Sample Questions for Exam 4 - Answers

Note - there are more questions in this sample than there will be in the real exam

Test significance at alpha = 0.05 unless otherwise indicated.

1. A psychologist is testing a group of 40 TV addicts with a word association game, in which the subject responds to a word with the first word that pops into his head. The psychologist wants to determine if these people are abnormal in their responses by comparing the frequency of each response to those in the "normal" population. Are these people a bit off-the-mark? (Data also in Excel worksheet.)
   

   Test word: "Blanket"
   
   

   Response	Frequency in general population (%)	Number of test subjects giving response
   Bed	55	10
   Warm	25	6
   Comfort	10	2
   Leroy	0	13
   Tuberculosis	0	7
   Other response	10	2

   
   This question was a bit trickier than I intended. Ideally, you will have some finite probability of any outcome included in the predicted category. If predicited categories have a count of zero, it's best to either leave them out of the analysis or treat the expected number as 0.5. The chi-square in the first case is 10.15 (DF = 3), and 407.15 (DF = 5) in the second. Either is significant.
   
   Another possible test would be log-likelihood.

2. You want to test the hypothesis that the likelihood of retinal detachment is related to eye color. You examine 100 people with detached retinas and 100 control subjects and obtain the counts shown below. Create a null hypothesis and test it using an appropriate method.
   

   Detached retina:	Blue eyes = 32	Brown eyes = 51	Other = 17
   Normal:	Blue eyes = 37	Brown eyes = 40	Other = 23

   
   The null hypothesis is that there is no difference in the proportion of normal and detached retinas among different eye colors. Test via a contingency table gives chi-square of 2.592 with DF=2, giving P=0.27 - no significant effect of eye color. Another test would be log-likelihood.

3. One hundred random people were surveyed and asked whether a) they smoked and b) they coughed heavily in the morning. Is there a difference in the proportion of coughers between smokers and non-smokers? Use any method desired to address this question. (Data also in Excel worksheet.)

   	Smoke?
   Cough?	Yes	No
   Yes	18	7
   No	12	63

   
   For Haber's correction (chi-square), the smallest f-hat is 7.5, giving d = 10.5 and D = 10. The final chi-square is 25.40 (DF = 1), a highly significant result. (The uncorrected chi-square is 28.00.) This means there is a difference in the proportion of coughers between smokers and non-smokers.

4. A study has just been completed examining the issue of mate choice as it affects that bizarre creature, the science geek. In an experiment representing a cross between Survivor and Club Med, 50 geeks and 100 normal people were left on a tropical island, and a year later researchers assayed who was romantically involved with whom. They found that, of the 75 couples, 40 were normal/normal, 20 were geek/geek, and 15 were normal/geek. Do geeks and normals appear to choose their mates at random, or is there a bias in their selection?
   

   The expected number of each type of couple is 33.4 normal/normal, 8.3 geek/geek, and 33.4 geek/normal. The resulting chi-square (other tests are possible) is 27.80, a highly significant difference, suggesting that each type prefers to mate with their own kind.

5. During an epidemic in a city, you are trying to determine whether new cases of the disease are appearing at random, or instead tend to cluster together. You have a map of the city showing how many cases have appeared in each square block. What hypothesis might you want to test, and how would you go about testing it?
   

   If the disease is appearing at random, then the number of cases per block out to have the distribution expected for random events (Poisson). To test this, take the total number of cases recorded and the number of blocks surveyed and divide the first by the second to get the average events per block, then create a table of expected counts per block if the appearances were random. Compare the observed counts to this random expectation using chi-square (or similar). If a difference is found, the events are not random.
   

6. You are studying the pattern of depolarization in a neuron from the brain of a banana slug. Over a two-hour period, you measure the number of depolarizations per minute (data in Excel worksheet). Can you reject the hypothesis that the timing of the depolarizations is random?

   First, you need to determine the average number of events per minute - 3.784. Then use this value to generate the expected distribution of events/min if the events are random. Comparison to the observed pattern gives a chi-square of 191.43 (DF = 9), which is highly significant, so the depolarizations are not random.
   

7. A new stop-smoking product, the Oral Patch, is being tested for effectiveness in keeping people from starting to smoke again while using the product. The patch, which is placed over the user's mouth, is being compared to a more traditional nicotine-laced skin patch. A "survival" table is created for the 100 people tested for each product, showing how many individuals started smoking or were lost each week. Using this data (in the Excel worksheet), finishing constructing the suvival tables and then compare the effectiveness of the two products.

   Comparison of the two life tables by the Mantel-Cox method (and ignoring lost individuals when calculating death rates) gives a chi-square of 3.898 and P = 0.048. The oral patch is signficantly more effective than the skin patch.