11.3 Goodness-of-fit test  (Page 2/3)

The null and alternate hypotheses are:

• $H_o$ : The absent days occur with equal frequencies, that is, they fit a uniform distribution.
• $H_a$ : The absent days occur with unequal frequencies, that is, they do not fit a uniform distribution.

If the absent days occur with equal frequencies, then, out of 60 absent days (the total in the sample: 15 + 12 + 9 + 9 + 15 = 60), there would be 12 absences on Monday, 12 on Tuesday, 12 on Wednesday, 12 on Thursday,and 12 on Friday. These numbers are the expected ( $E$ ) values. The values in the table are the observed ( $O$ ) values or data.

This time, calculate the ${\chi }^2$ test statistic by hand. Make a chart with the following headings and fill in the columns:

• Expected ( $E$ ) values (12, 12, 12, 12, 12)
• Observed ( $O$ ) values (15, 12, 9, 9, 15)
• $(O-E)$
• ${(O-E)}^2$
• $\frac{{(O-E)}^2}{E}$

The last column ( $\frac{{(O-E)}^2}{E}$ ) should have 0.75, 0, 0.75, 0.75, 0.75.
Now add (sum) the last column. Verify that the sum is 3. This is the ${\chi }^2$ test statistic.

To find the p-value, calculate $P({\chi }^2>3)$ . This test is right-tailed.
(Use a computer or calculator to find the p-value. You should get $\text{p-value}=0.5578$ .)

The $\mathrm{dfs}$ are the $\text{number of cells}-1=\mathrm{5\; -\; 1\; =\; 4}$ .

TI-83+ and TI-84: Press 2nd DISTR . Arrow down to ${\chi }^2$ cdf . Press ENTER . Enter (3,10^99,4) . Rounded to 4 decimal places, you should see 0.5578 which is the p-value.

Next, complete a graph like the one below with the proper labeling and shading. (You should shade the right tail.)

The decision is to not reject the null hypothesis.

Conclusion: At a 5% level of significance, from the sample data, there is not sufficient evidence to conclude that the absent days do not occur with equal frequencies.