A survey was done to determine if there is an associative relationship between t

Question

A survey was done to determine if there is an associative relationship between the size of a family and the type of car driven. A random sample of 1, 500 families of sizes 2, 3, 4 and 5 members, yielded the results shown in the attached table. Perform a test of independence at an alpha (alpha) level of 5%. Calculate: a. The Expected frequencies (f_e) b. The Chi-Square statistic, c. Degrees of freedom, d. Chi-Square critical value (use alpha = 05), e. State the null hypothesis (H_0), f. State the alternative hypothesis (H_1), g. What is the final conclusion?

Explanation / Answer

a) expected frequency is given in expected table.

b)chi square test stat =16.652

c) degree of freedom =(row-1)*(column-1) =(4-1)*(5-1)=12

d)critical value for 12 degree of freedom and 0.05 level =21.0261

e)null hypothesis:Ho: size of family and type of cars are independent.

f) alternate hypothesis:Ha:size of family and type of cars are dependent.
g)for test stat is less then critical value 21.0261; therefore we can not reject null hypothesis

hence we can not reject that size of family and type of cars are independent.

Observed O small medium large luxury SUV Total 2 members 62 69 74 85 55 345 3 members 81 68 80 73 71 373 4 members 72 82 61 79 84 378 5 members 83 88 68 75 90 404 Total 298 307 283 312 300 1500 Expected E=rowtotal*column total/grand total small medium large luxury SUV Total 2 members 68.540 70.610 65.090 71.760 69.000 345 3 members 74.103 76.341 70.373 77.584 74.600 373 4 members 75.096 77.364 71.316 78.624 75.600 378 5 members 80.261 82.685 76.221 84.032 80.800 404 Total 298 307 283 312 300 1500 chi square =(O-E)^2/E small medium large luxury SUV Total 2 members 0.624 0.0367 1.2197 2.4428 2.8406 7.164 3 members 0.642 0.911 1.317 0.271 0.174 3.315 4 members 0.128 0.278 1.492 0.002 0.933 2.833 5 members 0.093 0.342 0.887 0.971 1.048 3.340 Total 1.487 1.567 4.916 3.686 4.995 16.652

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