Question

Statistics.

Show your work.

2- It has been reported that 9% of U.S. households do not own a car, 33% own one car, 38% own 2 cars, and 20% own 3 or more cars. The car ownership in a random sample of 100 household in a community is summarized in the following table- Number of Cars Owned Number of Households 0 20 354 234 2 3 or more 22. Does the car ownership in this community differ from the nation?

Please be as neat as possible in answering the questions

Answer 1

This is a goodness of fit test for the given proportions against the observed frequencies. The expected frequencies for each of the value of number of cars owned are computed here as:

E(0) = 100*0.09 = 9
E(1) = 0.33*100 = 33
E(2) = 0.38*100 = 38
E(3 or more) = 0.2*100 = 20

The chi square test statistic here is computed as:

$\chi^2 = \sum \frac{(O_i - E_i)^2}{E_i}$

$\chi^2 = \frac{11^2}{9} + \frac{2^2}{33} + \frac{15^2}{38} + \frac{2^2}{20}$

$\chi^2 = 19.6867$

For k - 1 = 3 degrees of freedom, the p-value here is computed from the chi square distribution tables as:

$p = P(\chi^2_3 > 19.6867) = 0.0002$

As the given p-value here is 0.0002 which is very very low, therefore the test is significant here and we can reject the null hypothesis conclude that we have sufficient evidence that the car ownership in this community differs from the nation.