Identify the null and alternative hypotheses for a chi-square test of independen

Question

Identify the null and alternative hypotheses for a chi-square test of independence based on the information in the table. This test will have a significance level of a=0.025

Calculate the expected frequencies for each cell in the contingency table.

Calculate the chi-square test statistic.

using a=0.025, determine the chi-square critical value 2.

state your conclusions.

p-value?

Interpret the meaning of the p-value.

Consider the following contingency table of observed frequencies. Complete parts a. through e. below. Click iContingency table based on the se the correct answer a. Identify informatio below Column Variable OA, Ho Row Variable C1 18 O B. Ho 13

Explanation / Answer

Ans:

H0:both variables are independent.

Ha:both variables are not independent.

Expected count=row sum*column sum/overall sum

chi square test statistic=16.450

df=(3-1)*(3-1)=4

p-value=CHIDIST(16.450,4)=0.0025

As,p-value<0.025,we reject null hypothesis.

There is not sufficient evidence to conclude that both variables are independent .

Observed(O) C1 C2 C3 Total R1 18 5 18 41 R2 8 6 13 27 R3 8 15 5 28 Total 34 26 36 96 Expected(E) C1 C2 C3 Total R1 14.52 11.10 15.38 41 R2 9.56 7.31 10.13 27 R3 9.92 7.58 10.50 28 Total 34 26 36 96 (O-E)^2/E C1 C2 C3 R1 0.834 3.356 0.448 R2 0.255 0.236 0.816 R3 0.370 7.254 2.881 Total chi square score=16.450

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