Question

Part III. Goodness of Fit test. (15 pts) This question is exercise 15 from your textbook's chapter 13. Assume that a researcher wants to be sure that the sample in her study is not unrepresentative of the distribution of ethnic groups in her community. Her sample includes 300 whites, 80 african americans, 100 hispanics, 40 asians, and 80 "others". Ethnicity thus comprises a 5 category nominal dependent variable. In her community, according to census records, the population has 48% whites, 12% african americans, 18% hispanics, 9% asians, and 13% "others" Your task is to test the null hypothesis that her sample is representative, using the methods we have covered for the 1-way layout for a goodness of fit test. Note that this is NOT a contingency table or test of independence. Use the approach outlined in lecture with the "beverage purchase" illustration. The key to this question is carefully thinking about what the "expected frequencies should be, based on information above. Summarize here, and attach your complete set of computations on another Computed Chi sq value: Degrees of freedom Critical Value from the chi square distribution Decision: Reject or Accept

Answer 1

Assuming the level of significance to be 5%

The following table is obtained (fo-fe)2lfe (300-288)/2880.5 (80-72)/72 0.889 (100-108)/108 0.593 (40-54)/54 3.63 (80-78)178 0.051 Categories Whites African Americans HispanicS Asians Others Observed 300 80 100 Expected 600 0.48-288 600 0.12-72 600*0.18-108 600 0.09-54 600*0.13 78 80 600 Sum - 600 5.662

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Но : PI 0.15.P2-0. 12.P3 0.18. P4-0.09.75 = 0.13

Ha​: Some of the population proportions differ from the values stated in the null hypothesis

This corresponds to a Chi-Square test for Goodness of Fit.

(2) Rejection Region

Based on the information provided, the significance level is α=0.05, the number of degrees of freedom is df=5−1=4, so then the rejection region for this test is R={χ2:χ2>9.488}.

(3) Test Statistics

The Chi-Squared statistic is computed as follows:

0.50.889 + 0.593 + 3.63 + 0.051- 5.662 Ei

(4) Decision about the null hypothesis

Since it is observed that χ2=5.662≤χc2​=9.488, it is then concluded that the null hypothesis is not rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is not rejected. Therefore, there is NOT enough evidence to claim that some of the population proportions differ from those stated in the null hypothesis, at the α=0.05 significance level.

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