The following are the hypotheses for a test of the claim that college graduation statue and cola preference are independent.Ho: College graduation status and cola preference are independent.H,: College graduation status and cola preference are dependent.If the test statistic is y2 = 0.579 and the critical value is y2 = 5.991, what is your conclusion about the null hypothesis and about the%3Dclaim?...O A. Fail to reject the null hypothesis. There is sufficient evidence to warrant rejection of the claim that college graduation statusand cola preference are independent.O B. Reject the null hypothesis. There is sufficient evidence to warrant rejection of the claim that college graduation status and colapreference are independent.O C. Reject the null hypothesis. There is not sufficient evidence to warrant rejection of the claim that college graduation status andcola preference are independent.O D. Fail to reject the null hypothesis. There is not sufficient evidence to warrant rejection of the claim that college graduation statusand cola preference are independent.

We have given that The test statistic is χ2 = 0.579 The critical value is χ2 = 5.991

Answered: If the test statistic is y2 = 0.579 and…

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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Test the claim that the mean GPA of night students is larger than 3.1 at the 0.10 significance level.The null and alternative hypothesis would be: H0:p=0.775H0:p=0.775H1:p≠0.775H1:p≠0.775 H0:p=0.775H0:p=0.775H1:p>0.775H1:p>0.775 H0:μ=3.1H0:μ=3.1H1:μ>3.1H1:μ>3.1 H0:μ=3.1H0:μ=3.1H1:μ<3.1H1:μ<3.1 H0:p=0.775H0:p=0.775H1:p<0.775H1:p<0.775 H0:μ=3.1H0:μ=3.1H1:μ≠3.1H1:μ≠3.1 The test is: right-tailed two-tailed left-tailed Based on a sample of 75 people, the sample mean GPA was 3.14 with a standard deviation of 0.03The test statistic is: (to 2 decimals)The p-value is: (to 2 decimals)Based on this we fail to reject the null hypothesis. reject the null hypothesis.
Test the claim that the proportion of men who own cats is significantly different than 70% at the 0.1 significance level. The null and alternative hypothesis would be: Но: р 3 0.7 Но:д — 0.7 Но:д %3D 0.7 Но:р — 0.7 Но:р — 0.7 Но: д — 0.7 H:р +0.7 H:n # 0.7 Hi:р > 0.7 H:p > 0.7 H:р < 0.7 H:р < 0.7 The test is: two-tailed left-tailed right-tailed Based on a sample of 55 people, 75% owned cats The test statistic i: (to 2 decimals) The positive critical value is: (to 2 decimals) Based on this we: O Fail to reject the null hypothesis O Reject the null hypothesis
Test the claim that the proportion of men who own cats is significantly different than 20% at the 0.02 significance level. The null and alternative hypothesis would be: Ho:µ = 0.2 Ho:H = 0.2 Ho:p= 0.2 Ho:µ = 0.2 Ho:p= 0.2 Ho:p= 0.2 H1: µ > 0.2 H1:470.2 H1:p7 0.2 H1:µ 0.2 The test is: left-tailed right-tailed two-tailed Based on a sample of 4 people, 25% owned cats The test statistic is: (to 2 decimals) The positive critical value is: (to 2 decimals) Based on this we: O Reject the null hypothesis O Fail to reject the null hypothesis Check Answer
Test the claim that the mean GPA of night students is significantly different than 2.7 at the 0.04 significance level.The null and alternative hypotheses would be: H0:μ=2.7H0:μ=2.7H1:μ≠2.7H1:μ≠2.7 H0:p=0.675H0:p=0.675H1:p>0.675H1:p>0.675 H0:p=0.675H0:p=0.675H1:p<0.675H1:p<0.675 H0:μ=2.7H0:μ=2.7H1:μ>2.7H1:μ>2.7 H0:p=0.675H0:p=0.675H1:p≠0.675H1:p≠0.675 H0:μ=2.7H0:μ=2.7H1:μ<2.7H1:μ<2.7 The test is: left-tailed two-tailed right-tailed Based on a sample of 30 people, the sample mean GPA was 2.66 with a standard deviation of 0.16The p-value is: (to 3 decimals)The positive significance level is: (to 2 decimals)Based on this we: Reject the null hypothesis Fail to reject the null hypothesis
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I
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Looking at youth depression scores from a general sample of random children in the US, we want to determine if our sample is significantly different than the national average of scores of children with depression in the US. The national average of scores on a measure of depression is 5 points. We have no reason to assume it will be higher or lower than the US national average. The sample: 11 9 12 3 3 4 13 1. What is the null and alternative hypotheses? Do not specify any directionality. 2. What did you discover (using 95% Cl)? Provide an interpretation.
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Test the claim that the proportion of men who own cats is significantly different than the proportion of women who own cats at the 0.02 significance level. The null and alternative hypothesis would be: Но: HF Ho: UM = µF Ho:PM — pF Ho:рм PF Ho: им — MF Ho:; PF Нi: рм > иF Hi: рм # pF Hi:рм рF Hi: рм < pF Hi:рм # pғ The test is: right-tailed two-tailed left-tailed Based on a sample of 80 men, 25% owned cats Based on a sample of 40 women, 50% owned cats The test statistic is: (to 2 decimals) The p-value is: (to 2 decimals) Based on this we: O Reject the null hypothesis O Fail to reject the null hypothesis
A new treatment for a certain disease is claimed to be effective in more than 80 % of cases . To test this claim the treatment was given to 100 randomly selected patients . The treatment was effective in 85 out of the 100 patients . What is the conclusion at a 5 % level of significance ? A. Reject the null hypothesis .There is sufficient evidence to claim that the treatment is effective in more than 80 % of cases B. Reject the null hypothesis .There is not sufficient evidence to claim that the treatment is effective in more than 80 % of cases C. Fail to reject the null hypothesis. There is sufficient evidence to claim that the treatment is effective in more than 80 % of cases D. Fail to reject the null hypothesis. There is not sufficient evidence to claim that the treatment is effective in more than 80 % of cases

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