A USA Today story reported the college students living situation reflects the fo

Question

A USA Today story reported the college students living situation reflects the following percentages: A random sample of college students resulted In the following information: Is the distribution of this sample significantly different from the distribution reported in the newspaper? a. What are the null and alternate hypothesis for this test? b. Calculate the expected counts for this distribution. c. What are the conditions for this test? Are they met? d. Perform the Chi-square test for goodness of fit to answer the question above. Show your work. e. Write an appropriate conclusion for this test.

Explanation / Answer

null hypothesis: proportion is similar to magazine report

alternate hypothesis: it is not similar

cvondition for this test is that alll expected cell should be greater then 5 and data should not have correlation b/w columns or rows means that should be independent .Yes they are meeting

also as sum of probabilty is not equal to 100% we need to do a little add up into each to make it 100

from above chi stat =8.4

and for 3 degree of freedom p value =0.0384

as p vlaue is less then 0.05 level we reject null hypothesis; that proportion is equal as reported in magazine

observed Expected Chi square P O E=total*p =(O-E)^2/E Parent Home 0.46 484 446.248 3.19 Campus 0.26 220 253.616 4.46 Off campus 0.18 168 177.144 0.47 Own off -campus 0.09 96 90.992 0.28 968 968 8.40

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