Question

The problemesenpoon indicates that the samples samples. For these reasons, it is reasonable to condude that the conditions are met. Step 6 Recall that when the conditions for a single factor ANOVA F test for equality of three or more means are met. and the null hypothesis, Ho: , -, , , is true, we use the following test statistic where df, - k-1, df, -N-k MST is the mean square of treatments, MSE is the mean square of errors, SSTr is the treatment Sum of squares, and SSE is the error sum of squares. F= MST MSE T- To solve for the test statistic, we first must find SSTT, SSE, df, and df. Recall that we previously determined that N = 40, 3, 4, 5, -5.8, 5, = 6.2. 7. = 7.2. , - , -, -, -10,7 = 5.8. s, = 1.8856 5 -1.6193, 5, -1.3166, and s-1.4757. Additionally, there are four restrictive rating labels so k = 4. (Round your answer for SSE to one decimal place.) SST = 6,-)2+,-) + ... +nQ. - 7)? = 10(4 - 5.8)2 + 10(5.8 - 5.8)2 + 10(6.2 - 5.8)2 + 1017.2 - 58) - 53.6 SSE = (n. - 1)s, + (n -1), + ... + (n-1)s.? = (10 - 1)(1.8856)2 + (10 - 1X 1.6193 )+(10-1)(1.3166)2 + (10 - 1)(1.4757) = 908 df, = k-1 = X - 1 of = N-k -4 Subme Sko (you cannot come back Need Help? Read It Watch Type here to search O

Answer 1

Given :

k = 4 and N = 40

So

df1 = k - 1

= 4 - 1

= 3

Therefore df1 = 3

df2 = N - k

= 40 - 4

= 36

Therefore df2 = 36