2. The most appropriate graphs are
a. time series graph
b. scatter diagram
c. scatter diagram
d. scatter diagram
e. side-by-side boxplots
3. a. .912 = .2824
b. .96 = .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 difference-in-means or chi-square test.
A difference-in-means test gives:
where p-hat = (1641 + 2142)/(3332 + 3808). This difference is highly significant
(p < 0.000001) and seems substantial< 49.25% versus 56.25%.
A chi-square test gives chi-square = 34.96, which (except for rounding) is
equal to the Z-value squared. The p-value 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 chi-square test. There were a total of 773 deaths.
If a death during this 8-week period is equally likely to be in any week, the
expected value for each week is 773/8 = 96.625. The chi-square value is
With 8 - 1 = 7 degrees of freedom, the p-value 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;29:258-266.]
8. The first ball doesnt 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 t-value is much larger than 2.
b. Holding constant team batting average and earned run average (X1 and X2),
a million-dollar 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:
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.
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
Alternatively, we can use Bayes Rule: