Download Homework 9 Solutions 1. Note that we have the following summary

Homework 9 Solutions
1. Note that we have the following summary statistics:
19.8 a) It is important that the loggers were unaware of the effects of logging were being
assessed. The reason is that loggers may change the amount of logging if they
knew that they were being studied. It is also important that the plots were
randomly assigned to make sure that we are can rule out soil and nutrients as
confounders and just see the effect of logging on the average number of species in
a plot.
b)
Ho: µl = µu
Ho: µl < µu
µl - µu = 0
µl - µu < 0
µdiff = 0
µdiff < 0
There are 8 degrees of freedom (using the conservative procedure recommended
by the book). The pvalue (one-sided) is P(T < -2.185).
We reject the null hypothesis since we have .025 < pvalue < .05. The p-value is
small and conclude that there is an effect of logging on the average number of
species in a plot.
19.10 Below is a 90% confidence interval for the true difference in the average number
of species between unlogged and logged plots using t with 8 d.f.
We are 90% confident that on average the logged plots have between .57 and 7.09
less species compared to the unlogged plots.
2. a) Below is the stem and leaf plot for group a. We note that group a has 24 observations.
The mean score for group a is 24.42 and the standard deviation is 6.31. From the plot we see that
the distribution is skewed right.
1
2*
2.
3*
3.
4*
|
|
|
|
|
|
|
567
0011112334444
78889
1
5
4
Below is a table of the summary statistics for group a:
Variable |
Obs
Mean
Std. Dev.
Min
Max
-------------+-------------------------------------------------------score |
24
24.41667
6.310217
15
44
Below is the stem and leaf plot for group p. We note that group p also has 24
observations. The mean score for group p is 17.88 and the standard deviation is 4.03. From the
plot we see that the distribution is also skewed right.
1t
1f
1s
1.
2*
2t
2f
2s
|
|
|
|
|
|
|
|
223
45555
66777
889
00111
5
66
Below is a table of the summary statistics for group p:
Variable |
Obs
Mean
Std. Dev.
Min
Max
-------------+-------------------------------------------------------score |
24
17.875
4.025057
12
26
b)
Below is a table of output for the two-sample t test with unequal variances.
-----------------------------------------------------------------------------Group |
Obs
Mean
Std. Err.
Std. Dev.
[95% Conf. Interval]
---------+-------------------------------------------------------------------a |
24
24.41667
1.288068
6.310217
21.7521
27.08124
p |
24
17.875
.8216114
4.025057
16.17537
19.57463
---------+-------------------------------------------------------------------combined |
48
21.14583
.8937279
6.191929
19.34788
22.94378
---------+-------------------------------------------------------------------diff |
6.541667
1.527797
3.451552
9.631782
-----------------------------------------------------------------------------diff = mean(a) - mean(p)
t =
4.2818
Ho: diff = 0
Satterthwaite's degrees of freedom = 39.0578
Ha: diff < 0
Pr(T < t) = 0.9999
i.
ii.
iii.
iv.
v.
Ha: diff != 0
Pr(|T| > |t|) = 0.0001
Ha: diff > 0
Pr(T > t) = 0.0001
The 95% confidence interval for the mean number of symbols learned for the passive
group is (16.17, 19.57).
The 95% confidence interval for the mean number of symbols learned for the active
group is (21.75, 27.08).
The 95% confidence interval for the difference in the mean number of symbols of the
active group compared to the passive group is (3.45, 9.63).
The pvalue for the two sided test is Pr(|T| > |t|) = 0.0001. The alternative hypothesis
is Ha: µdiff not= 0.
The conservative d.f. recommended by the text is 23 for this problem. Stata tells us
that the more precise approximation has 39.06 degrees of freedom. The conservative
procedure would have a higher (but still very significant) p-value.
3.
Below is the summary information for the results of the first and second tries on
the verbal section of the SAT.
n
mean
s.d.
