10.2: Chi-Square Goodness of Fit
In probability, you calculated probabilities using both experimental and theoretical methods. There are times when it is important to determine how well the experimental values match the theoretical values. An example of this is if you wish to verify if a die is fair. To determine if observed values fit the expected values, you want to see if the difference between observed values and expected values is large enough to say that the test statistic is unlikely to happen if you assume that the observed values fit the expected values. The test statistic in this case is also the chi-square. The process is the same as for the chi-square test for independence.
Hypothesis Test for Goodness of Fit Test
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State the null and alternative hypotheses and the level of significance
\(H_{o}\): The data are consistent with a specific distribution
\(H_{A}\): The data are not consistent with a specific distribution
Also, state your \(\alpha\) level here. -
State and check the assumptions for the hypothesis test
- A random sample is taken.
- Expected frequencies for each cell are greater than or equal to 5 (The expected frequencies, E , will be calculated later, and this assumption means \(E \geq 5\)).
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Find the test statistic and p-value
Finding the test statistic involves several steps. First the data is collected and counted, and then it is organized into a table (in a table each entry is called a cell). These values are known as the observed frequencies, which the symbol for an observed frequency is O . The table is made up of k entries. The total number of observed frequencies is n . The expected frequencies are calculated by multiplying the probability of each entry, p , times n .
\(\text{Expected frequency( entry }i )=E=n^{*} p\)
Test Statistic :
\(\chi^{2}=\sum \dfrac{(O-E)^{2}}{E}\)
where O is the observed frequency and E is the expected frequency.
Again, the test statistic involves squaring the differences, so the test statistics are all positive. Thus a chi-squared test for goodness of fit is always right tailed.
p-value:
Using the TI-83/84: \(\chi \text { cdf }(\text { lower limit, } 1 \mathrm{E} 99, d f)\)
Using R: \(1-\text { pchisq }\left(\chi^{2}, d f\right)\)
Where the degrees of freedom is df = k - 1
4. Conclusion
This is where you write reject \(H_{o}\) or fail to reject \(H_{o}\). The rule is: if the p-value < \(\alpha\), then reject \(H_{o}\). If the p-value \(\geq \alpha\), then fail to reject \(H_{o}\),
5. Interpretation
This is where you interpret in real world terms the conclusion to the test. The conclusion for a hypothesis test is that you either have enough evidence to show \(H_{A}\) is true, or you do not have enough evidence to show \(H_{A}\) is true.
Example \(\PageIndex{1}\) goodness of fit test using the formula
Suppose you have a die that you are curious if it is fair or not. If it is fair then the proportion for each value should be the same. You need to find the observed frequencies and to accomplish this you roll the die 500 times and count how often each side comes up. The data is in Example \(\PageIndex{1}\). Do the data show that the die is fair? Test at the 5% level.
| Die values | 1 | 2 | 3 | 4 | 5 | 6 | Total |
|---|---|---|---|---|---|---|---|
| Observed Frequency | 78 | 87 | 87 | 76 | 85 | 87 | 100 |
Solution
1. State the null and alternative hypotheses and the level of significance
\(H_{o}\): The observed frequencies are consistent with the distribution for fair die (the die is fair)
\(H_{A}\): The observed frequencies are not consistent with the distribution for fair die (the die is not fair)
\(\alpha\) = 0.05
2. State and check the assumptions for the hypothesis test
- A random sample is taken since each throw of a die is a random event.
- Expected frequencies for each cell are greater than or equal to 5. See step 3.
3. Find the test statistic and p-value
First you need to find the probability of rolling each side of the die. The sample space for rolling a die is {1, 2, 3, 4, 5, 6}. Since you are assuming that the die is fair, then \(P(1)=P(2)=P(3)=P(4)=P(5)=P(6)=\dfrac{1}{6}\).
Now you can find the expected frequency for each side of the die. Since all the probabilities are the same, then each expected frequency is the same.
