Goodness-of-Fit Test

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

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 is at least five. The new results are in M03_Ch03_tbl003 and M03_Ch03_tbl004.

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

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

df = number of cells – 1 = 4 – 1 = 3.

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 tab_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 percent significance level.

The null and alternative hypotheses are as follows:

• 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
• $\frac{{\left(O\text{ – }E\right)}^2}{E}$

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.

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 percent level of significance, from the sample data, there is not sufficient evidence to conclude that the absent days do not occur with equal frequencies.

At the 1 percent significance level, does it appear that the distribution number of televisions of far western U.S. families is different from the distribution for the American population as a whole?

This problem asks you to test whether the far western U.S. 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 M03_Ch03_tbl011.

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 U.S. families is the same as the number of televisions distribution of the American population.

H_a: The number of televisions distribution of far western U.S. 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.

Calculate the test statistic: χ² = 29.65

Graph

Probability statement: p-value = P(χ² > 29.65) = .000006

Compare α and the p-value:

• α = .01
• p-value = 0.000006

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

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

Conclusion: At the 1 percent 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.

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 percent significance level.

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.

H₀: 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: χ² = 2.14.

Graph

Probability statement: p-value = P(χ² > 2.14) = 0.3430.

Compare α and the p-value:

• α = .05
• p-value = 0.3430

Make a decision: Since α < p-value, do not reject H₀.

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

An archeologist is calculating the distribution of the frequency of the number of artifacts she finds in a dig site. Based on previous digs, the archeologist creates an expected distribution broken down by grid sections in the dig site. Once the site has been fully excavated, she compares the actual number of artifacts found in each grid section to see if her expectation was accurate.

An economist is deriving a model to predict outcomes on the stock market. He creates a list of expected points on the stock market index for the next two weeks. At the close of each day’s trading, he records the actual points on the index. He wants to see how well his model matched what actually happened.

a goodness-of-fit test

A personal trainer is putting together a weight-lifting program for her clients. For a 90-day program, she expects each client to lift a specific maximum weight each week. As she goes along, she records the actual maximum weights her clients lifted. She wants to know how well her expectations met with what was observed.

$df=$ ______

3

State the null and alternative hypotheses.

χ² test statistic = ______

2.04

p-value = ______

At the 5 percent significance level, what can you conclude?

We decline to reject the null hypothesis. There is not enough evidence to suggest that the observed test scores are significantly different from the expected test scores.

If the ethnicities of patients followed the ethnicities of the total county population, fill in the expected number of cases per ethnic group.
Perform a goodness-of-fit test to determine whether the occurrence of disease cases follows the ethnicities of the general population of Santa Clara County.

H₀: _______

H₀: the distribution of disease cases follows the ethnicities of the general population of Santa Clara County.

H_a: _______

Is this a right-tailed, left-tailed, or two-tailed test?

right-tailed

degrees of freedom = _______

χ² test statistic = _______

2016.136

p-value = _______

Graph the situation. Label and scale the horizontal axis. Mark the mean and test statistic. Shade in the region corresponding to the p-value.

Let α = 0.05.

Decision: ________________

Reason for the decision: ________________

Conclusion (write out in complete sentences): ________________

Graph: Check student’s solution.

Decision: Reject the null hypothesis.

Reason for decision: p-value < alpha

Conclusion: The make-up of cases does not fit the ethnicities of the general population of Santa Clara County.

Does it appear that the pattern of disease cases in Santa Clara County corresponds to the distribution of ethnic groups in this county? Why or why not?

A six-sided die is rolled 120 times. Fill in the expected frequency column. Then, conduct a hypothesis test to determine if the die is fair. The data in id6180063 are the result of the 120 rolls.

The marital status distribution of the U.S. male population, ages 15 and older, is as shown in id7856744.

Suppose that a random sample of 400 U.S. males, 18 to 24 years old, yielded the following frequency distribution. We are interested in whether this age group of males fits the distribution of the U.S. adult population. Calculate the frequency one would expect when surveying 400 people. Fill in id7856744, rounding to two decimal places.

