In a study of cell phone usage and brain hemispheric dominance, an Internet survey was e-mailed to 6959 subjects randoreturned. Use a 0.01 significance level to test the claim that the return rate is less than 20%. Use the P-value method andIdentify the null hypothesis and alternative hypothesis.OA. Ho: p0.2H₁: p=0.2O C.. Ho: P = 0.2H₁: p=0.2O E. Ho: p<0.2H₂₁: p=0.2The test statistic is z =(Round to two decimal places as needed.)The P-value is(Round to three decimal places as needed.)Because the P-value is less than the significance level,4rejectOB. Ho: p=0.2H₁ p > 0.2OD. Ho:p>0.2H₁: p=0.2OF. Ho: p=0.2H₁ p<0.2the null hypothesis. There issufficient evide ance, an Internet survey was e-mailed to 6959 subjects randomly selected from an online group involved with ears. There were 1288 surveysthe return rate is less than 20%. Use the P-value method and use the normal distribution as an approximation to the binomial distribution.evel,کےCO B. Ho: p=0.2H₁: p>0.2OD. Ho:p>0.2H₁: p=0.2OF. Ho: p=0.2H₁: p<0.2reject the null hypothesis. There issufficient evidence to support the claim that the return rate is less than 20%.Next5:4110/05/20

Solution-: Here, we use proportion test (Z-test) Given: x=1288,n=6959,P0=0.20,α=0.01 We want to (a)…

