ANOVA: Single FactorSUMMARYGroupsTreatment 1Treatment 2Treatment 3Treatment 4ANOVASource of VariationBetween GroupsWithin GroupsTotalCount16191715SS65.2023409.9619475.1642Sum871116653Average Variance5.43759.06255.84218.36263.88234.61033.53333.5524df MSABCFP-valueFind the values that would go in boxes A, B, and C of this ANOVA table. Pleaseidentify each answer by typing the following below A = # , B = #, C = #

The value of n is computed as follows:

Answered: ANOVA: Single Factor SUMMARY Groups…

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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Summary of Fit RSquare 0.673175 0.659743 RSquare Adj Root Mean Square Error Mean of Response Observations (or Sum Wgts) 11.36538 106.8831 77 Analysis of Variance Sum of Source DF Squares Mean Square F Ratio Model 3 19422.399 6474.13 50.1203 Error 73 9429.549 129.17 Prob > F C. Total 76 28851.948 |t| Intercept 29.593628 7.186991 4.12 <.0001* Sugars 3.8623817 0.348158 11.09 <.0001* Sodium 0.0089653 0.016746 0.54 0.5940 Carbo 3.3058183 0.408135 8.10 <.0001* 1. Using the above JMP output, which of the predictors present concern to the overall model? Assume the response, y, is calories found in a cup of cereal.
Considering that the dependent variable (DV) is continuous, what ANOVA procedure would you like to use if there is a single within-subject Independent Variable (IV) with three repeated observations and a single between-subjects Independent Variable (IV) with two distinct groups? Group of answer choices None of the ANOVA methods can be applied One factor ANOVA Dependent t test Repeated measures ANOVA Split-Plot ANOVA
What is the alternative hypothesis for this research question? Group of answer choices There is no difference in mean job satisfaction across the three height groups. Mean job satisfaction will differ between at least two of the height groups. The mean job satisfaction for tall people is higher than the mean job satisfaction for short people. The mean of one group will differ from the average job satisfaction of the population.
esc ! Frequency 1 7 13 19 25 31 37 data a Based on the histogram above, what is the class width? Class width=6 What is the sample size? Sample size = Submit Question F1 Q @ 2 F2 W 20 #3 F3 E DDD 000 S4 $ F4 R 1 % 5 F5 6 MacBook Air T ^ 6 F6 ◄◄ & Y 7 F7 * 8
Analysis of Variance results: Where: Age > 5Responses: AgeFactors: Sex Response statistics by factor Sex n Mean Std. Dev. Std. Error female 121 10.490083 4.2222704 0.38384277 male 155 9.0806452 3.0091656 0.241702 ANOVA table Source DF SS MS F-Stat P-value Sex 1 134.98939 134.98939 10.466692 0.0014 Error 274 3533.79 12.897044 Total 275 3668.7794 Tukey HSD results (95% level) female subtracted from Difference Lower Upper P-value male -1.4094375 -2.2670907 -0.55178425 0.0014 In one complete sentence, interpret your confidence interval. Make sure in the interpretation it is clear the direction of the difference (i.e. which gender weighs more and by how much. Explain why a confidence interval provides more information to a tiger researcher than the results of a hypothesis test
ANOVA table Source df sum of sqaures Mean squares F Regression 1 4.9 4.9 13.38 Error 3 1.1 0.366 Total 4 6 i. name the sampling distribution of the F-test statistic ii should we reject the null or fail to reject the null iii. justify using the rejection region and p-value, draw the sampling distribution and shade area corresponding to p-value iii
Treatment II III 2 N = 24 G = 72 EX2 = 292 3 M = 2 M = 3 T = 16 T= 24 T= 32 SS = 16 SS = 24 SS = 20 Use an ANOVA with a = .05 to determine whether there is a significant mean difference between the two treatments. (Round to two decimal places where needed.) Source df MS Feritical Between treatments Within treatments Total F Distribution Numerator Degrees of Freedom = 6 Denominator Degrees of Freedom = 16 0.0 1.0 3.0 4.0 5.0 6.0 7.0 9.0 11.0 120 Conclusion: Fail to reject the null hypothesis; there are no significant differences among the three treatments. Reject the null hypothesis; there are no significant differences among the three treatments. Reject the null hypothesis; there are significant differences among the three treatments. o Fail to reject the null hypothesis; there are significant differences among the three treatments.
Fill in the ANOVA table using the data attached. Source df sum of squares MEan squares F Regression 1 Error 3 Total 4
Considering that the dependent variable (DV) is continuous, what ANOVA procedure would you like to use if there is a single between-subject Independent Variable (IV) with three distinct groups? Group of answer choices One-Way ANOVA Two-Way ANOVA Split-Plot ANOVA Repeated Measures ANOVA All of the Above
Chicken diet and weight: Compare the effects of six types of feed: casein, horsebean, linseed, meat meal, soybean, and sunflower. The ANOVA output below can be used to test for differences between the mean weights of chicks on different diets. df Sum Sq Mean Sq F value Pr(>F) feed 5 231129.16 46225.83 15.36 0.0000 residuals 65 195556.02 3008.55 Conduct a hypothesis test to determine if these data provide convincing evidence that the average weight of chicks varies across some (or all) groups. Figures and summary statistics are shown below. Test to see if there is a difference between mean weight of chicks between at least two of the diets using overall a = 0.05. 400 350- 300 - 250 200 150 - 100 casein horsebean linseed meatmeal soybean sunflower Mean SD casein 323.58 64.43 12 horsebean 160.2 38.63 10 linseed 218.75 52.24 12 meatmeal 276.91 64.9 11 soybean 246.43 54.13 14 sunflower 328.92 48.84 12 Weight (in grams)

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