In this type of hypothesis test, you determine whether the data **"fit"** a particular distribution or not. For example, you may suspect your unknown data fit a binomial distribution. You use a chi-square test (meaning the distribution for the hypothesis test is chi-square) to determine if there is a fit or not. **The null and the alternative hypotheses for this test may be written in sentences or may be stated as equations or inequalities.**

The test statistic for a goodness-of-fit test is:

where:

*O*= observed values (data)*E*= expected values (from theory)*k*= the number of different data cells or categories

**The observed values are the data values and the expected values are the values you would expect to get if the null hypothesis were true.** There are *n* terms of the form $\frac{{(O-E)}^{2}}{E}$.

The number of degrees of freedom is *df* = (number of categories – 1).

**The goodness-of-fit test is almost always right-tailed.** If the observed values and the corresponding expected values are not close to each other, then the test statistic can get very large and will be way out in the right tail of the chi-square curve.

## Note

The expected value for each cell needs to be at least five in order for you to use this test.

## Example 11.1

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 Table 11.1.

Number of absences per term | Expected number of students |
---|---|

0–2 | 50 |

3–5 | 30 |

6–8 | 12 |

9–11 | 6 |

12+ | 2 |

A random survey across all mathematics courses was then done to determine the actual number **(observed)** of absences in a course. The chart in Table 11.2 displays the results of that survey.

Number of absences per term | Actual number of students |
---|---|

0–2 | 35 |

3–5 | 40 |

6–8 | 20 |

9–11 | 1 |

12+ | 4 |

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

** H_{0}:** Student absenteeism

**fits**faculty perception.

The alternative hypothesis is the opposite of the null hypothesis.

** H_{a}:** Student absenteeism

**does not fit**faculty perception.

### Problem

a. Can you use the information as it appears in the charts to conduct the goodness-of-fit test?

### Solution

a. **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 Table 11.3 and Table 11.4.

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 |

### Problem

b. What is the number of degrees of freedom (*df*)?

### Solution

b. There are four "cells" or categories in each of the new tables.

*df* = number of cells – 1 = 4 – 1 = 3

## Try It 11.1

A factory manager needs to understand how many products are defective versus how many are produced. The number of expected defects is listed in Table 11.5.

Number produced | Number defective |
---|---|

0–100 | 5 |

101–200 | 6 |

201–300 | 7 |

301–400 | 8 |

401–500 | 10 |

A random sample was taken to determine the actual number of defects. Table 11.6 shows the results of the survey.

Number produced | Number defective |
---|---|

0–100 | 5 |

101–200 | 7 |

201–300 | 8 |

301–400 | 9 |

401–500 | 11 |

State the null and alternative hypotheses needed to conduct a goodness-of-fit test, and state the degrees of freedom.

## Example 11.2

### Problem

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 Table 11.7. 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.

Monday | Tuesday | Wednesday | Thursday | Friday | |
---|---|---|---|---|---|

Number of Absences | 15 | 12 | 9 | 9 | 15 |

### Solution

The null and alternative hypotheses are:

*H*: The absent days occur with equal frequencies, that is, they fit a uniform distribution._{0}*H*: The absent days occur with unequal frequencies, that is, they do not fit a uniform distribution._{a}

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} - $\frac{{(O\text{\u2013}E)}^{2}}{E}$

Now add (sum) the last column. The sum is three. 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

## Using the TI-83, 83+, 84, 84+ Calculator

Press `2nd DISTR`

. Arrow down to `χ`

. Press ^{2}cdf`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.

## Using the TI-83, 83+, 84, 84+ Calculator

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 has the 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 (the values 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. **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. Alternatively, you can press `STAT`

and press 4 (for `ClrList`

). Enter the list name and press `ENTER`

.

## Try It 11.2

Teachers want to know which night each week their students are doing most of their homework. Most teachers think that students do homework equally throughout the week. Suppose a random sample of 56 students were asked on which night of the week they did the most homework. The results were distributed as in Table 11.8.

Sunday | Monday | Tuesday | Wednesday | Thursday | Friday | Saturday | |
---|---|---|---|---|---|---|---|

Number of Students | 11 | 8 | 10 | 7 | 10 | 5 | 5 |

From the population of students, do the nights for the highest number of students doing the majority of their homework occur with equal frequencies during a week? What type of hypothesis test should you use?

## Example 11.3

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 Table 11.9.

Number of Televisions | Percent |
---|---|

0 | 10 |

1 | 16 |

2 | 55 |

3 | 11 |

4+ | 8 |

The table contains expected (*E*) percents.

A random sample of 600 families in the far western United States resulted in the data in Table 11.10.

Number of Televisions | Frequency |
---|---|

0 | 66 |

1 | 119 |

2 | 340 |

3 | 60 |

4+ | 15 |

Total = 600 |

The table contains observed (*O*) frequency values.

