Question 851:

Asked on March 5, 2010

Tags: Chi-Square Test of Independence , Chi-Square Goodnes of Fit Test

1

Answer:

No answer provided yet.We have counts of viewers for different TV stations and we can use the Chi-Square Goodness of Fit test to see if the differences in viewers suggest something different than equal viewership. The Null Hypothesis for this test is that there is no difference in percentages between viewer. The alternative hypothesis is that at least one TV station has a higher or lower percent of viewers.

Since there are three TV stations, we'd expect a 33.33% viewership breakdown for each TV station. We see a percentage breakdown of .40, .34 and .26. Are these differences more than we'd expect from chance alone?

The Chi-Square test statistic is found by using the formula  [ Observed -Expected ]^2 / Expected. Observed is the raw count and expected is the total count times the expected percent. Since there are 150 viewers, the expected count is 50 (.333 of 150). For example, on WNAE, we'd have the formula = (53-50)^2 / 50 = .18. We repeat this for the other two stations and add up the values to get .18+3.9+5.77 = 9.88, which is the chi-square value.

To evaluate this value we look it up in a Chi-Square table of values using 3-1=2 degrees of freedom (the number of categories minus 1). We get the p-value which is less than .007. Since the p-value is less than the alpha significance level of .05 we reject the Null Hypothesis. We conclude that there is a difference in the proportion of viewers for the  three entrances.