In a study of cell phone usage and brain hemispheric dominance, an Internet survey was e-mailed to 6968 subjects randomly selected from an online group involved with ears. There were 1288= surveys returned. Use a 0.01 significance level to test the claim that the return rate is less than 20%. Use the P-value method and use the normal distribution as an approximation to the binomialdistribution.@2WView an example Get more help.SIdentify the null hypothesis and alternative hypothesis.XA. Ho: p<0.2H₁: p=0.2OC. Ho: p*0.2H₁: p=0.2andOE. Ho: p=0.2H₁: p=0.2двThe test statistic is z=₁(Round to two decimal places as needed.)X380F3EDC$48RF%5V&TG1 N6BMacBook AirF6YI'H&7N44F7U√iOB. Ho: p=0.2H₁: p>0.2JOD. Ho:p>0.2H₁: p=0.2*8F. H₂:p=0.2H₁: p<0.2DIIIM1₂(9K(1,0)DOMoreFOO<I)OLF10PClear all.:;I{+[command option=Check answer11?1I4mF121xdelete

Given that Sample size n=6970 Favorable cases x =1285 Sample proportion p^=x/n =1288/6968 =0.1848

Answered: s returned. Use a 0.01 significance…

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 2.6 at the 0.01 significance level. The null and alternative hypothesis would be: Ho:p > 0.65 Ho:p 2.6 H1:p 0.65 H1:p+ 0.65 H1:µ > 2.6 H1:µ + 2.6 H:µ < 2.6 %3D The test is: two-tailed right-tailed left-tailed Based on a sample of 65 people, the sample mean GPA was 2.63 with a standard deviation of 0.07 The test statistic is: (to 2 decimals) The p-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 mean GPA of night students is larger than 2.8 at the .10 significance level. The null and alternative hypothesis would be: Ho: p=0.7 Ho: p = 0.7 Ho:μ = 2.8 Ho: H₁:p> 0.7 H₁:p 2.8 H₁: The test is: left-tailed two-tailed right-tailed o The test statistic is: 2.8 Ho:p = 0.7 Ho:μ = 2.8 0 <2.8 H₁:p 0.7 H₁: 2.8 Based on a sample of 40 people, the sample mean GPA was 2.83 with a standard deviation of 0.02 (to 2 decimals) The critical value is: = (to 2 decimals)
Use technology to find the P-value for the hypothesis test described below. The claim is that for a smartphone carrier's data speeds at airports, the mean is μ=13.00 Mbps. The sample size is n=28 and the test statistic is t=−1.288.
K The display provided from technology available below results from using data for a smartphone carrier's data speeds at airports to test the claim that they are from a population having a mean less than 5.00 Mbps. Conduct the hypothesis test using these results. Use a 0.05 significance level. Identify the null and alternative hypotheses, test statistic, P-value, and state the final conclusion that addresses the original claim. Click the icon to view the display from technology. D iew an example Get more help. 42 W S Assuming all conditions for conducting a hypothesis test are met, what are the null and alternative hypotheses? OA. Ho: 5.00 Mbps J D. Ho: -5.00 Mbps H₁: - Clear all : - ; X option F11 { [ ? + = 1 I Check answer F12 } ] delete retu
Use technology to find the P-value for the hypothesis test described below. The claim is that for 12 AM body temperatures, the mean is μ > 98.6 °F. The sample size is n = 3 and the test statistic is t = 1.234.
Test the claim that the mean GPA of night students is smaller than the mean GPA of day students at the 0.025 significance level. The null and alternative hypothesis would be: Ho: HN S HD Ho:pN 2 PD Ho:PN = PD Ho:UN = HD Ho: PN HD H:PN PD H1: µN Next Question hp
The test statistic in a right-tailed test is z=−1.83. Determine the P-value and decide whether, at the 5% significance level, the data provide sufficient evidence to reject the null hypothesis in favor of the alternative hypothesis. The P-value is _____________ (Round to three decimal places as needed.) This P-value _____________________ (does not provide or provides) sufficient evidence to reject the null hypothesis in favor of the alternative hypothesis because it is_____________ (greater than or less than) the significance level.
Use technology to find the P-value for the hypothesis test described below. The claim is that for the population of adult males, the mean platelet count is μ>210. The sample size is n=48 and the test statistic is t=1.798.
Use the random sample data to test the claim that 72% of students have internet access at home. Use 5% level of significance. Sample data: x = 158, n = 200 1. Identify the tail of the test. 2.Find the P-value 3.Will the null hypothesis be rejected? 4. Is the initial claim supported?
Test the claim that the proportion of people who own cats is smaller than 70% at the 0.1 significance level.The alternative hypothesis would be: p<0.7p<0.7 p≠0.7p≠0.7 μ>0.7μ>0.7 μ<0.7μ<0.7 μ≠0.7μ≠0.7 p>0.7p>0.7 The test is: left-tailed right-tailed two-tailed Based on a sample of 600 people, 66% owned catsThe 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
The test statistic in a left-tailed test is z=−0.59. Determine the P-value and decide whether, at the 10% significance level, the data provide sufficient evidence to reject the null hypothesis in favor of the alternative hypothesis. The P-value is ___________ (Round to four decimal places as needed.) This P-value ________________ (provides, does not provide) sufficient evidence to reject the null hypothesis in favor of the alternative hypothesis because it is_____________(greater than, less than) the significance level.

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