Suppose there is sufficient evidence to reject Ho: ₁ = ₂ = μ3 using a one-way ANOVA. The mean square error fromthe ANOVA is determined to be 22.8. The sample means are x₁ = 4.6, x₂ = 10.7, and x3 = 19.6, withn₁ = n₂ = n3 = 7. Use Tukey's test to determine which pairwise means are significantly different using a familywiseerror rate of α=0.01.Click here to view page 1 of the table of critical values for Tukey's Test for an alpha of 0.05.Click here to view page 2 of the table of critical values for Tukey's Test for an alpha of 0.05.Click here to view page 1 of the table of critical values for Tukey's Test for an alpha of 0.01.Click here to view page 2 of the table of critical values for Tukey's Test for an alpha of 0.01.Calculate the test statistic qo for each pairwise difference.912913923(Round to three decimal places as needed.)The critical value is.(Round to three decimal places as needed.)Based on this data,the hypothesis Ho: #₁ =1₂₂the hypothesis Ho: #₂ = 13-the hypothesis Ho: ₁ =3, and

Given: MSE=22.8x¯1=4.6x¯2=10.7x¯3=19.6α=0.01k=3 (Total number of treatments)n=7

Answered: Suppose there is sufficient evidence to…

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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One sample has n=7 and SS=35 and a second sample has n =17 and SS =45. What is the pooled variance for these two samples?
A company is comparing methods for producing pipes and wants to choose the method with the least variability. It has taken a sample of the lengths of the pipes using both methods as shown on the Table below. Conduct an F test at alpha-05 to determine if the results are significant. Ho - Variance 1 Variance 2. Method 1 4.7 3.7 3.2 3.1 3.9 4.8 3.1 5.1 4.9 5.3 2.8 4.2 Fcritical Fstatistic Method 2 3.7 4.1 3.5 5.5 4.1 4.7 4.9 3.5 3.9 3.7 4.8 5 5.3 4.4 4.6 Reject null hypothesis? (Yes, No) [Choose] [Choose] [Choose]
A sample of colored candies was obtained to determine the weights of different colors. The ANOVA table is shown below. It is known that the population distributions are approximately normal and the variances do not differ greatly. Use a 0.05 significance level to test the claim that the mean weight of different colored candies is the same. If the candy maker wants the different color populations to have the same mean weight, do these results suggest that the company has a problem requiring corrective action? Source: DF: SS: MS: Test Stat, F: Critical F: P-Value: Treatment: 6 0.072 0.012 2.4011 2.2059 0.0341 Error: 86 0.430 0.005 Total: 92 0.502 Should the null hypothesis that all the colors have the same mean weight be rejected? A. No, because the P-value is greater than the significance level. B. Yes, because the P-value is greater than the significance…
Two different types of catalysts are used in a chemical process. We want to conduct a hypothesis test to see if there is a difference in the yields at 5% significance level. Assume both populations are normaly distributed. To this end, we collect a simple random sample from each process. Assume equal variances. = = n-11, n=13,x,-90, x 91.4, 8, 4, 8=4.5
In the past, the time for Jeff's journey to work had mean 45.7 minutes and standard deviation 5.6 minutes. This year he is trying a new route. In order to test whether the new route has reduced his journey time, Jeff finds the mean time for a random sample of 30 journeys using the new route. He camies out a hypothesis test at the 2.5% significance level. Jeff assumes that, for the new route, the joumey time has a nomal distribution with standard deviation 5.6 minutes. (a) State appropriate null and alternative hypotheses for the test. (b) Determine the rejection region for the test.
Using traditional methods it takes 8.2 hours to receive a basic flying license. A new license training method using Computer Aided Instruction (CAI) has been proposed. A researcher used the technique on 26 students and observed that they had a mean of 8.0 hours with a variance of 2.89. Is there evidence at the 0.1 level that the technique reduces the training time? Assume the population distribution is approximately normal. Step 4 of 5: Determine the decision rule for rejecting the null hypothesis. Round your answer to three decimal places.
Suppose a recent study of Trucks and SUVs gave an average fuel efficiency of 19.2 mpg for Trucks and 21.4 mpg for SUVs. The sample standard deviation for Trucks was 3.2 mpg and 2.8 mpg for SUVs. These results were based on samples of 35 Trucks and 39 SUVs. At the 1% significance level, can we conclude that, on average, SUVs have better fuel efficiency than Trucks? Clearly state the null and alternative hypotheses, use the critical value approach (draw a detailed picture), and write out a detailed conclusion. Organize: Hypotheses: Detailed picture: Detailed conclusion:

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