The Chi Squared Test
- Page ID
- 64239
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Chi Square and Two Way Tables Two Way Tables Is there any relationship between political affiliation and the type of car a person drives? A survey was done and the following data was collected:
We computed the numbers in parentheses by calculating the expected counts. We multiplied the row marginal total by the column proportions. We could also compute this number by calculating \( \text{Expected Count} = \frac{ \text{(Row Margin Total)(Column Marginal Total)}}{\text{Grand Total}} \)
We have the hypothesis: H0: The true proportions are the same for all of the populations H1: The true proportions are not the same for all of the populations. We compute:
\( = \frac{(32 - 35)^{2}}{35} +\frac{(21 - 24)^{2}}{24} +\frac{(8 - 13)^{2}}{13} + ... = 6.87 \) The degrees of freedom is (num of rows - 1)(num of columns - 1) = (2 - 1)(3 - 1) = 2 Now the \(\chi^{2}\) that corresponds to 2 degrees of freedom and \(\alpha\) = .05 is 5.99 We can reject H0 and therefore accept H1 hence there is an association between political affiliation and the type of car a person drives. An applet that does the two way table computations can be found here
Chi Square For Univariate Data Recall that we use a t-statistic for a difference between proportions. If there are three or more Boolean variables, then we must use a different solution.
Example Suppose that we run a lunch special in our restaurant and want to determine if it makes a difference which day of the week to close. In other words are all days equally frequented by customers? We take a tally of the customers for each day and find:
We have the following hypotheses: H0: p1 = 1/7, p2 = 1/7, p3 = 1/7, p4 = 1/7, p5 = 1/7, p6 = 1/7, p7 = 1/7 H1: At least one of the p's is not 1/7 Let \(\alpha\) = .05 The test statistic that we will use is also called the chi square statistic and is also denoted by the Greek letter \(\chi^{2}\) . It is computed as follows: Notice that the total sample size is 210, hence if H0 is true, then the expected count for each day is \( \frac{210}{7} = 30 \) For each of the data, we compute: We compute: S (observed - expected)2/expected \( \frac{(30 - 30)^2}{30} = 0 \hspace{1cm} \frac{(33 - 30)^2}{30} = 0.3 \) \( \frac{(20 - 30)^2}{30} = 3.3 \hspace{1cm} \frac{(22 - 30)^2}{30} = 2.1 \hspace{1cm} \frac{(35 - 30)^2}{30} = 0.83 \) \( \frac{(40 - 30)^2}{30} = 3.3 \hspace{1cm} \frac{(30 - 30)^2}{30} = 0.3 \) Now we add these numbers to get: 0 + 0.3 + 3.3 + 2.1 + 0.83 + 3.3 + 0 = 9.8 Hence we have \(\chi^{2}\) = 9.8 The degrees of freedom is k - 1 = 7 - 1 (k is the number of samples) Now go to the chi square table then the critical value for the \(\chi^{2}\) with 6 degrees of freedom is 12.50. Since 9.8 < 12.59 we see that there is not enough evidence to conclude that the day of the week is a factor in lunch attendance. An applet that does goodness of fit computations can be found here
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