Question

At a gymnastics meet, three judges evaluate the balance beam performances of five gymnasts, using a scale from 1 to 10, where 10 is a perfect score. A statistician wants to see if the gymnasts (Factor A) differ in their performances and also wants to see if the scores vary significantly across judges (Factor B) in order to assess any bias in the judges. To do this the statistician uses a two-way ANOVA without interaction and the results are show below. At 0.05 level of significance, what is the critical value for testing differences in mean scores across the Gymnasts

Answer 1

Answer:

As   2.551 < 3.84 therefore we reject H0.

As 44.803 > 4.46 so we accept null hypothesis.

Step-by-step explanation:

The answer is attached.

There are three judges so v1 = 3-1= 2  and v2 = (4*2)= 8

There are five gymnasts so v1 = 5-1= 4 and v2=  (4*2)= 8

For alpha = 0.05 we find the value of F1 and F2 from the table.

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