Homework SourseA Fortune study found that the variance in the number of veh
A Fortune study found that the variance in the number of vehicles owned or leased by subscribers to Fortune magazine is 0 .94. Assume a sample of 12 subscribers to another magazine provided the following data on the number of vehicles owned or leased: 2, 1, 2, 0, 3, 2, 2, 1, 2, 1, 0, 1. a. Compute the sample variance in the number of vehicles owned or leased by the 12 subscriber s. b. Test the hypothesis: to determine whether the variance in the number of vehicles owned or leased by subscribers of the other magazine differs from for Fortune . At a 0.05 level of significance, what is your conclusion?
Solution
Getting the mean, X,
X = Sum(x) / n
Sum(x) = 17
As n = 12
Thus,
X = 1.416666667
Setting up tables,
x x - X (x - X)^2
2 0.583333333 0.340277778
1 -0.416666667 0.173611111
2 0.583333333 0.340277778
0 -1.416666667 2.006944444
3 1.583333333 2.506944444
2 0.583333333 0.340277778
2 0.583333333 0.340277778
1 -0.416666667 0.173611111
2 0.583333333 0.340277778
1 -0.416666667 0.173611111
0 -1.416666667 2.006944444
1 -0.416666667 0.173611111
Thus, Sum(x - X)^2 = 8.916666667
Thus, as
s^2 = Sum(x - X)^2 / (n - 1)
As n = 12
s^2 = 0.810606061 [answer, sample variance]
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Thus,
s = 0.900336637
Formulating the null and alternative hypotheses,
Ho: sigma = 0.94
Ha: sigma =/ 0.94
As we can see, this is a two tailed test.
Thus, getting the critical chi^2, as alpha = 0.05 ,
alpha/2 = 0.025
df = N - 1 = 11
chi^2 (crit) = 3.815748252 and 21.92004926
Getting the test statistic, as
s = sample standard deviation = 0.900336637
sigmao = hypothesized standard deviation = 0.94
n = sample size = 12
Thus, chi^2 = (N - 1)(s/sigmao)^2 = 10.09129319
As chi^2 is between the two critical values, we FAIL TO REJECT THE NULL HYPOTHESIS.
Thus, there is no significant evidence that the variance in the number of vehicles owned or leased by subscribers of the other magazine differs from for Fortune. [CONCLUSION]
