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 number of expected values inside each cell needs to be at least five in order to use this test.

### Example 11.4

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.

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

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

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 11.5

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 |

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 11.6

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.

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?

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?

### Example 11.7

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.