Answered: In a study of cell phone usage 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 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 80% at the 0.01 significance level.The null and alternative hypothesis would be: H0:μ=0.8H0:μ=0.8H1:μ>0.8H1:μ>0.8 H0:μ=0.8H0:μ=0.8H1:μ≠0.8H1:μ≠0.8 H0:p=0.8H0:p=0.8H1:p<0.8H1:p<0.8 H0:μ=0.8H0:μ=0.8H1:μ<0.8H1:μ<0.8 H0:p=0.8H0:p=0.8H1:p>0.8H1:p>0.8 H0:p=0.8H0:p=0.8H1:p≠0.8H1:p≠0.8 The test is: two-tailed left-tailed right-tailed Based on a sample of 35 people, 76% owned catsThe test statistic is: (to 2 decimals)The positive critical value is: (to 2 decimals)Based on this we: Reject the null hypothesis Fail to 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 proportion of men who own cats is significantly different than the proportion of women who own cats at the 0.05 significance level. The null and alternative hypothesis would be: H0:pM=pFH0:pM=pFH1:pM≠pFH1:pM≠pF H0:pM=pFH0:pM=pFH1:pM>pFH1:pM>pF H0:μM=μFH0:μM=μFH1:μM<μFH1:μM<μF H0:pM=pFH0:pM=pFH1:pM<pFH1:pM<pF H0:μM=μFH0:μM=μFH1:μM>μFH1:μM>μF H0:μM=μFH0:μM=μFH1:μM≠μFH1:μM≠μF The test is: left-tailed right-tailed two-tailed Based on a sample of 40 men, 25% owned catsBased on a sample of 40 women, 30% owned cats The test statistic is: (to 2 decimals) The positive critical 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 people who own cats is significantly different than 80% at the 0.1 significance level.The null and alternative hypothesis would be: H0:μ=0.8H0:μ=0.8H1:μ≠0.8H1:μ≠0.8 H0:p=0.8H0:p=0.8H1:p≠0.8H1:p≠0.8 H0:p≤0.8H0:p≤0.8H1:p>0.8H1:p>0.8 H0:p≥0.8H0:p≥0.8H1:p<0.8H1:p<0.8 H0:μ≤0.8H0:μ≤0.8H1:μ>0.8H1:μ>0.8 H0:μ≥0.8H0:μ≥0.8H1:μ<0.8H1:μ<0.8 The test is: right-tailed left-tailed two-tailed Based on a sample of 200 people, 87% owned catsThe p-value is: (to 2 decimals)Based on this we: Reject the null hypothesis Fail to reject the null hypothesis
Test the claim that the proportion of people who own cats is smaller than 60% at the 0.025 significance level. The null and alternative hypothesis would be: Ho: p=0.6 Ho:μ ≥ 0.6 Ho: p 0.6 The test is: left-tailed right-tailed two-tailed Based on a sample of 600 people, 53% owned cats The test statistic is: The p-value is: Based on this we: Ho:p>0.6 Ho: 0.6 Ho: μ 0.6 Reject the null hypothesis O Fail to reject the null hypothesis (to 2 decimals) (to 2 decimals) =
Test the claim that the proportion of people who own cats is significantly different than 20% at the 0.1 significance level. The null and alternative hypothesis would be: Ho:p 2 0.2 Ho:p = 0.2 Ho: u 0.2 Ho: u = 0.2 Ho:p 0.2 H1 : p 0.2 The test is: left-tailed two-tailed right-tailed Based on a sample of 700 people, 11% owned cats The p-value is: 0.1 x (to 2 decimals) Based on this we: O Fail to reject the null hypothesis O Reject the null hypothesis
Test the claim that the mean GPA of night students is significantly different than 2.3 at the 0.05 significance level.The null and alternative hypothesis would be: H0:μ=2.3H0:μ=2.3H1:μ>2.3H1:μ>2.3 H0:p=0.575H0:p=0.575H1:p>0.575H1:p>0.575 H0:μ=2.3H0:μ=2.3H1:μ<2.3H1:μ<2.3 H0:p=0.575H0:p=0.575H1:p≠0.575H1:p≠0.575 H0:μ=2.3H0:μ=2.3H1:μ≠2.3H1:μ≠2.3 H0:p=0.575H0:p=0.575H1:p<0.575H1:p<0.575 The test is: two-tailed left-tailed right-tailed Based on a sample of 80 people, the sample mean GPA was 2.27 with a standard deviation of 0.02The test statistic is: (to 2 decimals)The positive critical value is: (to 2 decimals)Based on this we: Fail to reject the null hypothesis Reject the null hypothesis
hopes that the results of a trial period will enable her to conclude with a level of significance of 0.05 that the new procedure increases the average profit of installing a carpet. What will be the consequences if she makes a Type I error on her hypothesis test? A Type I error would mean that Carpetland implements a new procedure that actually decreases profits. A Type I error would mean that Carpetland implements a new procedure that actually increases profits. A Type 1 error would mean that Carpetland does not implement a new procedure that increases profits. A Type I error would mean that Carpetland does not implement a new procedure that decreases profits.. A Type I error would mean that Carpetland implements a new procedure that actually does not increase profits. A Type I error would mean that Carpetland implements a new procedure that actually does not decrease profits.
Test the claim that the proportion of people who own cats is smaller than 30% at the 0.025 significance level. The null and alternative hypothesis would be: Ho: p=0.3 Ho: p = 0.3 Ho: p= 0.3 H₁: p 0.3 The p-value is: Ho: p = 0.3 Ho: p = 0.3 H₁: p0.3 H₁:p> 0.3 (to 2 decimals) (to 2 decimals)
Test the claim that the proportion of men who own cats is significantly different than 80% at the 0.1 significance level.The null and alternative hypothesis would be: H0:p=0.8H0:p=0.8H1:p≠0.8H1:p≠0.8 H0:p=0.8H0:p=0.8H1:p>0.8H1:p>0.8 H0:μ=0.8H0:μ=0.8H1:μ<0.8H1:μ<0.8 H0:μ=0.8H0:μ=0.8H1:μ>0.8H1:μ>0.8 H0:p=0.8H0:p=0.8H1:p<0.8H1:p<0.8 H0:μ=0.8H0:μ=0.8H1:μ≠0.8H1:μ≠0.8 The test is: two-tailed right-tailed left-tailed Based on a sample of 100 people, 87% owned catsThe test statistic is: (to 2 decimals)The positive critical value is: (to 2 decimals)Based on this we: Fail to reject the null hypothesis Reject the null hypothesis
Test the claim that the mean GPA of night students is significantly different than 3.3 at the 0.2 significance level. The null and alternative hypothesis would be: Ho:p0.825 H p 3.3 H p 0.825 H p <3.3 H:p 3.3 The test is: right-tailed two-tailed left-tailed Based on a sample of 80 people, the sample mean GPA was 3.33 with a standard deviation of 0.04 The p-value is: (to 2 decimals) Based on this we: Reject the null hypothesis Fail to reject the null hypothesis

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