\(\text{Expected Frequency} =E=n^{*} p=500 * \dfrac{1}{6} \approx 83.33\)
Test Statistic:
It is easier to calculate the test statistic using a table.
| O | E | O-E | \((O-E)^{2}\) | \(\dfrac{(O-E)^{2}}{E}\) |
|---|---|---|---|---|
| 78 | 83.33 | -5.22 | 28.4089 | 0.340920437 |
| 87 | 83.33 | 3.67 | 13.4689 | 0.161633265 |
| 87 | 83.33 | 3.67 | 13.4689 | 0.161633265 |
| 76 | 83.33 | -7.33 | 53.7289 | 0.644772591 |
| 85 | 83.33 | 1.67 | 2.7889 | 0.033468139 |
| 87 | 83.33 | 3.67 | 13.4689 | 0.161633265 |
| Total | 0.02 | \(\chi^{2} \approx 1.504060962\) |
The test statistic is \(\chi^{2} \approx 1.504060962\)
The degrees of freedom are df = k - 1 = 6 - 1 = 5
Using TI-83/84: \(p-\text {value}=\chi^{2} \operatorname{cdf}(1.50406096,1 E 99,5) \approx 0.913\)
Using R: \(p-\text {value}=1-\text { pchisq }(1.50406096,5) \approx 0.9126007\)
4. Conclusion
Fail to reject \(H_{o}\) since the p-value is greater than 0.05.
5. Interpretation
There is not enough evidence to show that the die is not consistent with the distribution for a fair die. There is not enough evidence to show that the die is not fair.
Example \(\PageIndex{2}\) goodness of fit test using technology
Suppose you have a die that you are curious if it is fair or not. If it is fair then the proportion for each value should be the same. You need to find the observed frequencies and to accomplish this you roll the die 500 times and count how often each side comes up. The data is in Example \(\PageIndex{1}\). Do the data show that the die is fair? Test at the 5% level.
Solution
1. State the null and alternative hypotheses and the level of significance
\(H_{o}\): The observed frequencies are consistent with the distribution for fair die (the die is fair)
\(H_{A}\): The observed frequencies are not consistent with the distribution for fair die (the die is not fair)
\(\alpha\) = 0.05
2. State and check the assumptions for the hypothesis test
- A random sample is taken since each throw of a die is a random event.
- Expected frequencies for each cell are greater than or equal to 5. See step 3.
3. Find the test statistic and p-value
Using the TI-83/84 calculator:
Using the TI-83:
To use the TI-83 calculator to compute the test statistic, you must first put the data into the calculator. Type the observed frequencies into L1 and the expected frequencies into L2. Then you will need to go to L3, arrow up onto the name, and type in \((L 1-L 2)^{\wedge} 2 / L 2\). Now you use 1-Var Stats L3 to find the total. See Figure \(\PageIndex{1}\) for the initial setup, Figure 11.2 .2 for the results of that calculation, and Figure \(\PageIndex{3}\) for the result of the 1-Var Stats L3.
Figure \(\PageIndex{1}\): Input into TI-83 Figure \(\PageIndex{2}\): Result for L3 on TI-83 Figure \(\PageIndex{3}\): 1-Var Stats L3 Result on TI-83The total is the chi-square value, \(\chi^{2}=\sum x \approx 1.50406\).
The p-value is found using \(p-\text {value}=\chi^{2} \operatorname{cdf}(1.50406096,1 E 99,5) \approx 0.913\), where the degrees of freedom is df = k - 1 = 6 - 1 = 5.
Using the TI-84:
To run the test on the TI-84, type the observed frequencies into L1 and the expected frequencies into L2, then go into STAT, move over to TEST and choose \(\chi^{2}\) GOF-Test from the list. The setup for the test is in Figure \(\PageIndex{4}\) .
Figure \(\PageIndex{4}\): Setup for Chi-Square Goodness of Fit Test on TI-84Once you press ENTER on Calculate you will see the results in Figure \(\PageIndex{5}\) .
Figure \(\PageIndex{5}\): Results for Chi-Square Test on TI-83/84The test statistic is \(\chi^{2} \approx 1.504060962\)
The p-value \(\approx 0.913\)
The CNTRB represent the \(\dfrac{(O-E)^{2}}{E}\) for each die value. You can see the values by pressing the right arrow.
Using R:
Type in the observed frequencies. Call it something like observed.
observed<- c(type in data with commas in between)
Type in the probabilities that you are comparing to the observed frequencies.
