# Answers to selected homework problems by P2GZaf

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```									             Answers to selected homework problems (Ch 16-21)

Chapter 16

4a.    2.064 (use df = 24)
b.     0.685

6.     3.939 to 7.061 mean percent changes in level

8a.    df = 14
b.     t is between 1.761 and 2.145. The probabilities for these entries are p = .05 and p
= 0.025
c.     P-value is between p = .025 and .05.
d.     Significant at the  = 0.05 level, not significant at the  = 0.01 level.

10.    t is approx –8.87 (from table), so P < .001. Conclusion: We would expect a
sample mean of 59.59 percent of nitrogen less than 0.1% of the time if the null
hypothesis were true. Thus there is very strong evidence for Ha, that Cretaceous
air is quite different from modern air.

14a.   H0:  = 0; Ha:  > 0. Use a one-sided alternative because researchers believe
that CO2 will increase the growth rate.
b.     Using x = 1.916, and S = 1.05, we find P = .0436. Thus, about 4% of the time we
would have a sample mean such as this when the actual mean was really 0.
c.     Because n = 3 is such a small sample, we can’t tell if this is Normally distributed.

16a.   H0:  = \$0, Ha:  > \$0; t = 43.47 and P = 0. Thus, we reject H0. The new policy
would increase credit card usage.
b.     \$312 to \$352
c.     The sample size is large, and we have a SRS.
c.     Choose another SRS of size 200 without the offer as a control group and repeat
the study comparing the mean increase.

18.    The mean number of unexcused absences for all workers is 8.10 to 11.66 days.

28.    Since 2-sided use p = .0025. The critical value is 3.174. So we would reject H0
for t < -3.174 or t > 3.174.

Chapter 17

6a.    t = 7.36 means that if df = k is reasonable large, the t(k) distribution is similar to
the Normal distribution, wo we can estimate the P values by this. Based on the
68-95-99.7 rule, we would almost never get a value as large as 7.36 from the
N(0,1) distribution, so we can say P is essentially 0. Thus, the result is
significant.
b.     df = 32.
24a.   H0: 1 = 2 (where 1 = men’s rage, and 2 = women’s rage)
Ha: 1 > 2
t = 6.129, P is essentially 0.
Thus, there is extremely strong evidence for Ha, that the male mean is
higher than the female mean.

b.     They contacted only people with telephones, so the entire population may include
poorer people, who may have more rage.

26.    H0: 1 = 2; Ha: 1 < 2; p = .02956. Thus there is strong evidence at the  =
0.03 level for Ha .
Note: this probably should be 2-sided, Ha: 1  2, since researchers wouldn’t
know beforehand. Then p = 0.06.

Chapter 17 on-line quiz:

(If you use your calculator, your answers are more precise than the options given in the
multiple choice for numbers 2, 3, and 4)

2.     34.6  38.59
3.     P  .0487
4.     4  5.85 yards

Chapter 18

4a.    The mean is p = .14 and s = .0155 (approx)
b1.    probability is 0.02% --not likely
b2.    about 26%

6.     The population must be at least 10 times as large as the sample, and hers is not.

16a.   m = .0460 (approx)
b.     n  635

18.    H0: p = .5; Ha: p  .5
So z = -2.40 (approx), and P = .0162 (approx). Thus, strong evidence for Ha.

22.    Approx. 0.4918 to 0.6706

28a.   m = 0.0306 –Yes
b.     We don’t have sample sizes for each gender, but we can figure it out.
c.     Greater than 0.03 since the sample size will be smaller.

Chapter 19
2.     Approx 0.09152 to 0.29081.

6.          ˆ     ˆ        ˆ     ˆ
H0: p t = p c ; Ha: p t > p c ; z = 4.82 (could be negative if you set this up as
control – treatment). Z is off the chart, so p will be tiny, and we declare that the
drug is successful.

8a.    approx 0.60933 to 0.65071.
b.     0.36933 to 0.45064.

