Test the claim that the proportion of men who own cats is smaller than the proportion of women who owncats at the .01 significance level.The null and alternative hypothesis would be:Ho: HMH1: HMµf Ho:µF H1: µM < µF H1:PM < PF H1:PMµF Ho:PMPF H1:HM # µF H1:PM # PFµF Ho: µMPF Ho: рм= PF Ho::HMрм —PFThe test is:left-tailed right-tailed two-tailedBased on a sample of 80 men, 35% owned catsBased on a sample of 80 women, 45% owned catsThe test statistic is:(to 2 decimals)The critical value is:(to 2 decimals)Based on this we:Reject the null hypothesisO Fail to reject the null hypothesis

Name: Test the claim that the proportion of men who own cats is smaller tha
Uploaded: 2024-03-20T07:07:00.000Z
Channel: Bartleby
Description: Test the claim that the proportion of men who own cats is smaller than the proportion of women who owncats at the .01 significance level.The null and alternative hypothesis would be:Ho: HMH1: HMµf Ho:µF H1: µM < µF H1:PM < PF H1:PMµF Ho:PMPF H1:HM #

Given: The proportion of men who own cats is given as:…

Math

Statistics

Test the claim that the proportion of men who own cats is smaller than the proportion of women who own cats at the .01 significance level. The null and alternative hypothesis would be: Ho: HM = µF Ho:µM µF Ho:PM PF Ho:PM PF Ho:µM = µF Ho:pM = PF Ні: рм > MF Hi:pм < рF Hi:рм < pr Hi:pм > pr Hi:им pp Hi:pм + pr The test is: left-tailed right-tailed two-tailed Based on a sample of 80 men, 35% owned cats Based on a sample of 80 women, 45% owned cats The test statistic is: (to 2 decimals) The 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 smaller than the proportion of women who own cats at the .01 significance level. The null and alternative hypothesis would be: Ho: HM = µF Ho:µM µF Ho:PM PF Ho:PM PF Ho:µM = µF Ho:pM = PF Ні: рм > MF Hi:pм < рF Hi:рм < pr Hi:pм > pr Hi:им pp Hi:pм + pr The test is: left-tailed right-tailed two-tailed Based on a sample of 80 men, 35% owned cats Based on a sample of 80 women, 45% owned cats The test statistic is: (to 2 decimals) The critical value is: (to 2 decimals) Based on this we: Reject the null hypothesis Fail to reject the null hypothesis

