You can also download a PDF copy of this homework assignment.
Instructions: The following exercises will test your understanding of goodness-of-fit tests and tests of independence using the \(X^2\) test statistic. For each test conduct four steps (state hypotheses, compute the test statistic, compute the p-value, and make a decision). For goodness-of-fit tests in this homework assignment the degrees of freedom will always be one less than the number of possible outcomes, whereas for tests of independence the degrees of freedom will always be \((r-1)(c-1)\) where \(r\) and \(c\) represent the number of rows and columns in the table of counts. Use a significance level of \(\alpha\) = 0.05 for all tests.
| Outcome | Shape | Color |
|---|---|---|
| RRYY | round | yellow |
| rRYY | round | yellow |
| RRyY | round | yellow |
| rRyY | round | yellow |
| rRYY | round | yellow |
| rrYY | wrinkled | yellow |
| rRyY | round | yellow |
| rryY | wrinkled | yellow |
| RRyY | round | yellow |
| rRyY | round | yellow |
| RRyy | round | green |
| rRyy | round | green |
| rRyY | round | yellow |
| rryY | wrinkled | yellow |
| rRyy | round | green |
| rryy | wrinkled | green |
| Traits | Probability | Observed | Expected |
|---|---|---|---|
| round and yellow | 9/16 | 315 | |
| wrinkled and yellow | 3/16 | 101 | |
| round and green | 3/16 | 108 | |
| wrinkled and green | 1/16 | 32 |
The probabilities come from the first table (e.g., 9 of 16 outcomes result in an offspring with peas that are round and yellow).1 The table also shows the observed counts from one of Mendel’s experiments to determine if the Law of Independent Assortment applies to the traits of shape and color in pea plants. The expected counts are not given, so you will need to compute them for what follows. Conduct a goodness-of-fit test to determine whether or not the probabilities given above fit the data.
| Level | Probability |
|---|---|
| 0 | 0.0625 |
| 1 | 0.2500 |
| 2 | 0.3750 |
| 3 | 0.2500 |
| 4 | 0.0625 |
Timmy’s friends examined 100 of the Wizards & Wyverns characters he created, each with three attributes, for a total of 300 attributes. Of these 300 attributes, 15 were at Level 0, 59 were at Level 1, 112 were at Level 2, 81 were at Level 3, and 33 were at Level 4. These are the observed counts based on a sample size of 300. Conduct a goodness-of-fit test to determine if there is evidence that Timmy is cheating by using the null hypothesis that the probabilities given above are correct (note that these probabilities were derived under the assumption that Timmy was not cheating).
Recall the example from lecture of my daughter playing the game Pounce. This was one of the examples I used when I introduced the concept of a significance test. Suppose she plays 50 trials of Pounce. On each trial she is presented with three words, one of the three words is spoken, and then she needs to choose the correct word. Suppose she is correct 25 times, and thus incorrect 25 times. If she guesses the correct word her probability of being correct is 1/3, and her probability of being incorrect is 2/3. If \(p\) is the probability of a correct response then we can conduct a significance test of the hypotheses \(H_0\!: p = 1/3\) versus \(H_a\!: p > 1/3\). The test statistic is \[ z = \frac{\hat{p} - p}{\sqrt{p(1-p)/n}}, \] where \(\hat{p} = 25/50\), \(p = 1/3\), and \(n = 50\). This yields a test statistic of \(z = 2.5\) and a p-value of approximately 0.006. Another approach would be to use a goodness-of-fit test. Let \(p_c = 1/3\) and \(p_w = 2/3\) be the probabilities of a correct and a wrong response from Milena assuming she is guessing. Conduct a goodness-of-fit test assuming these probabilities for the null hypothesis. You should find that the value of the \(X^2\) test statistic is equal to the square of the \(z\) test statistic (i.e., \(z^2\) = 6.25), and that the p-value is (approximately, due to rounding) twice as large as that given above (i.e., 0.012, this is because the goodness-of-fit test is implicitly two-sided whereas the p-value given above was for a one-sided test).
