Question

Assignment Gradebook Chapter 13, Section 13.2, Problem 07 According to the records of the National Safety Council, eccidental deaths in the United States during a recent year had the following distribution according to the principal types of accidents. Motor Vehicle! Falls Drowning! Burns Poison Other 46% 15%| 4% 4% |23% 8% In a particular geographical region, the accidental deaths, classified according to the principal types of accidents, yielded the following frequency distribution. Motor Vehicle Falls Drowning Burns Poison Other Total 438 110 69 26 77 212 932 Do these data show a significantly different distribution of accidental deaths as compared to that for the entire United States, Test at α-o.。. Choose the correct answer No, these data do not show a significantly different distribution of accidental deaths as compared to thet for the entire United States Yes, these data show a significantly different distribution of accidental deaths as compared to that for the entire United States OensborWork . Click if you would like to show work for this question:

Answer 1

Answer is:

Yes, these data show a significantly different distribution of accidental deaths as compared to that for the entire United States

Yes.

Well we need to perform a chi square test of goodness of fit in order to answer this question:

The following table is obtained (fo-fe)2fe (438-428.72)1428.72-0.201 (110-139.8)2/139.8 6.352 (69-37.28)2137.28 26.989 (26-37.28)2137.28 3.413 (77-74.56)2174.56 0.08 (212-214.36)-/214.36 0.026 Categories Motor Vehicles Falls Drowning Burns Poison Other Observed 438 110 69 26 Expected 932 0.46-428.72 932 0.15-139.8 932 0.04-37.28 932 0.04-37.28 932 0.08-74.56 932 0.23-214.36 212 932 932 37.061 Sum -

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho : PI = 0.16.P2 0. 15.P3 = 0.04.24 0.04.75 = 0.08. P6 = 0.23

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=0.01, the number of degrees of freedom is df=6−1=5, so then the rejection region for this test is R={χ2:χ2>15.086}.

(3) Test Statistics

The Chi-Squared statistic is computed as follows:

0.2016.352 26.989 + 3.413+0.08+0.026 37.061

(4) Decision about the null hypothesis

Since it is observed that χ2=37.061>χc2​=15.086, it is then concluded that the null hypothesis is rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that some of the population proportions differ from those stated in the null hypothesis, at the α=0.01 significance level.

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