Call it something like null.probs.
null.probs <- c(type in probabilities with commas in between)
chisq.test(observed, p=null.probs) – the command for the hypothesis test
For this example (Note since you are looking to see if the die is fair, then the probability of each side of a fair die coming up is 1/6.)
observed<-c(78, 87, 87, 76, 85, 87)
null.probs<-c(1/6, 1/6, 1/6, 1/6, 1/6, 1/6)
chisq.test(observed, p=null.probs)
Output:
Chi-squared test for given probabilities
data: observed
X-squared = 1.504, df = 5, p-value = 0.9126
The test statistic is \(\chi^{2}=1.504\) and the p-value = 0.9126.
4. Conclusion
Fail to reject \(H_{o}\) since the p-value is greater than 0.05.
5. Interpretation
There is not enough evidence to show that the die is not consistent with the distribution for a fair die. There is not enough evidence to show that the die is not fair.
Homework
Exercise \(\PageIndex{1}\)
In each problem show all steps of the hypothesis test. If some of the assumptions are not met, note that the results of the test may not be correct and then continue the process of the hypothesis test.
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According to the M&M candy company, the expected proportion can be found in Example \(\PageIndex{3}\). In addition, the table contains the number of M&M’s of each color that were found in a case of candy (Madison, 2013). At the 5% level, do the observed frequencies support the claim of M&M?
Blue Brown Green Orange Red Yellow Total Observed Frequencies 481 371 483 544 372 369 2620 Expected Proportion 0.24 0.13 0.16 0.20 0.13 0.14 Table \(\PageIndex{3}\): M&M Observed and Proportions -
Eyeglassomatic manufactures eyeglasses for different retailers. They test to see how many defective lenses they made the time period of January 1 to March 31. Example \(\PageIndex{4}\) gives the defect and the number of defects. Do the data support the notion that each defect type occurs in the same proportion? Test at the 10% level.
Defect type Number of defects Scratch 5865 Right shaped - small 4613 Flaked 1992 Wrong axis 1838 Chamfer wrong 1596 Crazing, cracks 1546 Wrong shape 1485 Wrong PD 1398 Spots and bubbles 1371 Wrong height 1130 Right shape - big 1105 Lost in lab 976 Spots/bubble - intern 976 Table \(\PageIndex{4}\): Number of Defective Lenses -
On occasion, medical studies need to model the proportion of the population that has a disease and compare that to observed frequencies of the disease actually occurring. Suppose the end-stage renal failure in south-west Wales was collected for different age groups. Do the data in Example \(\PageIndex{5}\) show that the observed frequencies are in agreement with proportion of people in each age group (Boyle, Flowerdew & Williams, 1997)? Test at the 1% level.
Age Group 16-29 30-44 45-59 60-75 75+ Total Observed Frequency 32 66 132 218 91 539 Expected Proportion 0.23 0.25 0.22 0.21 0.09 Table \(\PageIndex{5}\): Renal Failure Frequencies -
In Africa in 2011, the number of deaths of a female from cardiovascular disease for different age groups are in Example \(\PageIndex{6}\) ("Global health observatory," 2013). In addition, the proportion of deaths of females from all causes for the same age groups are also in Example \(\PageIndex{6}\). Do the data show that the death from cardiovascular disease are in the same proportion as all deaths for the different age groups? Test at the 5% level.
Age 5-14 15-29 30-49 50-69 Total Cardiovascular Frequency 9 16 56 433 513 All Cause Proportion 0.10 0.12 0.26 0.52 Table \(\PageIndex{6}\): Deaths of Females for Different Age Groups -
In Australia in 1995, there was a question of whether indigenous people are more likely to die in prison than non-indigenous people. To figure out, the data in Example \(\PageIndex{7}\) was collected. ("Aboriginal deaths in," 2013). Do the data show that indigenous people die in the same proportion as non-indigenous people? Test at the 1% level.
Prisoner Dies Prisoner Did Not Die Total Indigenous Prisoner Frequency 17 2890 2907 Frequency of Non-Indigenous Prisoner 42 14459 14501 Table \(\PageIndex{7}\): Death of Prisoners -
A project conducted by the Australian Federal Office of Road Safety asked people many questions about their cars. One question was the reason that a person chooses a given car, and that data is in Example \(\PageIndex{8}\) ("Car preferences," 2013).
Safety Reliability Cost Performance Comfort Looks 84 62 46 34 47 27 Table \(\PageIndex{8}\): Reason for Choosing a Car
- Answer
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For all hypothesis tests, just the conclusion is given. See solutions for the entire answer.
1. Reject Ho
3. Reject Ho
5. Reject Ho