CHALLENGEACTIVITY402562.2600368.qx3zqy78.2.3: Hypothesis testing for two population proportions (one-tailed z-test).Jump to level 1A university department chair believes that the afternoon section of a large survey class has a consistently lowerproportion of students that pass the class compared to the morning section. The chair checks a random sampleof student records from each section. The results of the data collected are shown below.Morning sectionAfternoon sectionSuccessesObservationsp-hat_1346 Successes400 Observations0.865 p-hat_2Confidence level95%-0.5548p-value0.29123262P= Ex 1.234Morning section samples: ₁Ex: 9Afternoon section samples: 1₂Sample proportion for morning section samples: p1Sample proportion for afternoon section samples: p₂–O500LEE2 №7 || |||Ex: 1.234DDDO

Answered: Image /qna-images/answer/ee5ffe14-5750-4f77-9135-e689368968ca.jpg

Answered: ALLENGE IVITY 62.2600368.qx3zqy7 8.2.3:…

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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Please help me with this question.
. The term sample usually refers to a sample that ___ - Consists of people with chemical dependency problems - Uses the same group of individuals with a before/after measurement - Requires a dependent variable for hypothesis testing - Is randomly selected from two dependent populations
Violation of which assumption below for the two-factor ANVOA is not a cause for concern with large sample sizes? a. The populations from which the samples are selected must have equal variances. b. The populations from which the samples are selected must be normal. c. A violation of any assumption below would be a concern, even with large sample sizes. d. The observations within each sample must be independent.
Read the following prompt and complete question 2
Need sub part B answered only
A sociologist believed that the average family size in Saskatchewan was larger than the average family size in British Columbia. Random samples from the two provinces yielded the following results: s? Saskatchewan (Pop 1) British Columbia (Pop 2) 3.45 1.34 60 2.89 1.32 52 Test at a = 0.05 whether there is sufficient evidence to support the claim. As well, calculate the p-value for this test.
Class 1 Final Sample Size: 30 Sample Variance: 80.55 Sample Mean: 44.2917 Class 2 Final Sample Size:30 Sample Variance: 55.29 Sample Mean: 36.4833 The assignment says to use the Z test for means and assume the hypothesized difference is 0. 1. Is there a significant difference in the means of the two finals at the .01 level of significance? 2. Why or why not?
In a random sample of males, it was found that 27 write with their left hands and 210 do not. In a random sample of females, it was found that 55 write with their left hands and 454 do not. Use a 0.05 significance level to test the claim that the rate of left-handedness among males is less than that among females. Complete parts (a) through (c) below. Question content area bottom Part 1 a. Test the claim using a hypothesis test. Consider the first sample to be the sample of males and the second sample to be the sample of females. What are the null and alternative hypotheses for the hypothesis test? A. H0: p1=p2 H1: p1>p2 B. H0: p1≥p2 H1: p1≠p2 C. H0: p1≠p2 H1: p1=p2 D. H0: p1=p2 H1: p1≠p2 E. H0: p1=p2 H1: p1<p2 F. H0: p1≤p2 H1: p1≠p2 Part 2 Identify the test statistic. z=enter your response here (Round to two decimal places as needed.) Part 3 Identify the P-value.…
work shown without excel would be so appreciated!
At a 0.05, test to see if the population variances from which the following samples were drawn are equal. The test statistic is Group 1 n₁ = 21 Group 2 n₂ = 19 $1 = 20 S2 = 18 (A) 1.56 (B) 1.23 (C) 1.26 (D) 1.11
with work
I In 2000, 6.1% of people used marijuana. This year, a company wishes to use their employment drug screening to test a claim. They take a simple random sample of 2848 job applicants and find that 145 individuals fail the drug test for marijuana. They want to test the claim that the proportion of the population failing the test is lower than 6.1%. Use .10 for the significance level. Round to three decimal places where appropriate. Hypotheses: Ho:p=6.1%Ho:p=6.1% H1:p<6.1%H1:p<6.1% Test Statistic: z = Critical Value: z = p-value: Conclusion About the Null: Reject the null hypothesis Fail to reject the null hypothesis Conclusion About the Claim: There is sufficient evidence to support the claim that the proportion of the population failing the test is lower than 6.1% There is NOT sufficient evidence to support the claim that the proportion of the population failing the test is lower than 6.1% There is sufficient evidence to warrant rejection of the claim that the…

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