Question

At a certain university, 40% of the students come from Orange County, 20% come from Los Angeles County, 20% come from another county in California, 10% come from another state, and 10% come from a different country. Zoe wants to see if the students in her class fit or do not fit this profile. She takes a sample: 100 come from Orange County, 35 come from Los Angeles County, 30 come from another county in California, 20 come from another state, and 15 come from another country. Please calculate the test statistic, state the critical value, and come to a conclusion concerning the make-up of the class. Let α = .05.

Answer 1

Total Frequency = 100 + 35 + 30 + 20 + 15 = 200

Therefore using this, the expected frequencies here are computed as:
E(Orange County) = 0.4*200 = 80
E(Los Angeles County) = 0.2*200 = 40
E(Another California county) = 0.2*200 = 40
E(Another state) = 0.1*200 = 20
E(Another country) = 0.1*200 = 20

Using these the chi square test statistic here is computed as:

For n - 1 = 4 degrees of freedom, we get the p-value from the chi square distribution tables here:

For 4 degrees of freedom, we have form the chi square distribution tables here:

Therefore 9.49 is the critical value here.

As the p-value here is 0.0524 > 0.05 which is the level of significance, therefore the test is not significant here and we cannot reject the null hypothesis here. Therefore we dont have sufficient evidence here that the distribution is different from the expected distribution.