2. The most appropriate graphs are
a. time series graph
b. scatter diagram
c. scatter diagram
d. scatter diagram
e. sidebyside boxplots
3. a. .9^{12} = .2824
b. .9^{6} = .5314
c. Using the binomial distribution,
4. The null hypothesis is that the probability of throwing a double does not depend on whether or not a double was thrown in the previous inning. We could use a differenceinmeans or chisquare test.
A differenceinmeans test gives:
where phat = (1641 + 2142)/(3332 + 3808). This difference is highly significant (p < 0.000001) and seems substantial< 49.25% versus 56.25%.
A chisquare test gives chisquare = 34.96, which (except for rounding) is equal to the Zvalue squared. The pvalue is the same.
5. Maybe there was an increase in smuggling between 1998 and 2000. [David Cole and John Lamberth, “The Fallacy of Racial Profiling,” The New York Times, May 13, 2001.]
6. We can use a chisquare test. There were a total of 773 deaths. If a death during this 8week period is equally likely to be in any week, the expected value for each week is 773/8 = 96.625. The chisquare value is
With 8  1 = 7 degrees of freedom, the pvalue is 0.1633, not sufficient to reject the null hypothesis at the 5% level. In addition, there were more deaths in the the four weeks before Passover than during the four weeks after Passover, the opposite of what was predicted.
7. Maybe the boys who used computers to play games did so BECAUSE they tended to exercise less, engage in fewer recreational activities, and have less social support than their peers, not vice versa. [Natalie Engler, “Boys Who Use Computers May Be More Active: Study, ” Reuters Health, October 22, 2001. (SOURCE: Journal of Adolescent Health 2001;29:258266.]
8. The first ball doesn’t matter. The second ball must land in one of the three remaining boxes (which has a 3/4 probability of occurring). If it does, then the third ball must land in one of the two remaining boxes (which has a 2/4 probability of occurring). If it does, then the fourth ball must land in the remaining box (which has a 1/4 probability of occurring). Using the multiplication rule, the probability of winning is (3/4)(2/4)(1/4) = 6/64.
9. As the number of binomial trials increases, it is increasingly certain that the success proportion x/n will be close to the success probability p, but less likely that it will be exactly equal to p. Thus the probability
a. decrease
b. increases
c. decrease
10. A guesser has a 1/n chance of getting the correct answer. The expected value is (+1)(1/n) + (x)((n  1)/n). This is equal to 0 if 0 = 1  x(n  1), or x = 1/(n  1). For example, if there are n = 5 possible answers, x = 1/4. Thus A guess will get 1 point one out of five times and lose a quarter point four out of five times.
11. Adjusting the bar heights for the fact that the last interval is twice as wide as the first two intervals, a histogram looks like this:
12. The sample mean is 46.00 and the standard deviation is 11.85, and the appropriate t value with 100 – 1 = 99 degrees of freedom is 1.98, giving a 95% confidence interval of
13. a. Yes, the tvalue is much larger than 2.
b. Holding constant team batting average and earned run average (X1 and X2), a milliondollar increase in the payroll is predicted to reduce the number of games won by 7.
c. The coefficient b3 measures the effect of a higher payroll on winning for teams that have equal offenses and defenses. That is, if two teams have equally good offenses and defenses, will the team with the higher payroll win? But the real question is whether teams with higher payrolls have better offenses and defenses.
14. Using the binomial distribution to test the null hypothesis that the probability of making a winning wager is 0.50, the probability of picking more than 901 winners in 1733 games is 0.0463.
15. Here are the p values:
a. As in Exercise 14 use the binomial distribution to determine that if the probability of making a winning wager is 0.50, the probability of picking at least 902 winners in 1733 games is 0.0463 and the probability of the probability of picking at least 883 winners in 1743 games is 0.2991. The probability of both occurring is 0.0463(0.2991) = 0.0138.
b. Similarly, if the probability of making a winning wager is 0.50, the probability of picking at least 1785 winners in 3476 games is 0.057
16. The p value in (a) is smaller. Both calculations involve 1785 winners in 3476 games, but (a) restricts the possible ways in which you can get this many winners.
17. Using a contingency table with 300 experiments, of which 100 involve a coin with two heads, 100 a coin with two tails, and 100 a coin with one head and one tail:
lands heads

lands tails

total


two heads

100

0

100

two tails

0

100

100

one head, one tail

50

50

100

total

150

150

300

Thus, after the coin lands heads, the probability that it is a coin with two
heads is 100/150 = 2/3, the probability that it is a coin with two tails is
0/150 = 0, and the probability that it is a coin with one head and one tail
is 50/150 = 1/3.
Alternatively, we can use Bayes’ Rule:
18. These data do not reject the null hypothesis.
19. We can use least squares regression, using a t statistic to test the null hypothesis that the slope coefficient is 0.
20. These data do not show that 95 percent (or even a majority) of shoppers are between the ages of 37.7 and 42.5. If the ages have a normal distribution with a mean of 40.1 and a standard deviation of 8.6, the fraction of the shoppers between 37.7 and 42.5 years of age is