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For the population of employees, do the days for the highest number of absences occur with equal frequencies during a five day work week? Test at a 5% significance level.

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 χ 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
  • ( O - E ) 2 E

The last column ( ( 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 χ 2 test statistic.

To find the p-value, calculate P ( χ 2 > 3 ) . This test is right-tailed.
(Use a computer or calculator to find the p-value. You should get p-value = 0.5578 .)

The dfs are the number of cells - 1 = 5 - 1 = 4 .

TI-83+ and TI-84: Press 2nd DISTR . Arrow down to χ 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.)

Blank nonsymmetrical chi-square curve for the test statistic of the days of the week absent.

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.

TI-83+ and some TI-84 calculators do not have a special program for the test statistic for the goodness-of-fit test. The next example (Example 11-3) hasthe calculator instructions. The newer TI-84 calculators have in STAT TESTS the test Chi2 GOF . To run the test, put the observed values (the data) into a first list and the expected values (thevalues you expect if the null hypothesis is true) into a second list. Press STAT TESTS and Chi2 GOF . Enter the list names for the Observed list and the Expected list. Enter the degrees of freedom and press calculate or draw . Make sure you clear any lists before you start. See below.
To Clear Lists in the calculators: Go into STAT EDIT and arrow up to the list name area of the particular list. Press CLEAR and then arrow down. The list will be cleared. Or, you can press STAT and press 4 (for ClrList ). Enter the list name and press ENTER .
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One study indicates that the number of televisions that American families have is distributed (this is the given distribution for the American population) as follows:

Number of Televisions Percent
0 10
1 16
2 55
3 11
over 3 8

The table contains expected ( E ) percents.

A random sample of 600 families in the far western United States resulted in the following data:

Number of Televisions Frequency
Total = 600
0 66
1 119
2 340
3 60
over 3 15

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Source:  OpenStax, Collaborative statistics. OpenStax CNX. Jul 03, 2012 Download for free at http://cnx.org/content/col10522/1.40
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