Example

Absenteeism of college students from math classes is a major concern to math instructors because missing class appears to increase the drop rate. Suppose that a study was done to determine if the actual student absenteeism rate follows faculty perception. The faculty expected that a group of 100 students would miss class according to this table.

A random survey across all mathematics courses was then done to determine the actual number (observed) of absences in a course. The chart in this table displays the results of that survey.

Determine the null and alternative hypotheses needed to conduct a goodness-of-fit test.

H₀: Student absenteeism fits faculty perception.

The alternative hypothesis is the opposite of the null hypothesis.

H_a: Student absenteeism does not fit faculty perception.

1. Can you use the information as it appears in the charts to conduct the goodness-of-fit test?
2. What is the number of degrees of freedom (df)?

Solution:

1. No. Notice that the expected number of absences for the “12+” entry is less than five (it is two). Combine that group with the “9–11” group to create new tables where the number of students for each entry are at least five. The new results are in the two tables below.
   

   Number of absences per term	Expected number of students
   0–2	50
   3–5	30
   6–8	12
   9+	8

   Number of absences per term	Actual number of students
   0–2	35
   3–5	40
   6–8	20
   9+	5

2. There are four “cells” or categories in each of the new tables.
   df = number of cells – 1 = 4 – 1 = 3

Example

Employers want to know which days of the week employees are absent in a five-day work week. Most employers would like to believe that employees are absent equally during the week. Suppose a random sample of 60 managers were asked on which day of the week they had the highest number of employee absences. The results were distributed as in the table below. 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.

Day of the Week Employees were Most Absent

Solution:

The null and alternative hypotheses are:

• H₀: 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 χ² 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
• [latex]\displaystyle\frac{{({O}-{E})}^{{2}}}{{E}}[/latex]

Now add (sum) the last column. The sum is three. This is the χ² test statistic.

To find the p-value, calculate P(χ² > 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

Press 2nd DISTR. Arrow down to χ2cdf. Press ENTER. Enter(3,10^99,4). Rounded to four decimal places, you should see 0.5578, which is the p-value.

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

The decision is not to 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.

Example

One study indicates that the number of televisions that American families have is distributed (this is the given distribution for the American population) as in the table.

The table contains expected (E) percents.

A random sample of 600 families in the far western United States resulted in the data in this table.

The table contains observed (O) frequency values.

At the 1% significance level, does it appear that the distribution “number of televisions” of far western United States families is different from the distribution for the American population as a whole?

Solution:

This problem asks you to test whether the far western United States families distribution fits the distribution of the American families. This test is always right-tailed.

The first table contains expected percentages. To get expected (E) frequencies, multiply the percentage by 600. The expected frequencies are shown in this table.

Therefore, the expected frequencies are 60, 96, 330, 66, and 48. In the TI calculators, you can let the calculator do the math. For example, instead of 60, enter 0.10*600.

H₀: The “number of televisions” distribution of far western United States families is the same as the “number of televisions” distribution of the American population.

H_a: The “number of televisions” distribution of far western United States families is different from the “number of televisions” distribution of the American population.

Distribution for the test: [latex]\displaystyle\chi^{2}_{4}[/latex] where df = (the number of cells) – 1 = 5 – 1 = 4.

Note: [latex]df\neq600-1[/latex]

Calculate the test statistic:χ² = 29.65

Graph

Probability statement:p-value = P(χ² > 29.65) = 0.000006

Compare α and the p-value:

α = 0.01
p-value = 0.000006

So, α > p-value.

Make a decision: Since α > p-value, reject H_o.

This means you reject the belief that the distribution for the far western states is the same as that of the American population as a whole.

Conclusion: At the 1% significance level, from the data, there is sufficient evidence to conclude that the “number of televisions” distribution for the far western United States is different from the “number of televisions” distribution for the American population as a whole.

Press STAT and ENTER. Make sure to clear lists L1, L2, and L3 if they have data in them (see the note at the end of Example 2). Into L1, put the observed frequencies 66, 119,349, 60, 15. Into L2, put the expected frequencies .10*600, .16*600, .55*600,.11*600, .08*600. Arrow over to list L3 and up to the name area L3. Enter (L1-L2)^2/L2 and ENTER. Press 2nd QUIT. Press 2nd LIST and arrow over to MATH. Press 5. You should see "sum" (Enter L3). Rounded to 2 decimal places, you should see 29.65. Press 2nd DISTR. Press 7 or Arrow down to 7:χ2cdf and press ENTER. Enter (29.65,1E99,4). Rounded to four places, you should see 5.77E-6 = .000006 (rounded to six decimal places), which is the p-value.