| Applicant | yes | no | Total |
|---|---|---|---|
| male | 21 | 3 | 24 |
| female | 14 | 10 | 24 |
| Total | 35 | 13 | 48 |
In an earlier homework you conducted a significance test of the null hypotheses \(H_0\!: p_m-p_f=0\) where \(p_m\) and \(p_f\) denote the probability that it will be decided to promote a male or female applicant, respectively. The test statistic was \[ z = \frac{\hat{p}_m-\hat{p}_f}{\sqrt{\hat{p}(1-\hat{p})(1/n_m + 1/n_f)}}, \] where \(\hat{p}_m = 21/24\), \(\hat{p}_f = 14/24\), \(\hat{p} = 35/48\), \(n_m = 24\), and \(n_f = 24\). The value of the test statistic was \(z \approx 2.274\) and the p-value (assuming a two-sided test) is then approximately 0.023. By definition if \(p_m=p_f\) (as assumed by the null hypothesis above) then the gender of the applicant and the decision are independent. Thus a test of this null hypothesis can also be done using the \(X^2\) test statistic for a test of independence. Conduct a test of independence for applicant gender and the promotion decision using the data given above. You should find that your \(X^2\) test statistic is equal to the square of the \(z\) test statistic given earlier (approximately, due to rounding error) and that the p-value is the same.3
| Temp | male | female | Total |
|---|---|---|---|
| 27.2 | 2 | 25 | 27 |
| 27.7 | 17 | 7 | 24 |
| 28.3 | 26 | 4 | 30 |
| 28.4 | 19 | 8 | 27 |
| 29.9 | 27 | 1 | 28 |
| Total | 91 | 45 | 136 |
| Temp | male | female | Total |
|---|---|---|---|
| 27.2 | 18.07 | 8.93 | 27 |
| 27.7 | 16.06 | 7.94 | 24 |
| 28.3 | 20.07 | 9.93 | 30 |
| 28.4 | 18.07 | 8.93 | 27 |
| 29.9 | 18.74 | 9.26 | 28 |
| Total | 91 | 45 | 136 |
Conduct a test of independence of incubation temperature and sex. Note that the expected counts are given, but you might find it useful to check that you know how to compute them based on the observed counts.
| CHD | Low | Medium | High | Total |
|---|---|---|---|---|
| yes | 23 | 35 | 51 | 109 |
| no | 1356 | 603 | 416 | 2375 |
| Total | 1379 | 638 | 467 | 2484 |
Conduct a test of independence of snoring frequency and coronary heart disease.
The null hypothesis that the probabilities given in the table (i.e., the 9:3:3:1 ratio) are correct, and the alternative hypothesis is that they are incorrect. The value of the test statistic is \(X^2 \approx 0.47\). The p-value (based on 3 degrees of freedom) is approximately 0.925. Thus we would not reject the null hypothesis. The Law of Independent Assortment appears to fit these traits. In modern terms there does not appear to be linkage between shape and color.
The null hypothesis is that the probabilities of each attribute level are as given in the first table, while the alternative hypothesis is that these probabilities are incorrect. The value of the test statistic is \(X^2 \approx 15.476\), and the p-value (using a degrees of freedom of 4) is approximately 0.004. This would lead us to reject the null hypothesis and conclude that the probabilities assuming that Timmy is not cheating do not fit the observed counts. There is some evidence that he is (at least sometimes) cheating when determining the attributes for his characters.
The null hypothesis is that the probabilities \(p_c = 1/3\) and \(p_w = 2/3\) are correct. The value of the test statistic is \(X^2 = 6.25\), which yields a p-value (using one degree of freedom) of approximately 0.012. This would lead us to reject the null hypothesis, and thus conclude that Milena was not guessing.
The null hypothesis is that the gender of the applicant and the promotion decision are independent, while the alternative hypothesis is that they are not independent. The value of the test statistic is \(X^2 \approx 5.169\), which yields a p-value (based on one degree of freedom) of approximately 0.023. As noted in the problem these are related to the results obtained using the \(z\) test statistic. The decision is to reject the null hypothesis and conclude that the gender of the applicant and the promotion decision are not independent (i.e., they are associated).
The null hypothesis is that incubation temperature and turtle sex are independent, and the alternative hypothesis is that they are not independent. The value of the test statistic using the given expected values is \(X^2 \approx 59.823\) (if the expected counts are computed more precisely the test statistic comes out to about \(X^2 \approx 59.799\)). The p-value (based on 4 degrees of freedom) is very small so the decision is to reject the null hypothesis that incubation temperature and turtle sex are not independent (i.e., there is an association between these two variables).
The null hypothesis is that coronary heart disease (yes or no) and snoring frequency (low, medium, or high) are independent. The alternative hypothesis is that they are not independent. The value of the test statistic is \(X^2 \approx 73.66\), and the p-value (using a degrees of freedom of 2) is very small. This leads us to reject the null hypothesis that coronary heart disease and snoring frequency are independent and conclude that there is evidence of an association.
These probabilities can also be derived by assuming that the dominant and recessive traits have probabilities of 3/4 and 1/4, respectively, and multiplying probabilities together assuming independence. For example, \(P(\text{wrinkled and yellow}) = P(\text{wrinkled})P(\text{yellow}) = 3/4 \times 1/4 = 3/16\).↩︎
Rosen, B. & Jerdee, J. (1974). Influence of sex role stereotypes on personnel decisions. Journal of Applied Psychology, 59, 9–14.↩︎
This relationship between the two approaches only applies two tests of independence with counts in a table with two rows and two columns. If there is more than two rows or more than two columns (as in the following problems) then only the \(X^2\) test statistic can be used.↩︎
I have not been able to track down the original source of these data. I assume that they are real but I am not sure.↩︎
Norton, P. G. & Dunn, E. V. (1985). Snoring as a risk factor for disease: An epidemiological survey. British Medical Journal, 291, 630–632.↩︎