### Problem

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 Table 11.11.

Number of Televisions | Percent | Expected Frequency |
---|---|---|

0 | 10 | (0.10)(600) = 60 |

1 | 16 | (0.16)(600) = 96 |

2 | 55 | (0.55)(600) = 330 |

3 | 11 | (0.11)(600) = 66 |

over 3 | 8 | (0.08)(600) = 48 |

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.

*H _{0}*: 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: ${\chi}_{4}^{2}$ where *df* = (the number of cells) – 1 = 5 – 1 = 4.

## Note

*df* ≠ 600 – 1

**Calculate the test statistic:** *χ*2 = 29.65

**Graph:**

**Probability statement:** *p*-value = *P*(*χ*^{2} > 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.

## Using the TI-83, 83+, 84, 84+ Calculator

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 11.2). Into `L1`

, put the observed frequencies `66`

, `119`

, `340`

, `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.

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 (the values 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.

## Try It 11.3

The expected percentage of the number of pets students have in their homes is distributed (this is the given distribution for the student population of the United States) as in Table 11.12.

Number of Pets | Percent |
---|---|

0 | 18 |

1 | 25 |

2 | 30 |

3 | 18 |

4+ | 9 |

A random sample of 1,000 students from the Eastern United States resulted in the data in Table 11.13.

Number of Pets | Frequency |
---|---|

0 | 210 |

1 | 240 |

2 | 320 |

3 | 140 |

4+ | 90 |

At the 1% significance level, does it appear that the distribution “number of pets” of students in the Eastern United States is different from the distribution for the United States student population as a whole? What is the *p*-value?

## Example 11.4

### Problem

Suppose you flip two coins 100 times. The results are 20 *HH*, 27 *HT*, 30 *TH*, and 23 *TT*. Are the coins fair? Test at a 5% significance level.

### Solution

This problem can be set up as a goodness-of-fit problem. The sample space for flipping two fair coins is {*HH*, *HT*, *TH*, *TT*}. Out of 100 flips, you would expect 25 *HH*, 25 *HT*, 25 *TH*, and 25 *TT*. This is the expected distribution. The question, "Are the coins fair?" is the same as saying, "Does the distribution of the coins (20 *HH*, 27 *HT*, 30 *TH*, 23 *TT*) fit the expected distribution?"

**Random Variable:** Let *X* = the number of heads in one flip of the two coins. *X* takes on the values 0, 1, 2. (There are 0, 1, or 2 heads in the flip of two coins.) Therefore, the **number of cells is three**. Since *X* = the number of heads, the observed frequencies are 20 (for two heads), 57 (for one head), and 23 (for zero heads or both tails). The expected frequencies are 25 (for two heads), 50 (for one head), and 25 (for zero heads or both tails). This test is right-tailed.

** H_{0}**: The coins are fair.

** H_{a}**: The coins are not fair.

**Distribution for the test:**
${\chi}_{2}^{2}$
where
*df* = 3 – 1 = 2.

**Calculate the test statistic:** *χ*^{2} = 2.14

**Graph:**

**Probability statement:** *p*-value = *P*(*χ*^{2} > 2.14) = 0.3430

**Compare α and the p-value:**

*α*= 0.05*p*-value = 0.3430

*α* < *p*-value.

**Make a decision:** Since *α* < *p*-value, do not reject *H _{0}*.

**Conclusion:** There is insufficient evidence to conclude that the coins are not fair.

## Using the TI-83, 83+, 84, 84+ Calculator

Press `STAT`

and `ENTER`

. Make sure you
clear lists `L1`

, `L2`

, and `L3`

if they have data in them. Into `L1`

, put the observed
frequencies `20`

, `57`

, `23`

. Into `L2`

, put the expected frequencies `25`

, `50`

, `25`

. 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 two decimal places, you
should see `2.14`

. Press `2nd DISTR`

. Arrow down to `7:χ2cdf`

(or press `7`

). Press
`ENTER`

. Enter `2.14,1E99,2)`

. Rounded to four places, you should see `.3430`

, which
is the p-value.

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 (the values 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.

## Try It 11.4

Students in a social studies class hypothesize that the literacy rates across the world for every region are 82%. Table 11.14 shows the actual literacy rates across the world broken down by region. What are the test statistic and the degrees of freedom?

MDG Region | Adult Literacy Rate (%) |
---|---|

Developed Regions | 99.0 |

Commonwealth of Independent States | 99.5 |

Northern Africa | 67.3 |

Sub-Saharan Africa | 62.5 |

Latin America and the Caribbean | 91.0 |

Eastern Asia | 93.8 |

Southern Asia | 61.9 |

South-Eastern Asia | 91.9 |

Western Asia | 84.5 |

Oceania | 66.4 |