Table entries in lightface type are critical values for the .05 level of significance. Boldface type values are for the .01 level of significance.At the significance level a = 0.01, the main effect due to factor A isand the interaction effect between the two factors isthe main effect due to factor B is You perform an ANOVA to test that there are no main effects of factor A, no main effects of factor B, and no interaction between factors A and B.Some of the results are presented in the following ANOVA table.SourceBetween treatmentsFactor AFactor BA X B interactionWithin treatmentsTotalANOVA TableSS6.56670.066712.1667df529MS4.03331.2334F17.295.29Work through the following steps to complete the preceding ANOVA table.1. The main effect for factor A evaluates the mean differences between the levels of factor A. The main effect for factor B evaluates the meandifferences between the levels of factor B. Select the correct values for the sums of squares for factors A and B in the ANOVA table.2. Select the correct value for the within-treatments sum of squares in the ANOVA table.3. Select the correct degrees of freedom for all the sums of squares in the ANOVA table.4. Select the correct values for the mean square due to A X B interaction, the within treatments mean square, and the F-ratio for the A X B interaction.5. Use the results from the completed ANOVA table and the F distribution table (click on the following dropdown menu to access the table) to makethe following conclusions.

ANOVA test is used to test whether there are no main effects of factor A, no main effects of factor…

Answered: Some of the results are presented in…

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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questiun = Question Heip Course Note Packet Chapter 9 Page 7,9 and 10 The data table below contains the amounts that a sample of nine customers spent for lunch (in dollars) at a fast-food restaurant. Complete parts (a) through (d). Click here to view page 1 of the table of critical values of t. Click here to view page 2 of the table of critical values of t. 4.20 5.07 5.93 6.49 7.31 7.63 8.44 8.59 9.89 a a. At the u.UT Tevei or signinIcance, is tnere evidence tnar tne mean amount spent Tor iuncn is aitrerent trom 56.50? State the null and alternative hypotheses. (Type integers or decimals. Do not include the S symbol in your answer.) Identify the critical value(s). The critical value(s) is(are) (Round to four decimal places as needed. Use a comma to separate answers as needed.) Determine the test statistic The test statistic is (Round to four decimal places as needed.) State the conclusion. Ho. There is v evidence conclude that the mean amount spent for lunch is different from $6.50. b.…
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what is the value in cell D? Round to 2 decimal places. How many total observations were there in the study? Hint: use df(total) to figure out the number of total observations. The following computer output is for an analysis of variance in which yields (bu/acre) of different varieties of oats were compared. Answer the following questions based on this ANOVA table. Source df SS MS F ratio Prob Between groups 2 76.90 38.45 D 0.63 Within groups 11 859.81 78.16 Total 13 936.17
Refer to the ANOVA table for this regression. Source SS d.f. MS Regression 1,164,578 5 232,916 Residual 1,500,689 45 33,349 Total 2,665,267 50 (a) State the degrees of freedom for the F test for overall significance.
Consider the following ANOVA table below. Some parts are blank intentionally. The number of observations is 249. Source Model (Regression) Residual (Error) Total The MSE is: SS 1,332 DF MS 921 Round your answer to two decimal places.
A company manufactures printers and fax machines at plants located in Atlanta, Dallas, and Seattle. To measure how much employees at these plants know about quality management, a random sample of 6 employees was selected from each plant and the employees selected were given a quality awareness examination. The examination scores for these 18 employees are shown in the following table. The sample means, sample variances, and sample standard deviations for each group are also provided. Managers want to use these data to test the hypothesis that the mean examination score is the same for all three plants. Sample mean Sample variance Sample standard deviation Plant 1 Atlanta 84 76 81 76 70 81 78 25.2 5.02 Plant 2 Plant 3 Dallas Seattle 72 74 72 74 70 88 75 42.8 6.54 60 64 62 68 76 66 66 32.0 5.66

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