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

See similar textbooks

Similar questions

Test the claim that the proportion of men who own cats is significantly different than the proportion of women who own cats at the 0.2 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: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 The test is: two-tailed right-tailed left-tailed Based on a sample of 60 men, 30% owned catsBased on a sample of 40 women, 40% 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 Check AnswerQuestion 14
You wish to test the following daim (Ha) at a significance level of a = 0.10. H,:P1 P2 You obtain 615 successes in a sample of size n1 = 704 from the first population. You obtain 490 successes in a sample of size n2 = 581 from the second population. critical value = [three decimal accuracy] test statistic = [three decimal accuracy] The test statistic is... O in the critical region O not in the critical region This test statistic leads to a decision to... O reject the null accept the null fail to reject the null As such, the final condusion is that... O There is sufficient evidence to support that the first population proportion is greater than the second population proportion. O There is not sufficient evidence to support that the first population proportion is greater than the second population proportion.
Test the claim that the proportion of men who own cats is larger than 24% at the .025 significance level. The null and alternative hypothesis would be: Ho:p = 0.24 Ho: = 0.24 Ho: u = 0.24 Ho:p = 0.24 Ho:u = 0.24 Ho:p 0.24 H1:p 0.24 H1:µ > 0.24 H1:µ # 0.24 H:p 0.24 The test is: right-tailed two-tailed left-tailed ㅇ Based on a sample of 60 people, 16 2/3% owned cats The p-value is: (to 2 decimals) Based on this we: O Fail to reject the null hypothesis O Reject the null hypothesis Submit Question • Previous DEC 11 étv
0. A high school principal claims that the variance in the GPA of students in the graduating class is Tess than 0.39. You test this claim by sampling 31 seniors and find their GPA have a variance or 0.21. Ho:o? Ha:o2 < 0.39 (Claim) = 0.39 Is there enough evidence to reject Ho and support the principal's claim at a 0.10 level of significance? ohs A. No, the test statistic x? B. No, the test statistic x2 = 16.154 is not in the rejection region with critical value 40.256 C. Yes, the test statistic x? D. No, the test statistic x2 = 21.434 is not in the rejection region with critical value 20.599 E. Yes, the test statistic x? = 16.154 is in the rejection region with critical value 20.599 = 20.599 is not in the rejection region with critical value 21.434 %3D 20.599 is in the rejection region with critical value 16.154 %D
A hypothesis will be used to test that a population mean equals 14 against the alternative that the population mean is less than 14 with known variance o. What is the critical value for the test statistic Zo for the significance level of 0.010? Round your answer to two decimal places (e.g. 98.76). Za = i
Test the claim that the mean GPA of night students is larger than 3.3 at the .10 significance level.The null and alternative hypothesis would be: H0:μ=3.3H0:μ=3.3H1:μ>3.3H1:μ>3.3 H0:p=0.825H0:p=0.825H1:p<0.825H1:p<0.825 H0:p=0.825H0:p=0.825H1:p≠0.825H1:p≠0.825 H0:μ=3.3H0:μ=3.3H1:μ<3.3H1:μ<3.3 H0:μ=3.3H0:μ=3.3H1:μ≠3.3H1:μ≠3.3 H0:p=0.825H0:p=0.825H1:p>0.825H1:p>0.825 The test is: right-tailed two-tailed left-tailed Based on a sample of 80 people, the sample mean GPA was 3.35 with a standard deviation of 0.06The test statistic is: (to 2 decimals)The critical value is: (to 2 decimals)Based on this we: Reject the null hypothesis Fail to reject the null hypothesis Im not understanding this material.
Suppose the nutrition information on the package of Matilde's favorite brand of chips states that a serving size of 25 g equals about 12 chips. That means, on average, each chip should weigh 2.08 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 = 0.10 to test the null hypothesis, Ho: u = 2.08, against the alternative hypothesis, H1: µ # 2.08, 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 P-value 45 2.04 0.14 -2.076 0.044 Based on these results, complete the following sentences to state the decision and conclusion of the test.…
Test the claim that the proportion of men who own cats is significantly different than the proportion of women who own cats at the 0.2 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 The test is: right-tailed left-tailed two-tailed Based on a sample of 40 men, 25% owned catsBased on a sample of 20 women, 30% 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
My answer is as follows: T critical = 2.70 T value= 7.775 P< a (0.01) Therefore we reject the null. Is this correct?
ou wish to test the following claim (Ha) at a significance level of α=0.005. Ho:μ1=μ2Ha:μ1≠μ2 You believe both populations are normally distributed, but you do not know the standard deviations for either, and you have no reason to believe the variances of the two populations are equal. You obtain a sample of size n1=18 with a mean of M1=81.6 and a standard deviation of SD1=8.2 from the first population. You obtain a sample of size n2=23 with a mean of M2=88.9 and a standard deviation of SD2=5.8 from the second population. What is the p-value for this sample? (Report answer accurate to four decimal places.) p-value =
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: H0:μN=μDH0:μN=μDH1:μN≠μDH1:μN≠μD H0:pN≤pDH0:pN≤pDH1:pN>pDH1:pN>pD H0:μN≤μDH0:μN≤μDH1:μN>μDH1:μN>μD H0:pN≥pDH0:pN≥pDH1:pN<pDH1:pN<pD H0:pN=pDH0:pN=pDH1:pN≠pDH1:pN≠pD H0:μN≥μDH0:μN≥μDH1:μN<μDH1:μN<μD The test is: left-tailed two-tailed right-tailed The sample consisted of 35 night students, with a sample mean GPA of 3.34 and a standard deviation of 0.06, and 35 day students, with a sample mean GPA of 3.36 and a standard deviation of 0.03.The 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
A test of Ho: H=0 versus H₁: >0 is performed using a significance level of a=0.05. The P-value is 0.07. Part: 0/3 Part 1 of 3 (a) Is Ho rejected? Since P Part: 1/3 Part 2 of 3 Part: 2/3 ▾a, we do not reject Ho at the a= 0.05 level. (b) If the true value of μ is 0, is the result a Type I error, a Type II error, or a correct decision? The result is a correct decision Part 3 of 3 X X S (c) If the true value of His 4, is the result a Type I error, a Type II error, or a correct decision? S Español