9. Solve the ANOVA tableLevels of A=4Levels of B=4Sample size per condition (n) = 7SourceBetween Columns ABetween Rows BInteraction ABWithin CellsTotalSSdfMSF75.00210.006.002.99

Step 1: Begin with our incomplete ANOVA tableSourceSSdfMSFBetween columns A 2.99Between Rows B…

Answered: 9. Solve the ANOVA table Levels of A=4…

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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A paint company has large outlet stores in two cities. Ten samples of the weekly sales (in tons) for the two cities are as follows. City A: n = 10, X10 = 148.5, S1 = = 4.3. City B: n = 10, Y 10 = 144.2, S2 = 3.8. At a = 0.05 level of significance, test if the weekly sales are different. The following ¨answers" have been proposed. (a) The null hypothesis is that the weekly sales are the same and the alternative is that the they are not the same. By using an F test, we reject the null hypothesis. (b) The null hypothesis is that the weekly sales are the same and the alternative is that the they are not the same. By using an F test, we do not reject the null hypothesis. (c) The null hypothesis is that the weekly sales are the same and the alternative is that the they are not the same. By using a i test, we do not reject the null hypothesis. (d) The null hypothesis is that the weekly sales are the same and the alternative is that the they are not the same. By using a i test, we reject the…
The manager of a political party wants to find out the association between the campaign spending (Large, medium and small) and the vote obtained (Majority, Average and minority). Use Chi-squared hypothesis test for association to test that there is an association between the variable in campaign spending and the variable in vote obtained with five percent level of significance. The manager uses MS excel to calculate the Chi-squared stat value and the result is 8.874. Critical value: At 1% = 13.277 :At 5% = 9.488 :At 10% = 7.779 1. State the decision rule 2. Conclusion and decision 3. State the alternative hypothesis and the null hypothesis
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Ques = Homework: Ch 8 - Sampling Distributions Critical values of the correlation coefficient ars. Com - X Int. Critical Values for Correlation Coefficient Data table 3 0,997 4 0.950 5 0.878 6 0.811 7 0.754 Year Rate of Return of Company Rate of Return of Company ble. Chod 6 2 2007 0.203 0.402 8 0.707 9 0.666 10 0.632 2008 0.310 0.510 2009 0.267 0.410 11 0.602 12 0.576 13 0.553 14 0.532 15 0.514 16 0.497 2010 0.195 0.426 2011 -0.101 -0.060 2012 - 0.130 -0.151 2013 -0.234 -0.357 2014 0.264 0.328 17 0.482 18 0.468 19 0.456 20 0.444 21 0.433 2015 0.090 0.207 2016 0.030 -0.014 2017 0.128 0.113 2018 - 0.035 0.027 22 0.423 23 0.413 24 0.404 25 0.396 Print Done 26 0.388 27 0.381 Tne correlauon coeicient is- (Round to three decimal places as needed.) Print Done
32.
Polyunsaturated fatty acids in the diet favorably affect several risk factors for cardiovascular disease. The principal dietary polyunsaturated fat is linoleic acid. Suppose we randomly selected twelve adults for our study. We took their blood pressure measurements (systolic blood pressure) at the start and recorded them for each adult. Then we placed required these adults to consume 23g/day of safflower oil which is high in linoleic acid for four weeks. At the end of the four weeks, we took another systolic blood pressure(SBP) measurement. The paired data is provided below: Subject 1 2 3 4 5 6 7 8 9 10 11 12 Baseline SBP 119.67 100.00 123.56 109.89 96.22 133.33 115.78 126.39 122.78 117.44 111.33 92.39 1-month SBP 117.33 98.78 123.83 107.67 95.67 128.89 113.22 121.56 126.33 110.39 107.00 93.06 Use the appropriate nonparametric test at alpha level of 0.05, to test whether the diet heavy in linoleic acid increases systolic blood pressure (SBP). Use Steps 0-5
The ANOVA table of the above data is as follows: ANOVA Source of Variation SS df MS 2.083 P-value Factor A J 1 0.725 Marketing startegies 0.750 1 0.750 0.832 Interaction 90.750 K 90.750 0.043 Within 125.333 15.667 Total 218.917 10. Calculate J and L 11. State HO and H1 hypotheses of interaction effect based on the above problem. 12. At a = 0.1, is there any significant effect? What is your decision? Support your answer. 13. What is your conclusion?
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A test was conducted on the differences in academic performance of three groups of students subjected to different teaching strategies. Below is the ANOVA table of the results of data analysis. Source F- MS Ratio critical value F- P- of df SS variation Among 3 2100 MST groups ??? 3.18 0.036 Within 45 10000 MSE group

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9. Solve the ANOVA table Levels of A=4 Levels of B=4 Sample size per condition (n) = 7 Source Between Columns A Between Rows B Interaction AB Within Cells Total SS df MS F 75.00 210.00 6.00 2.99