Download Lesson 8 Homework Solutions

Solutions - Comparing Two Groups
10.1 Wealth gap:
a) The response variable is net worth and the explanatory variable is race.
b) The two groups that are the categories of the explanatory variable are white and black
households.
c) The samples of white and black households were independent. No household could
be in both samples.
10.2 Sampling sleep:
a) The samples on weekdays and weekends should be treated as dependent samples
because every person is in both samples.
b) When compared with other people from another year, the samples should be treated
as independent. No one person is in both samples.
10.3 Binge drinking:
a) The estimated difference between the population proportions in 2005 and 1993 is
0.07. The proportion of students who reported bingeing at least 3 times within the
past 2 weeks has apparently increased between 1993 and 2005.
b) The standard error is the standard deviation of the sampling distribution of differences
between the sample proportions.
se =
pˆ 1 1  pˆ 1  pˆ 2 1  pˆ 2 
=

n1
n2
0.3121  0.312  0.3821  0.382 =0.0429
159
485
c) 0.07 – (1.96)(0.0429) = -0.01
0.07 + (1.96)(0.0429) = 0.15
(-0.01, 0.15)
We can be 95% confident that the population mean change in proportion is between 0.01 and 0.15. This confidence interval contains zero; thus, we do not have enough
evidence to conclude that there was an increase in the population proportion of UW
students who reported binge drinking at least 3 times in the past 2 weeks between
1993 and 2005.
d) The assumptions are that the data are categorical (reported binge drinking at least 3
times in the past 2 weeks vs. did not), that the samples are independent and are
obtained randomly, and that there are sufficiently large sample sizes. Specifically,
each sample should have at least ten “successes” and ten “failures.”
10.11 Hormone therapy for menopause:
a) Assumptions: Each sample must have at least ten outcomes of each type. The data
must be categorical, and the samples must be independent random samples.
Notation: p is the probability that someone developed cancer.
H 0 : p1  p2 ; H a : p1  p2
b) The test statistic is 1.03, and the P-value is 0.303 (rounds to 0.30). If the null
hypothesis were true, the probability would be 0.30 of getting a test statistic at least as
extreme as the value observed. It is plausible that the null hypothesis is correct and
that there are the results for the hormone therapy group are not different from the
results for the placebo group.
c) We cannot reject the null hypothesis.
10.16 Housework for women and men:
a) The study estimated that, on the average, women spend 32.6 – 18.1 = 14.5
b) The standard error for comparing the means is
s1
s
18.22  12.92  0.297
 2 
n1
n2
6764
4252
The standard error is so small compared to the sample standard deviations for the two
groups because the sample sizes are so large.
c) We can be 95% confident that the difference between the population mean scores of
women and men falls between 13.9 and 15.1. Because zero is not in the interval, we
can conclude that there is a mean difference between the populations. It appears that
the population mean for women is higher than the population mean for men is.
d) The assumptions are that the data are quantitative, both samples are independent and
random, and there is approximately a normal population distribution for each group.
2
2
se 
10.25 Females or males more nicotine dependent?:
s1
s
3.62  2.92  0.364
 2 
n1
n2
150
182
The standard error is the standard deviation of the difference between sample from
different studies using these sample sizes.
x  x2   0  2.8  1.6  0  3.297 (rounds to 3.30)
b) t  1
se
0.364
P-value: 0.001
If the null hypothesis were true, the probability would be 0.001 of getting a test
statistic at least as extreme as the value observed. We have very strong evidence that
there is a difference between men’s and women’s population mean HONC scores.
The females seem to have a higher population mean HONC score than the males.
c) The HONC scores were probably not normal for either gender. The standard
deviations are bigger than the means, an indication of skew. The lowest possible
value of 0 is 2.8/3.6 = 0.778 standard deviations below the mean for females, and
1.6/2.9 = 0.552 standard deviations below the mean for males, indicating skew to the
right in both cases. This does not affect the validity of our inference greatly because
of the robustness of the two-sided test for the assumption of a normal population
distribution for each group.
2
a) se 
2
10.27 TV watching and gender:
a) H 0 : 1   2 ; H a : 1   2
b) The test statistic is 1.26 and the P-value is 0.21. If the null hypothesis were true, the
probability would be 0.21 of getting a test statistic at least as extreme as the value
observed. With a significance level of 0.05, we cannot reject the null hypothesis.
Because the probability is 0.21 of observing our test statistic or one more extreme,
due to random variation, we have insufficient evidence that there is a gender
difference in TV watching.
c) Yes, since we failed to reject the null hypothesis of no difference between the
population means, 0 is a plausible value for this difference.
d) The distribution of TV watching is not likely normal. The standard deviations are
almost as large as the means (the lowest possible value of 0 is 2.99/2.34 = 1.28
standard deviations below the mean for females and 2.86/2.22 = 1.29 standard
deviations below the mean for males), an indication of skew to the right. This does
not affect the validity of our inference greatly because of the robustness of the twosided test for the assumption of a normal population distribution for each group.
Inferences assume a randomized study and normal population distributions.
10.34 Bulimia CI:
a) The standard error for comparing the means:
s
n1  1s1 2  n2  1s 2 2
se  s
b)
n1  n2  2

