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Between Y 39.62 8.70 Total use the tool to find the critical F value, set both the numerator F distnbution. Move the orange l
F distribution. Move the orange line until the area in the tail is equivalent to the alpha level you are investigating. 3000
O 3.098 O 4.718 O 4.718 Now use the tool to evaluate the F-ratio. you the The nul h
İ:-CENGAGE I MNDTAP O 4.938 O 4.718 O 3.009 At the a- .01 level of significance, the boundary of the critical region for this
Between Y 39.62 8.70 Total use the tool to find the critical F value, set both the numerator F distnbution. Move the orange line until the area in the tail is equivalent of freedom; this will show you the appropriate to the alpha level you are investigating. here to s
F distribution. Move the orange line until the area in the tail is equivalent to the alpha level you are investigating. 3000 At the a - .os level of significance, the boundary of the critical region for this ANOVA is: O 3.098 O 4.938 O 4.718 О 3.009 At the a01 level of the boundary of the critical region for this ANOVA is: Type here to search
O 3.098 O 4.718 O 4.718 Now use the tool to evaluate the F-ratio. you the The nul h
İ:-CENGAGE I MNDTAP O 4.938 O 4.718 O 3.009 At the a- .01 level of significance, the boundary of the critical region for this ANOVA is O3.009 о 3.098 O 4.718 O 4.938 Now use the tool to evaluate the F-ratio. (Hint: Select the icon with one orange Ine and one purple line. You can set the orange line at the crtical boundary and move the purple ine to the F-ratio in the ANOVA table given.) To use the tool to find the F-ratio, set both the numerator and the denominator degrees of freedom; this will show you the appropriate F distribution. Move the orange line until the area in the tail is equivalent to the alpha level you are investigating. At the a -0.01 level of O The nul h can be rejected
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Answer #1

df(Between) = 4-1 = 3

df( Within) = 27-3 =24

F(Crit) = F(0.05,3,24) = 3.009 (Use Excel function "FINV(0.05,3,24))

F(Crit) = F(0.01,3,24) = 4.718 (Use Excel function "FINV(0.01,3,24))

F= 8.70 (which is greater than both the critical value)

So, the null hypothesis can be rejected

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