<14:51 0←Note: -Chegg Expert Hiringhttps://experthiring.cheggindia.com/Subject TestAnswerYou are attempting question 1 out ofIn a randomized block design, a treatment has four levels with seven blocks. Use thisinformation and a = 0.01, complete the following table and write the decision about thenull hypothesis.Source of Variation SSTreatment199.48Blocks|ErrorTotalAnswer Choices:b.a. F=4.90, fail to reject the null hypothesisF-3.90, fail to reject the null hypothesisF= 2.90, fail to reject the null hypothesisF-3.90, reject the null hypothesisd.|||OAOBOC265.24306.59ODdfSubmitQ 74%OMSF< :<

Number of Treatment = 4 Number of blocks = 7 Significance level, α=0.01 Source of Variation SS df…

Answered: In a randomized block design, a…

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 smaller than 90% at the 0.01 significance level. The null and alternative hypothesis would be: Ho: µ 0.9 Ho:p 0.9 Ho:µ = 0.9 Họ:p H1:µ > 0.9 H1:p 0.9 Hị: µ < 0.9 H1:µ + 0.9 H1:p# 0.9 0.9 The test is: two-tailed left-tailed right-tailed Based on a sample of 200 people, 87% owned cats The test statistic is: (to 2 decimals) The critical value is: (to 2 decimals)
Listed below are the lead concentrations in ug/g measured in different traditional medicines. Use a 0.05 significance level to test the claim that the mean lead concentration for all such medicines is less than 13 ug/g. Assume that the sample is a simple random sample. 12.5 7.5 4 14.5 3 12 22 3 19.5 4.5 D Assuming all conditions for conducting a hypothesis test are met, what are the null and alternative hypotheses? Ο Α. H: μ= 13 μg/g H1: µ> 13 ug/g B. H0 μ 13 μg/g H4:µ#13 µg/g C. Ho: µ = 13 µg/g H1:µ13 μg/g H1:H<13 µg/g
the answer for part b was given to me by a tutor 3.79 and the system shows that is incorrect, please help where the red x is.
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: Ho: PM = PF Ho: MM = F Ho: PM = PF H₁ PM PF H₁: μM μF H₁ PM > PF O O The test is: left-tailed two-tailed O O Based on a sample of 80 men, 25% owned cats Based on a sample of 20 women, 35% owned cats The test statistic is: The p-value is: right-tailed O Based on this we: O Reject the null hypothesis O Fail to reject the null hypothesis Ho: PM = MF H₁: MM > MF (to 2 decimals) (to 2 decimals) Ho: MM = F Ho:PM = PF H₁: MM < MF H₁:PM < PF O O
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) =
Find the value of the standard score, z, and determine whether to reject the null hypothesis at a 0.01 significance level. Is the alternative hypothesis supported? Ho p= 18.5 meters, H p18.5 meters, n= 144, x = 19.1 meters, o = 1.3 meters %3D The value of the standard score is (Round to two decimal places as needed.)
Use the given information to find the P-value. Also, use a 0.05 significance level and state the conclusion about the null hypothesis (reject the null hypothesis or fail to reject the null hypothesis). The test statistic in a two-tailed test is z=1.95. A. 0.9744; fail to reject the null hypothesis B. 0.0512; fail to reject the null hypothesis C. 0.0512; reject the null hypothesis D. 0.0256; 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.
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State the Result: A hypothesis test was conducted at the alpha = 0.01 level of significance. The test resulted in a p-value of 0.044.
A claim is given. Select the corresponding null hypothesis and using a significance level of α = 0.05 and the given p-value, determine if you reject the null hypothesis or fail to reject the null hypothesis. Claim: Americans spend on average less than 8 minutes in the shower. Data: In a sample of Americans (n = 1,803), the average time spent in the shower is 7.8 minutes with a p-value of 0.0546. A. H0 :μ=8and we reject the null. B. H0 :μ=8and we fail reject the null. C. H0 :μ>8and we reject the null. D. H0 :μ>8and we fail reject the null. E. H0 :μ<8and we reject the null. F. H0 :μ<8and we fail reject the null.
For your study on the food consumption of teenage males, you randomlyselected 20 males and asked each how many 12-ounce servings of soda he drinks eachday. The results are listed below. At alpha = 0.1, is there enough evidence to support theclaim that teenage males drink fewer than five 12-ounce servings of soda per day? Noneed to check the criteria –just do the hypothesis test.3.3 4.1 3.5 2.7 3.4 3.3 4.8 3.4 5.2 3.22.5 3.8 5.0 1.9 3.9 5.3 2.9 1.8 4.2 2.2

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