» 3.1%. So z = -16.4/3.1 » -5, and P » 0. The difference is real. (Students who used the calculator seemed to forget what the arithmetic was all about.)HOMEWORK #9 ANSWERS
Chapter 27
2. (a) The SE for the difference is 5.9%, so z = 2.4/5.9 » 0.4; looks like chance.
(b) The SE for the difference is 1.8; the observed difference is 4.9; so z = 4.9/1.8 »
2.7 and P » 0.3 of 1%. The difference looks real.
Comment. There is more information in the sample average than in the number of positive terms, at least for this example.
6. This is just like the previous exercise. In the calculator group, 7.2% get the right
answer; in the pencil-and-paper group, 23.6%. The SEs are 1.6% and 2.7%. The difference between the percentages is -16.4%, and the SE for the difference is conservatively estimated as » 3.1%. So z = -16.4/3.1 » -5, and P » 0. The difference is real. (Students who used the calculator seemed to forget what the arithmetic was all about.)
9. This is like the radiation-surgery example in section 4. In the positive group, the
percentage accepted is 28/53 x 100% » 52.8%; in the negative group, 14.8%. The SEs are 6.9% and 4.8%. The difference is 38% and the SE for the difference is 8.4%. So z = 38/8.4 » 4.5 and P » 0%. There is a big difference between the two groups, and the difference cannot be explained by chance: journals prefer the positive articles.
Chapter 28
3. Use the method of section 4. The expecteds are as follows:
|
|
Married |
Widowed, divorced or separated |
Never married |
|
Employed |
654.1 |
109.3 |
132.6 |
|
Unemployed |
67.9 |
11.3 |
13.8 |
|
Not in labor force |
62.0 |
10.4 |
12.6 |
chi 2 » 30 on 4 degrees of freedom, which is off the end of the table; by computer, P < 5/106. This is not chance variation. The married men do better at getting jobs. (Or, men with jobs do better at getting married: the chi 2 -test will not tell you which is the cause and which is the effect.)
5. (a) With 10 degrees of freedom, P will be bigger. Reason: that curve has more
area to the right of 15.
(b) The P-value is bigger when
chi 2 = 15. Reason: the area under the curve to theright of 15 is bigger than the area to the right of 20.
7. Use the method of sections 1-2:
chi 2 » 0.2 on 2 degrees of freedom, P » 90%, agood fit.
Chapter 29
4. Yes: data snooping (p.549).
Chapter 29 Special Review Exercises
1. (a) This is an observational study. It is the prisoners themselves who decide whether to stay in the program or drop out.
(b) The treatment group consists of the prisoners who finish boot camp. The control group consists of those who drop out.
(c) Those who stayed the course might have been quite different to start with--
better motivated, more self-disciplined--than those who dropped out.
(d) (i) The treatment group consists of all those who volunteer for the program,
whether or not they complete it. The control group consists of those who do not volunteer.
(ii) The recidivism rate in the two groups is similar, suggesting that the program
has little effect--although it is an effective screening device. (The ones who finish are motivated and have enough self-discipline to go straight when they are released.) [Comment. If anything, this comparison is biased in favor of treatment, because those who volunteer would seem more likely to succeed in civilian life than those who do not volunteer.]
(e) Most completed. The recidivism rate for the combined group is 36%; the rate
for those who did not complete is 74%; for those who completed, the rate was 29%. If most dropped out, the rate for the combined group would be nearly 74%. If most complete, the rate for the combined group would be nearly 29%--and it is. A more exact answer is possible. Let x be the fraction who complete. Then 29x + 74(1 - x) = 36, so x
» 0.84.18. Using a telephone survey tends to exclude the homeless or recently homeless. On
this basis, the 3% is a little too low. Also, people might not want to admit having been homeless; that also makes the 3% too low. On the other hand, if people were homeless, they might remember that as being more recent than it really was. (Vivid experiences tend to be brought forward in memory.) That would make the 3% too high.
33. Model: there is one ticket in the box for each household in the country, marked 1 if
the household experienced a burglary within the last 12 months, and 0 otherwise.
There are 100 million tickets in this box. The Survey data are like 50,000 draws made at random from the box. (There is essentially no difference between drawing with or without replacement.) If the FBI data are accurate, the percentage of 1's in the box is 3.2%: this is the null hypothesis. In the null hypothesis, the 4.9% is higher than the 3.2% just because of sampling error. The alternative hypothesis: the percentage of 1's in the box is bigger than 3.2%.
If the null hypothesis is right, the percentage of 1's has an EV (expected value) of 3.2%, and the SE is 0.08 of 1%. (The SD of the box should be computed using the 3.2% specified by the null hypothesis.) The percentage of 1's in the sample is 4.9%. The difference between 4.9% and 3.2% is almost impossible to explain as chance variation: z
» 20. Many burglaries are not reported to the police.