Test the claim that the proportion of men who own cats is significantly different than the proportionof women who own cats at the 0.01 significance level.The null and alternative hypothesis would be:Ho: PMµF Ho:PMPF Ho:UMpF Ho: им — pF Ho:рм — pғ Но:рмНі: рм # pr Hi:рм > pF Hі: рм > иF Hі:рм < pr Hi:рм # pr Hі: рм < pF= µFThe test is:left-tailed right-tailed two-tailedBased on a sample of 80 men, 35% owned catsBased on a sample of 80 women, 50% owned catsThe test statistic is:(to 2 decimals)The p-value is:|(to 2 decimals)Based on this we:O Reject the null hypothesisO Fail to reject the null hypothesis

Given,nM=80 , pM^=0.35nF=80 , pF^=0.50α=0.01

Answered: Test the claim that the proportion of…

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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Suppose the nutrition information on the package of Matilde's favorite brand of chips states that a serving size of 28 g equals about 12 chips. That means, on average, each chip should weigh 2.33 g. Matilde decides to test the accuracy of this serving size information. She plans to conduct a one-sample t-test with a significance level of a = 0.05 to test the null hypothesis, Họ: H = 2.33, against the alternative hypothesis, H1: µ ± 2.33, where u is the average weight of a chip. Matilde selects a random sample of unbroken chips to weigh. She does not know the population standard deviation nor the distribution of chip weights, but she has confirmed that her sample does not contain any outliers. The summary statistics for her test are shown in the following table. Sample size Sample mean Sample standard deviation Test statistic Probability value n P-value 60 2.37 0.12 2.367 0.021 Based on these results, complete the following sentences to state the decision and conclusion of the test.…
A test of H₁: μ=7 versus H₁: u<7 is performed using a significance level of a=0.01. The P-value is 0.20. Part: 0/3 Part 1 of 3 (a) Is H rejected? Since P ≤ Part: 1 / 3 Part 2 of 3 (b) If the true value of u is 1, is the result a Type I error, a Type II error, or a correct decision? a, we do not reject Ho at the a= 0.01 level. The result is a Type I error Part: 2/3 Part 3 of 3 X 3 (c) If the true value of u is 7, is the result a Type I error, a Type II error, or a correct decision? The result is a [Choose one) ▾ Español
Test the claim that the proportion of men who own cats is smaller than the proportion of women who own cats at the .05 significance level.The null and alternative hypothesis would be: H0:μM=μFH1:μM<μF H0:pM=pFH1:pM<pF H0:pM=pFH1:pM>pF H0:pM=pFH1:pM≠pF H0:μM=μFH1:μM≠μF H0:μM=μFH1:μM>μF The test is: right-tailed left-tailed two-tailed Based on a sample of 20 men, 40% owned catsBased on a sample of 60 women, 55% 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
Test the claim that the proportion of men who own cats is smaller than the proportion of women who own cats at the .10 significance level.The null and alternative hypothesis would be: 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 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 The test is: left-tailed two-tailed right-tailed Based on a sample of 40 men, 35% owned catsBased on a sample of 20 women, 45% 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
Test the claim that the proportion of men who own cats is smaller than 30% at the .10 significance level. The null and alternative hypothesis would be: Ho: p=0.3 H₁:μ0.3 O Ho: p= 0.3 H₁:p> 0.3 Ο Η : μ = 0.3 Η : μ 0.3 ○ Ho: p= 0.3 H₁: p<0.3 Based on a sample of 65 people, 27% owned cats The test statistic is The critical value is Based on this we O Reject the null hypothesis O Fail to reject the null hypothesis (to 3 decimals) (to 2 decimals)
Test the claim that the proportion of men who own cats is smaller than the proportion of women who own cats at the .05 significance level.The null and alternative hypothesis would be: H0:μM=μFH0:μM=μFH1:μM≠μFH1:μM≠μF H0:pM=pFH0:pM=pFH1:pM>pFH1:pM>pF 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:μM=μFH0:μM=μFH1:μM>μFH1:μM>μF Incorrect The test is: left-tailed two-tailed right-tailed Incorrect Based on a sample of 20 men, 35% owned catsBased on a sample of 80 women, 40% owned catsThe test statistic is: Incorrect (to 2 decimals)The critical value is: Incorrect (to 2 decimals)Based on this we: Reject the null hypothesis Fail to reject the null hypothesis Incorrect Submit QuestionQuestion 6
Carry out the hypothesis test for the mean of a normally distributed population given the following information: Ho: µ ≥2.5 Ha: µ ≠2.5 Sample size = 30 mean = 1.8, variance =1.7, α = .05
Test the claim that the proportion of men who own cats is smaller than the proportion of women who own cats at the .005 significance level.The null and alternative hypothesis would be: 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: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 The test is: right-tailed left-tailed two-tailed Based on a sample of 60 men, 25% owned catsBased on a sample of 80 women, 50% owned catsThe test statistic is: (to 2 decimals)The 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 mean GPA of night students is significantly different than the mean GPA of day students at the 0.02 significance level. (a) The null and alternative hypothesis would be: H0:πN=πD H1:πN<πD H0:μN=μD H1:μN<μD H0:μN=μD H1:μN>μD H0:πN=πD H1:πN>πD H0:πN=πD H1:πN≠πD H0:μN=μD H1:μN≠μD (b) The test is: right-tailed left-tailed two-tailed The sample consisted of 45 night students, with a sample mean GPA of 3.02 and a standard deviation of 0.08, and 45 day students, with a sample mean GPA of 3.01 and a standard deviation of 0.03. (c) The t-score is: (Round to 2 decimal places.) (d) The p-value is: (Round to 4 decimal places.) (e) Based on this we conclude: null is plausible strong support for the alternative
You wish to test the following claim (HaHa) at a significance level of α=0.05 Ho:μ1=μ2 Ha:μ1≠μ2You believe both populations are normally distributed, but you do not know the standard deviations for either. However, assume that the variances of the two populations are equal. You obtain a sample of size n1=23 with a mean of ¯x1=74.5 and a standard deviation of SD1=20.4 from the first population. You obtain a sample of size n2=11with a mean of ¯x2=89.8and a standard deviation of SD2=7.3 from the second population.What is the test statistic for this sample? (Report answer accurate to three decimal places.) test statistic = What is the p-value for this sample? (Report answer accurate to four decimal places.) p-value = The p-value is... less than (or equal to) αα greater than αα This test statistic leads to a decision to... reject the null accept the null fail to reject the null As such, the final conclusion is that... There is sufficient evidence to warrant rejection of…
In a random sample of males, it was found that 25 write with their left hands and 217 do not. In a random sample of females, it was found that 58 write with their left hands and 450 do not. Use a .05 significance level to test the claim that the rate of left-handedness among males is less than that among females. Identify the test statistic.z=?
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

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