13  14.41  17  110.24
13  17  2
 2.782
1 1
1 1

 2.782

 1.025
n1 n2
13 17
x1  x2   t.025 se 
(2.0 – 4.8) – (2.048)(1.025) = - 4.90
(2.0 – 4.8) + (2.048)(1.025) = - 0.70
The confidence interval of (- 4.90, - 0.70) indicates that we can be 95% confident that
the population mean difference is between – 4.90 and -0.70. Because 0 does not fall
in this interval, we can conclude that, on average, the sexually abused students had
lower family cohesion than the non-abused students did.
10.36 Surgery vs. placebo for knee pain:
a) The confidence interval is (- 8.32, 8.72). We can be 95% confident that the population
mean difference is between – 8.32 and 8.72. Because 0 falls in this interval, it is
plausible that there is no difference between the debridement arthroscopic surgery
and placebo groups in terms of population mean knee pain scores.
b) 1) We assume independent random samples from the two groups, an approximately
normal population distribution for each group (particularly if the sample sizes are
small), and equal population standard deviations.
2) H 0 : 1   2 ; H a : 1   2
3) T-value: 0.05
4) P-value: 0.963 (rounds to 0.96)
5) If the null hypothesis were true, the probability would be 0.96 of getting a test
statistic at least as extreme as the value observed. It is plausible that the null
hypothesis is correct and that there is no mean population difference between pain
scores for the debridement arthroscopic surgery and placebo groups.
c) Based on the confidence interval and test, we would not conclude that the
arthroscopic surgery works better than placebo. According to the confidence interval,
a difference of 0 is plausible, and according to the significance test, we would get a
difference at least this extreme 0.963 of the time if the null hypothesis were true.
10.47 Does exercise help blood pressure?:
a) The three “before” observations and the three “after” observations are dependent
samples because the same patients are in both samples.
b) The sample mean of the “before” scores is 150, of the “after” scores is 130, and of the
difference scores is 20. The difference between the means for the “before” and “after”
scores is the same as the mean of the difference scores.
c) From software, the standard deviation of the difference scores is 5.00.
s
5
se  d 
 2.887
n
3
xd  t.025 se 
20 – (4.303)(2.887) = 7.577
20 + (4.303)(2.887) = 32.423
95% confidence interval: (7.577, 32.423) which rounds to (7.6, 32.4)
We can be 95% confident that the difference between the population means is
between 7.6 and 32.4. Because 0 is not included in this interval and because all
differences are positive, we can conclude that there is a decrease in blood pressure
after patients go through the exercise program.