1. The data fit the distribution.
2. The data do not fit the distribution.
3. 3
4. chi-square distribution with df = 3
5. 19.27
6. 0.0002
7. Check student’s solution.
8. 1. Alpha = 0.05
   2. Decision: Reject null hypothesis.
   3. Reason for decision: p-value < alpha
   4. Conclusion: Data do not fit the distribution.

Perform a goodness-of-fit test to determine whether the local results follow the distribution of the U.S. overall student population based on ethnicity.

Perform a goodness-of-fit test to determine whether the local results follow the distribution of U.S. AP examinee population, based on ethnicity.

1. H₀: The local results follow the distribution of the U.S. AP examinee population.
2. H_a: The local results do not follow the distribution of the U.S. AP examinee population.
3. df = 5
4. chi-square distribution with df = 5
5. chi-square test statistic = 13.4
6. p-value = 0.0199
7. Check student’s solution.

8. 1. Alpha = 0.05
   2. Decision: Reject null when a = 0.05.
   3. Reason for decision: p-value < alpha
   4. Conclusion: Local data do not fit the AP examinee distribution.
   5. Decision: Do not reject null when a = 0.01
   6. Conclusion: There is insufficient evidence to conclude that local data do not follow the distribution of the U.S. AP examinee distribution.

The city of South Lake Tahoe, California, has an Asian population of 1,419 out of a total population of 23,609. Suppose that a survey of 1,419 self-reported Asians in the borough of Manhattan in the New York City area yielded the data in id7017535. Conduct a goodness-of-fit test to determine if the self-reported subgroups of Asians in Manhattan fit that of the South Lake Tahoe area.

Conduct a goodness-of-fit test to determine if the actual college majors of graduating females fit the distribution of their expected majors.

1. H₀: The actual college majors of graduating females fit the distribution of their expected majors.
2. H_a: The actual college majors of graduating females do not fit the distribution of their expected majors.
3. df = 10
4. chi-square distribution with df = 10
5. test statistic = 11.48
6. p-value = 0.3211
7. Check student’s solution.
8. 1. Alpha = 0.05
   2. Decision: Do not reject null hypothesis when a = 0.05 and a = 0.01.
   3. Reason for decision: p-value > alpha
   4. Conclusion: There is insufficient evidence to conclude that the distribution of actual college majors of graduating females do not fit the distribution of their expected majors.

Conduct a goodness-of-fit test to determine if the actual college majors of graduating males fit the distribution of their expected majors.

In a goodness-of-fit test, the expected values are the values we would expect if the null hypothesis were true.

true

In general, if the observed values and expected values of a goodness-of-fit test are not close together, then the test statistic can get very large and on a graph will be way out in the right tail.

Use a goodness-of-fit test to determine if high school principals believe that students are absent equally during the week.

true

The test to use to determine if a six-sided die is fair is a goodness-of-fit test.

In a goodness-of-fit test, if the p-value is 0.0113, in general, do not reject the null hypothesis.

false

A sample of 212 commercial businesses was surveyed for recycling one commodity; a commodity here means any one type of recyclable material such as plastic or aluminum. fs-idm103073456 shows the business categories in the survey, the sample size of each category, and the number of businesses in each category that recycle one commodity. Based on the study, on average half of the businesses were expected to be recycling one commodity. As a result, the last column shows the expected number of businesses in each category that recycle one commodity. At the 5 percent significance level, perform a hypothesis test to determine if the observed number of businesses that recycle one commodity follows the uniform distribution of the expected values.

fs-idm106414640 contains information from a survey of 499 participants classified according to their age groups. The second column shows the percentage of obese people per age class among the study participants. The last column comes from a different study at the national level that shows the corresponding percentages of obese people in the same age classes in the United States. Perform a hypothesis test at the 5 percent significance level to determine whether the survey participants are a representative sample of the USA obese population.

The hypotheses for the goodness-of-fit test are:

• H₀: Surveyed obese fit the distribution of expected obese.
• H_a: Surveyed obese do not fit the distribution of expected obese.

Use a chi-square distribution with df = 4 to evaluate the data.

• The test statistic is χ² = 9.85
• The p-value = 0.0431
• At the 5% significance level, α = 0.05. For this data, p < α.
• At the 5% level of significance, from the data, there is sufficient evidence to conclude that the surveyed obese do not fit the distribution of expected obese.