21. For an equiprobable chi-squared goodness of fit test with an HO of Normally

Question

21. For an equiprobable chi-squared goodness of fit test with an HO of Normally distributed with the same mean and std as the sample for 26 obs., you would have how many cells. a. 4 b. 5 c. 6 d. 7 23. If you had an equiprobable chi-squared goodness of fit test with an HO of Normally distributed with the same sample mean and std and you had 4 cells, what would be the critical value at alpha 0.05 a. 3.841 b. 5.999 c. 7.814 d. 9.488 24. If your test from the above problem yielded the following data, you would? Cell Observed 8 6 4 2 Expected 5 5 5 5 a. Cannot reject Ho b. Reject Ho c. Neither

Explanation / Answer

(a)answer is c.6

using the Surge's rule, number of class k = 1 +3.22 log10( n)=1+3.22*log(26)=5.56 ( next integer is 6)

(b) answer is c.7.814

there are 4 cells , so critical value =chi-square(alpha, n-1)=chi-square(0.05,3)=7.814

( using ms-excel==CHIINV(0.05,3))

(c) cannot reject H0

calculated chi-square=sum((O-E)2/E)=4 is less than critical chi-square(0.05,3)=7.814, so fail to reject H0

O E (O-E) (O-E)2/E 8 5 3 1.8 6 5 1 0.2 4 5 -1 0.2 2 5 -3 1.8 sum= 20 20 0 4

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