Chapter 9 Homework
Set A
9.8 Time status versus gender for the 20-24 age category. Refer to Exercise 9.7. The table below breaks down the 20-24 age category by gender.
GenderStatus / Male / Female / Total
Full-time / 2719 / 2991 / 5710
Part-time / 535 / 680 / 1215
Total / 3254 / 3671 / 6925
a) Compute the marginal distribution for gender – find the corresponding proportions. Display the results graphically.
(b) Compute the conditional distribution of status (this statement “of status” implies that gender will used as the whole –denominator of fraction) for males and for females. Display the results graphically and comment on how these distributions differ.
(c) If you wanted to test the null hypothesis that there is no difference between these two conditional distributions, what would the expected cell counts be for the full-time status row of the table?
Full-time / 2683.082 / 3026.918 / 5710
(d) Computer software gives X2 = 5.17. Use Excel’s =chidist(f, df) calculator for chi-square to find the P-value ( or use http://www.stat.tamu.edu/~west/applets/chisqdemo.html) and state your conclusions at the 5% level.
P(X2 > 5.17) = chidist(5.17, 1)
= 0.0230
The result shows that we do have evidence that there is an association between the two variables. You might be asking why there would be since the two bargraphs look identical for male and female. The answer lies in the power of this test. Our sample size is 6,925, thus even if there is a small insignificant difference, this test can pick up on that small difference. So the result is significant at 5%, but it does not mean that the result offers a practical difference, meaning a large difference.
9.10 Waking versus bedtime symptoms. As part of the study on ongoing fright symptoms due to exposure to horror movies at a young age, the following table was presented to describe the lasting impact the movies have had during bedtime and waking life:
Waking SymptonsBedtime Symptons / Yes / No / Total
Yes / 36 / 33 / 69
No / 33 / 17 / 50
Total / 69 / 50 / 119
(a) What percent of the students have lasting waking-life symptoms?
69/119 = 0.5798 57.98% of students in the study had waking-life symptons.
(b) What percent of the students have both waking' life and bedtime symptoms?
36/119 = 0.3025 30.25% of students suffered from both waking and bedtime symptons.
(c’) Create the conditional distribution of bedtime symptoms (this statement “of bedtime” implies that waking will used as the whole –denominator of fraction). Create a bar graph. Does the data suggest there is an association between waking symptoms and bedtime symptoms?
Bedtime Symptons / Yes / No / Total
Yes / 0.52173913 / 0.66 / 0.579832
No / 0.47826087 / 0.34 / 0.420168
Total / 1 / 1 / 1
I can see that the percentage in the No column seem to be further apart than the yes column.
(c) Test whether there is an association between waking-life and bedtime symptoms. State the null and alternative hypotheses, the X2 statistic, and the P-value.
The null hypothesis says that there is no association between the bedtime and waking symptoms.
The alternative says there is an association.
P(X2 > 2.275) = chidist(2.275, 1)
= 0.1315
The result shows that we do not have evidence that there is a n asssociation (relationship) between the waking symptoms and the bedtime symptoms.
You might be asking why since the graph
shows a difference in the second set of bar graphs. The answer most likely lies in the sample size. This test lacks power to detect the difference between the two, meaning there probably is a difference but not that much of one, and the test lacks the power to detect that difference.
9.11 New treatment for cocaine addiction. Cocaine addiction is difficult to overcome. Addicts have been reported to have a significant depletion of stimulating neurotransmitters and thus continue to take cocaine to avoid feelings of depression and anxiety. A 3-year study with 72 chronic cocaine users compared an antidepressant drug called desipramine with lithium and a placebo. (Lithium is a standard drug to treat cocaine addiction. A placebo is a substance containing no medication, used so that the effect of being in the study but not taking any drug can be seen.) One-third of the subjects, chosen at random, received each treatment.14 Following are the results:
Cocaine RelapseTreatment / Yes / No
Desipramine / 10 / 14
Lithium / 18 / 6
Placebo / 20 / 4
(a) Compare the effectiveness of the three treatments in preventing relapse using percents and a bar graph. Write a brief summary.
ConditionalCocaine Relapse
Treatment / Yes / No
Desipramine / 0.416667 / 0.583333
Lithium / 0.75 / 0.25
Placebo / 0.833333 / 0.166667
The table shows the conditional distribution of cocaine relapse based on (the denominator) the type of treatment they received. I can see that the percentage of people who relapsed is different for each group. It looks like the placebo group had the biggest percentage of relapase while the desipramine group had the least occurance of relapse.
(b) Can we comfortably use the chi-square test to test the null hypothesis that there is no difference between treatments? Explain.
Expected / CountCocaine Relapse
Treatment / Yes / No / Total
Desipramine / 16 / 8 / 24
Lithium / 16 / 8 / 24
Placebo / 16 / 8 / 24
Total / 48 / 24 / 72
Page 536, in the summary explains that we need at least an expected cell count average of 5 or greater and we meet that criteria; our average is 12.
(c) Perform the significance test and summarize the results.
The null hypothesis states there is no relationship between the treatment groups and relapse.
The alternative says there is an association.
P(X2 > 10.49998) = chidist(10.49998, 2)
= 0.00542
Clearly the result indicates there is a relationship, since our p-value is 0.00542.
Notice that our sample size is 72, yet the result is statistically significant unlike the result from the previous problem, which had 119.
The reason is that the differences are larger and it also occurred more than once since the treatment had a factor level of 3.