### Sleepwalking Study: Proportion Hypothesis TestAccording to a reputable magazine, 35% of people in a certain large country have sleepwalked at least once in their lives. Suppose a random sample of 200 people showed that 48 reported sleepwalking. The null and alternative hypotheses for a test that would test whether the proportion of people who have sleepwalked is 0.35 are \( H_0: p = 0.35 \) and \( H_A: p \neq 0.35 \), respectively. The conditions for such a test are satisfied. Use the technology output provided to determine the test statistic and p-value for the test, and provide a conclusion. Use a significance level of 0.05.#### Hypothesis Test Details:1. **Determine the Test Statistic**: \( z = \) (Round to two decimal places as needed)2. **Determine the p-value**: \( \text{p-value} = \) (Round to three decimal places as needed)3. **Conclusion**: What is the conclusion to this test? [Dropdown] \( H_0 \) [Dropdown] sufficient evidence to conclude that the proportion of people who have sleepwalked [Dropdown] 0.35.#### Technology OutputThe technology output provides the following results:- **Hypothesis Test Results**: - \( p \): Proportion of successes - \( H_0: p = 0.35 \) - \( H_A: p \neq 0.35 \)- **Data**: - Proportion Count: 48 - Total Sample: 200 - Sample Proportion: 0.24 - Standard Error (Std. Err.): 0.03372684 - Z-Statistic (Z-Stat.): -3.2614970 - P-value: 0.0011Use this information to complete the hypothesis test.

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Description: ### Sleepwalking Study: Proportion Hypothesis TestAccording to a reputable magazine, 35% of people in a certain large country have sleepwalked at least once in their lives. Suppose a random sample of 200 people showed that 48 reported sleepwalking.

According to a reputable magazine. 35% of people in a certain large country have sleepwalked at least once in their lives Suppose a random sample of 200 people showed that 48 reported sleepwalking. Let p be the proportion of people who have sleepwalked. The null and alternative hypotheses for a test that would test whether the proportion of people who have sleepwalked is 0.35 are H0: p = 0 35 and Ha: p ≠0.35 respectively. The conditions for such a test are satisfied.We are given the technology output provided to determine the test statistic and p-value for the test. Moreover, a significance level, α=0.05. The technology output is as follows: Hypothesis test result: p: Proportion of successes H0 : p=0.35H1 : p≠0.35 Proportion Count Total Sample Prop. Std. Err. Z-Stat P-value p 48 200 0.24 0.03372684 -3.261497 0.0011 This gives the value of test statistic is the Z-stat in the output table, that is, Z=-3.2614970≈-3.26. The p-value is is the P-value in the output table, that is, p-value = 0.0011≈0.001. The significance level (α) is, 0.05. The decision rule based on p-value is given by: "If p-value ≤ α then we reject null hypothesis otherwise we fail to reject null hypothesis." Here we get p-value = 0.001 < α = 0.05, hence we reject null hypothesis. Conclusion: There is sufficient evidence to support the claim that the proportion of people who have sleepwalked is different from 0.35.

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Statistics

According to a reputable magazine, 35% of people in a certain large country have sleepwalked at least once in their lives. Suppose a random sample of 200 people showed that 48 reported sleepwalking The null and alternative hypotheses for a test that would test whether the proportion of people who have sleepwalked is 0.35 are Ho p=0 35 and H, p# 0.35, respectively. The conditions for such a test are satisfied Use the technology output provided to determine the test statistic and p-value for the test, and provide a conclusion. Use a significance level of 0.05 Click the icon to view the technology output Determine the test statistic Z= (Round to two decimal places as needed) Determine the p-value p-value = (Round to three decimal places as needed.) !! What is the conclusion to this test? 0.35. Ho There sufficient evidence to conclude that the proportion of people who have sleepwalked

According to a reputable magazine, 35% of people in a certain large country have sleepwalked at least once in their lives. Suppose a random sample of 200 people showed that 48 reported sleepwalking The null and alternative hypotheses for a test that would test whether the proportion of people who have sleepwalked is 0.35 are Ho p=0 35 and H, p# 0.35, respectively. The conditions for such a test are satisfied Use the technology output provided to determine the test statistic and p-value for the test, and provide a conclusion. Use a significance level of 0.05 Click the icon to view the technology output Determine the test statistic Z= (Round to two decimal places as needed) Determine the p-value p-value = (Round to three decimal places as needed.) !! What is the conclusion to this test? 0.35. Ho There sufficient evidence to conclude that the proportion of people who have sleepwalked

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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