12.          ˆ    ˆ        ˆ     ˆ
H0: p b = p w ; Ha: p b < p w ; P = .0124. Since the probability that pb = pw given
this ample proportion is only 1.24%, we conclude that black parents are less likely
to rate schools favorably.

Chapter 20

2a.    C or better with < 2 hours: approx 55%
2-12 hours: 74.7%
> 12 hours: 37.5%

graph should have % on y-axis and three bars.

b.     Some, but not too much time in extra curricular activities seems to be beneficial.

4a.
<2        2-12     > 12      TOTAL
hours     hours    hours
GRADE
C or up 13.8        62.71    5.51     82
D or F  6.22        28.29    2.49     37
TOTAL 20            91       8        119

b.     Compare the expected counts (above) with the observed counts in problem 20.2.
We expected more in the < 2 and > 12 for passing than what was observed and
less in the 2-12 hours category. So again we see that moderate activity seems
beneficial.

10a.   2 = 10.827, P = .013, df = 3. Thus, there is strong evidence for a relationship
between gender and major.

b.     one expected count is less than 5, which is only 1/8, so the chi-square procedure is
acceptable.

c.     Chart of percents:
Female           Male
Accounting     30.2           34.8
Admin          40.4           24.8
Econ           2.2            3.7
Finance        27.1           36.7

The biggest difference is in admin. A higher percentage of women chose this
major. Greater proportions of men chose other fields, especially Finance.

d.     The largest chi-square components are the two from the “Administration” row.
Many more women than we expect (91 actual, 76.36 expected) chose this major,
whereas only 40 men chose this (54.64 expected)

e.     46.5% did not respond.
16.    2 = 41.08, P = 0.0000000258. Thus, we reject H0 and claim that the differences
between US and Canada are statistically significant. We can also see this because
how far away 2 is from the mean of the distribution, which is df = 4.

Chapter 21

2a.    r2 = 0.9915 (rounded). Thus time explains almost all of the change in length.
b.     estimate of     = -2.3948 cm
 = 0.15848 cm/min
 = 0.80588 cm
c.     The regression equation is y = -2.3948 + 0.15848 x

4.     The Minitab output give b = 0.158483 and SEb = 0.003661. There were n = 18
observations, so df = 16. For a 95% confidence interval, we use t* = 2.120. So
0.158483  2.120 SEb = 0.1507 to 0.1662 cm/min.

The slope  represents the average rate in cm/min at which icicles grow under the
conditions of this experiment.

10.    Find the LinReg is y = -22.97x + 260.56; r = -.8428 with n = 19. Table F (p. 661)
tells us P < 0.0005. So there is strong evidence for Ha:  < 0 (note: one-sided).

18a.   The 95% CI for the slope is seen in cells F12 and G12 in Fig 21.9. So the interval
is .0033 to .0138. This can be verified by: b  t*SEb = .0086  (2.145)(.0025).
(Note: Use Table C to find t* for confidence level 95% and df = 16-2 = 14.)

b.     A 90% CI for  is .0086  (1.761)(.0025) = .0042 to .0130.

Chapter 22

2a.    Ho: all age groups have the same mean road-rage measurement.
Ha: at least one group has a different mean.
b.     F = 34.96, P < .01 implies the F test is quite significant, giving strong evidence
that the means are different. The sample means suggest that road rage decreases
with age.

8a.    To use ANOVA make sure the largest sample standard deviation is no more than
33 .1944
twice the smallest. For 22.3 we calculate           1.16 –ok
24 .8106

b.     9.970/8.341  1.20 –ok

22a.   The means for each of the four groups:
Weeks   xi     Si
2     123.8 4.6043
4     123.6 6.5422
8     134.4 9.5289
16     116.4 16.0873
The mean does not consistently decrease over time.

b.     Considering that 16.0873/4.6043 is greater than 2, ANOVA should not be used.

c.     The 2-sample t test does not require that the standard deviations be equal,
ANOVA does.

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