The following problems are copied from the chapter 16 exercises from Introduction to Modern Statistics First Edition by Mine Çetinkaya-Rundel and Johanna Hardin (https://openintro-ims.netlify.app/inference-one-prop.html)

Married at 25. A study suggests that the 25% of 25 year olds have gotten married. You believe that this is incorrect and decide to collect your own sample for a hypothesis test. From a random sample of 25 year olds in census data with size 776, you find that 24% of them are married. A friend of yours offers to help you with setting up the hypothesis test and comes up with the following hypotheses. Indicate any errors you see.

\(H_0 \hat p = 0.24\)

\(H_a \hat p \ne 0.24\)

ANSWER: The hypothesis should be for the population as a whole, not for the sample population.

  1. Legalization of marijuana, bootstrap test. The 2018 General Social Survey asked a random sample of 1,563 US adults: “Do you think the use of marijuana should be made legal, or not?” 60% of the respondents said it should be made legal. (NORC 2018) Consider a scenario where, in order to become legal, 55% (or more) of voters must approve.
  1. What are the null and alternative hypotheses for evaluating whether these data provide convincing evidence that, if voted on, marijuana would be legalized in the US.

ANSWER: \(H_0 p = 0.55\)

\(H_a p \ < .55\)

  1. A parametric bootstrap simulation (with 1,000 bootstrap samples) was run and the resulting null distribution is displayed in the histogram below. Find the p-value using this distribution and conclude the hypothesis test in the context of the problem. See the book for the histrogram.

ANSWER: \(\alpha= .05\)

\(z=(\hat p - p_0)/ \sqrt (p_0(1-p_0)/n)\)

\(z=(.60 - .55)/ \sqrt (.55(1-.55)/1563)= 3.97\)

p hat= .6 . Given the alpha value of .05 and the p value of 1 we fail to reject the null hypothesis as the p value of 1 is greater than the alpha value of .05. Therefore we can state with confidence that weed would be legalized as the null hypothesis states that 55% of people would legalize weed.

  1. National Health Plan, parametric bootstrap. A Kaiser Family Foundation poll for a random sample of US adults in 2019 found that 79% of Democrats, 55% of Independents, and 24% of Republicans supported a generic National Health Plan. There were 347 Democrats, 298 Republicans, and 617 Independents surveyed. (K. F. Foundation 2019)

A political pundit on TV claims that a majority of Independents support a National Health Plan. Do these data provide strong evidence to support this type of statement? One approach to assessing the question of whether a majority of Independents support a National Health Plan is to simulate 1,000 parametric bootstrap samples with p = 0.5 as the proportion of Independents in support.

See the book for the histogram displaying the parameterized bootstrap distribution.

  1. The histogram above displays 1000 values of what?

ANSWER: Parameterized bootstrap sample proportions of those who support a national health plan.

  1. Is the observed proportion of Independents consistent with the parametric bootstrap proportions under the setting where p = 0.5?

ANSWER: Yes as .55 falls into the distribution chart

  1. In order to test the claim that a majority of Independents support a National Health Plan. what are the null and alternative hypotheses?

ANSWER: \(H_0 p = 0.50\)

\(H_a p \ < .50\)

  1. Using the parametric bootstrap distribution, find the p-value and conclude the hypothesis test in the context of the problem.

ANSWER: \(\alpha= .05\)

\(z=(\hat p - p_0)/ \sqrt (p_0(1-p_0)/n)\)

\(z=(.55 - .5)/ \sqrt (.5(1-.5)/1000)= 3.16\)

phat=.5 . Independence is guaranteed in the problem and both the answers have more than 10 so we can use the central limit theorem. Given the alpha value of .05 and the p value of .999 we fail to reject the Null hypothesis as the p value is above the alpha value and therefore we can conclude that a majority of Independents support a National Health Plan.

Date and time completed: Thu Oct 13 22:01:29 2022