Econ 57 Fall 2000 Final Exam Answers

1. The probability of five heads is(1/2)^5 = 1/32; the probability of five tails is (1/2)^5 = 1/32. Therefore, the probability of five heads or five tails is 2/32 = 0.0625.

2. Using Bayes' Rule,

3. The natural null hypothesis is that every day of the week is equally likely to have an ozone Health Advisory: thus there is a 5/7 probability that such an advisory will occur on a weekday and a 2/7 probability that it will occur on a weekend. Using the binomial model, if p = 2/7 the probability of 8 or more successes in 12 trials is 0.00687. On the other hand, the binomial model is not really appropriate, since ozone buildups tend to carry over from one day to the next, so that each day is not an independent trial.

4. Yes, suppose the data look like this:

5. The difference in the heart attack rates (4/412 = 0.0097 versus 17/358 = 0.04749) does seem substantial; those on the low-fat diet were nearly five times more likely to experience a second heart attack. For a statistical test, the pooled proportion is (4 + 17)/(412 + 358) = 0.02727, and the Z value is

The 2-sided p value is 0.0014

6. A test of the null hypothesis that the average Pomona College student gains 15 pounds during the first year at college surveyed 100 students and obtained a sample mean of 4.82 pounds with a standard deviation of 5.96 pounds. Explain why you either agree or disagree with each of these conclusions:
a. Actually, they use the standard deviation of the sample to estimate the standard deviation of the population. The use of the t distribution in place of the normal distribution takes into account the fact that this is an estimate, rather than the actual value.
b. With this minuscule p value, the result is highly significant. The probability of observing this large a difference between the sample mean and the population mean specified by the null hypothesis is virtually 0.
c. Certainly the results of another survey would be somewhat different; however, the calculated p value takes into account the sample size. If this was indeed a random sample, it is unlikely that the mean of another sample would be very different and, in particular, would not reject the null hypothesis.

7. If F value of 0.0 (no evidence whatsoever against the null hypothesis), P = 1.

8. Here are the expected values were the injuries by sport independent of gender:

 Females Males Total Basketball 251.63 306.37 558 Soccer 176.32 214.68 391 Swimming 87.93 107.07 195 Tennis 73.96 90.04 164 Track 173.16 210.84 384 Total 763 929 1692
Females are injured far more often than expected (were the null hypothesis true) in swimming, and slightly more often than expected in tennis. The chi-square value is 132.28; with (5 - 1)(2 - 1) = 5 degrees of freedom, the p value is 0.000001.

9. For the poll i, label the number surveyed as ni and the fraction of those surveyed who prefer Bush ai = xi/ni. The overall fraction preferring Bush is

A 95% confidence interval is
For these particular data,
and a 95% confidence interval is
His actual percentage was 48%.