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17.2: One-Way Tables

Last updated

Apr 23, 2022
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- 17.1: Chi Square Distribution
- 17.3: Testing Distribution Demonstration

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Learning Objectives

Describe what it means for there to be theoretically-expected frequencies

Compute expected frequencies
Compute Chi Square
Determine the degrees of freedom

The Chi Square distribution can be used to test whether observed data differ significantly from theoretical expectations. For example, for a fair six-sided die, the probability of any given outcome on a single roll would be $1/6$ . The data in Table $\PageIndex{1}$ were obtained by rolling a six-sided die $36$ times. However, as can be seen in Table $\PageIndex{1}$ , some outcomes occurred more frequently than others. For example, a " $3$ " came up nine times, whereas a " $4$ " came up only two times. Are these data consistent with the hypothesis that the die is a fair die? Naturally, we do not expect the sample frequencies of the six possible outcomes to be the same since chance differences will occur. So, the finding that the frequencies differ does not mean that the die is not fair. One way to test whether the die is fair is to conduct a significance test. The null hypothesis is that the die is fair. This hypothesis is tested by computing the probability of obtaining frequencies as discrepant or more discrepant from a uniform distribution of frequencies as obtained in the sample. If this probability is sufficiently low, then the null hypothesis that the die is fair can be rejected.

Table $\PageIndex{1}$ : Outcome Frequencies from a Six-Sided Die
Outcome	Frequency
1	8
2	5
3	9
4	2
5	7
6	5

The first step in conducting the significance test is to compute the expected frequency for each outcome given that the null hypothesis is true. For example, the expected frequency of a " $1$ " is $6$ since the probability of a " $1$ " coming up is $1/6$ and there were a total of $36$ rolls of the die.

$\text{Expected frequency} = (1/6)(36) = 6$

Note that the expected frequencies are expected only in a theoretical sense. We do not really "expect" the observed frequencies to match the "expected frequencies" exactly.

The calculation continues as follows. Letting $E$ be the expected frequency of an outcome and $O$ be the observed frequency of that outcome, compute

$\frac{(E-O)^2}{E}$

for each outcome. Table $\PageIndex{2}$ shows these calculations.

Table $\PageIndex{2}$ : Outcome Frequencies from a Six-Sided Die
Outcome	E	O
1	6	8	0.667
2	6	5	0.167
3	6	9	1.500
4	6	2	2.667
5	6	7	0.167
6	6	5	0.167

Next we add up all the values in Column 4 of Table $\PageIndex{2}$ .

$\sum \frac{(E-O)^2}{E} = 5.333$

This sampling distribution of $\sum \frac{(E-O)^2}{E}$ is approximately distributed as Chi Square with $k-1$ degrees of freedom, where $k$ is the number of categories. Therefore, for this problem the test statistic is

$\chi _{5}^{2}=5.333$

which means the value of Chi Square with $5$ degrees of freedom is $5.333$ .

From a Chi Square calculator it can be determined that the probability of a Chi Square of $5.333$ or larger is $0.377$ . Therefore, the null hypothesis that the die is fair cannot be rejected.

This Chi Square test can also be used to test other deviations between expected and observed frequencies. The following example shows a test of whether the variable "University GPA" in the SAT and College GPA case study is normally distributed.

The first column in Table $\PageIndex{3}$ shows the normal distribution divided into five ranges. The second column shows the proportions of a normal distribution falling in the ranges specified in the first column. The expected frequencies ( $E$ ) are calculated by multiplying the number of scores ( $105$ ) by the proportion. The final column shows the observed number of scores in each range. It is clear that the observed frequencies vary greatly from the expected frequencies. Note that if the distribution were normal, then there would have been only about $35$ scores between $0$ and $1$ , whereas $60$ were observed.

Table $\PageIndex{3}$ : Expected and Observed Scores for $105$ University GPA Scores
Range	Proportion	E	O
Above 1	0.159	16.695	9
0 to 1	0.341	35.805	60
-1 to 0	0.341	35.805	17
Below -1	0.159	16.695	19

The test of whether the observed scores deviate significantly from the expected scores is computed using the familiar calculation.

$\chi _{3}^{2} = \sum \frac{(E-O)^2}{E} = 30.09$

The subscript " $3$ " means there are three degrees of freedom. As before, the degrees of freedom is the number of outcomes minus $1$ , which is $4 - 1 = 3$ in this example. The Chi Square distribution calculator shows that $p < 0.001$ for this Chi Square. Therefore, the null hypothesis that the scores are normally distributed can be rejected.

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