0.11 Ch 11: the chi-square distribution

This chapter is concerned with three chi-square applications: goodness-of-fit; independence; and single variance. We rely on technology to do the calculations, especially for goodness-of-fit and for independence. However, the first example in the chapter (the number of absences in the days of the week) has the student calculate the chi-square statistic in steps. The same could be done for the chi-square statistic in a test of independence.

The chi-square distribution generally is skewed to the right. There is a different chi-square curve for each df. When the df's are 90 or more, the chi-square distribution is a very good approximation to the normal. For the chi-square distribution, $\mu$ = the number of df's and $\sigma$ = the square root of twice the number of df's.

Goodness-of-fit test

In a past issue of the magazine GEICO Direct , there was an article concerning the percentage of teenage motor vehicle deaths and time of day. The following percentages were given from a sample.

For the purpose of this example, suppose another sample of 100 produced the same percentages. We hypothesize that the data from this new sample fits a uniform distribution. The level of significance is 1% ( $\alpha =0\text{.}\text{01}$ ).

• $H_o$ : The number of teenage motor vehicle deaths fits a uniform distribution.
• $H_a$ : The number of teenage motor vehicle deaths does not fit a uniform distribution.

The distribution for the hypothesis test is $X_7^2$

The table contains the observed percentages. For the sample of 100, the observed (O) numbers are 17, 8, 8, 6, 10, 16, 15 and 19. The expected (E) numbers are each 12.5 for a uniform distribution (100 divided by 8 cells). The chi-square test statistic is calculated using

${\sum }_8\frac{(0-E{)}^2}{E}$

$=\frac{(\text{17}-\text{12}\text{.}5{)}^2}{\text{12}\text{.}5}+\frac{(8-\text{12}\text{.}5{)}^2}{\text{12}\text{.}5}+\frac{(8-\text{12}\text{.}5{)}^2}{\text{12}\text{.}5}+\frac{(6-\text{12}\text{.}5{)}^2}{\text{12}\text{.}5}+\frac{(\text{10}-\text{12}\text{.}5{)}^2}{\text{12}\text{.}5}+\frac{(\text{16}-\text{12}\text{.}5{)}^2}{\text{12}\text{.}5}+\frac{(\text{15}-\text{12}\text{.}5{)}^2}{\text{12}\text{.}5}+\frac{(\text{19}-\text{12}\text{.}5{)}^2}{\text{12}\text{.}5}$

$=\text{13}\text{.}6$

If you are using the TI-84 series graphing calculators, ON SOME OF THEM there is a function in STAT TESTS called $x^2$ GOF-Test that does the goodness-of-fit test. You first have to enter the observed numbers in one list (enter as whole numbers) and the expected numbers (uniform implies they are each 12.5) in a second list (enter 12.5 for each entry: 100 divided by 8 = 12.5). Then do the test by going to $x^2$ GOF-Test.

If you are using the TI-83 series, enter the observed numbers in list1 and the expected numbers in list2 and in list3 (go to the list name), enter (list1-list2)^2/list2. Press enter. Add the values in list3 (this is the test statistic). Then go to 2nd DISTR $x^2$ cdf. Enter the test statistic (13.6) and the upper value of the area (10^99) and the degrees of freedom (7).