufficient evidence that the proportion of U.S. adults who planned to watch the sporting event in 2018 was less than in 2014. b. The 90% confidence interval is (0.3492, 0.3824).
This supports the hypothesis test because the entire interval is below 0.46, suggesting the true 2018 proportion is lower than the 2014 proportion.

View Full Solution - Free
Was this helpful?

Studdy solves anything!

banner

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

ParentsInfluencer programContactPolicyTerms
TwitterInstagramFacebookTikTokDiscord