Dropdown menu word options: Variance due to chance Variance due to both chance a

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Variance due to chance

Variance due to both chance and caffeine

Opinions about whether caffeine enhances test performance differ. You design a study to test the impact of drinks with different caffeine contents on students' test-taking abilities. You choose 21 students at random from your introductory psychology course to participate in your study. You randomly assign each student to one of three drinks, each with a different caffeine concentration, such that there are seven students assigned to each drink. You then give each of them a plain capsule containing the precise quantity of caffeine that would be consumed in their designated drink and have them take an arithmetic test 15 minutes later. The students receive the following arithmetic test scores: You plan to use an ANOVA to test the impact of drinks with different caffeine contents on students' test-taking abilities. What is the null hypothesis? The population mean test scores for all three treatments are not all equal. The population mean test score for the water population is different from the population mean test score for the black tea population. The population mean test scores for all three treatments are different. The population mean test scores for all three treatments are equal. Calculate the degrees of freedom and the variances for the following ANOVA table: The formula for the F-ratio is: F=MS_between/MS_within Using words (chosen from the dropdown menu), the formula for the F-ratio can be written as: Using the data from the ANOVA table given, the F-ratio can be written as: Thus: Use the Distributions tool to find the critical region for alpha=.01. At the alpha=.01 level of significance, what is your conclusion? You can reject the null hypothesis; caffeine does appear to affect test performance. You cannot reject the null hypothesis; caffeine does appear to affect test performance. You cannot reject the null hypothesis; caffeine does not appear to affect test performance. You can reject the null hypothesis; caffeine does not appear to affect test performance.

Explanation / Answer

H0: mu1=mu2=mu3 [mean test scores for all three treatments are equal) (option d)

ANOVA table.

SSB=SST-SSW=1225.24-744.86=480.38

SSB(df)=k-1=3-1=2

SSW(df)=N-k=21-3=18

SST (df)=N-1=21-1=20

MSB=SSB/k-1=480.38/2=240.19

MSW=SSW/N-K=744.86/18=41.38

F=MSB/MSW=240.19/41.38=5.80

P value:0.0114

p value is not less than alpha=0.01. Fail to reject H0. Option b)

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