Attempt 1
427 500
92
Attempt 2
427 529
97
a)
The two samples have the same people in them, so the test scores will be related
for the two samples. Some people tend to do better on tests compared to others. We have
a matched pairs situation.
b)
Ho: µ2 = µ1
Ho: µ2 > µ1
µ2 - µ1 = 0
µ2 - µ1 > 0
µdiff = 0. Mean score on both attempts is equal.
µdiff > 0. Population mean on attempt 2 is greater
population mean on attempt 1
Compute the one-sample t-statistic as
The pvalue is P(t > 10.16). We should use the t-distribution with 100 d.f. in our table.
All we learn there is that P < .0005 but that is small enough! Since the p-value is small,
we reject the null hypothesis and conclude that the two attempts have different means.
c)
This study only shows that people who have coaching tend to do better on their
second attempt than their first attempt. However these data don’t provide any thing to
compare this finding to ... e.g., how would people do on their second attempt without any
coaching? (In fact it turns out that people generally do better on their 2nd attempt, with or
without coaching, because they have learned something about how to do the test.) Since
we do not have a control group we do not know if the improvement is due to the coaching
or due to the fact that the students have learned from taking the SAT before.
4.
Below is a table for the mean change (test 2 – test 1) in verbal SAT scores and
the standard deviation of the changes for the coached and uncoached students.
n
mean
s.d.
Group C: Coached
427 29
59
Group U: Uncoached
1233 21
52
Ho: µc = µu µc - µu = 0 µdiff = 0
The mean improvement in verbal SAT scores is the same for the populations of coached
and uncoached students.
Ha: µc > µu µc - µu > 0 µdiff > 0
The mean improvement in verbal SAT scores is larger for the population of coached
students compared to the population of uncoached students.
Compute the two-sample t-statistic as
The pvalue is P(t > 2.49) = 1- P(t < 2.49) (with 426 d.f.) = .0066 (from a calculator).
Since the pvalue is small, we reject the null hypothesis and conclude that the mean
improvement in verbal SAT scores is larger for the coached students compared to the
uncoached students. The results of our sample suggest a statistically significant difference
in population means.
b) Using two one-sample confidence intervals is not as appropriate as a two-sample
method. In this case the fact that the very upper limit of one one-sample CI overlaps with
the very lower limit of the other does not mean it’s plausible the two means are the same.
It takes two implausible happenings – the first population mean needs to be near the
tippy-top of its interval and the second population mean needs to be near the bottom of its
interval. The two-sample method gives precise inference for the difference in the two
means and correctly tells us that it is not very likely that the two means are the same.
c) The test in (a) indicates the difference between coached/uncoached students is
statistically significant. This question asks whether the difference – 8 points – is
important in practice. Admissions officers would tell you that it would be better to spend
your time on an extracurricular activity rather than spend time on a coaching course that
will raise your score 8 points! (I’m not sure they are telling the truth but that’s what they
would tell you.)
d) This is an observational study. It was a survey of students who took the SAT twice
and then recorded whether they had coaching or not. Because it is observational, we still
can’t conclude that the significant difference is due to coaching (though this analysis is
much better than the one in question 3). There may be remaining confounders that
explain the difference – students who got coaching may be wealthier, have higher HS
grades, be more dedicated, etc.
5. a) The treatments are the different seat belt laws (more specifically the fact that
enforcement is different). The response is whether or not the driver is wearing their
seatbelt. This is an observational study since the seatbelt laws already existed in the two
cities.
b)
Only Hispanic females are used to control for possible confounders. For example,
suppose females are more likely to wear seatbelts. In addition, suppose Boston has more
females than New York. Then it would be possible to see a higher proportion of people
not wearing their seatbelts in New York compared to Boston. This may be due to the fact
that New York has less females and not due to the seatbelt laws. So we concentrate on
only one group, female Hispanics, in order to control for the effect of gender and race.
c)
A 95% confidence interval for the true proportion of female Hispanic drivers who
wear their seatbelts in NYC is between:
d)
A 95% confidence interval for the difference in the proportion of female Hispanic
drivers who wear their seatbelts in NYC compared to Boston is given by:
e)
Since 0 is not in the confidence interval calculated in part d, we would conclude
that the proportion of female Hispanic drivers who wear their seatbelts is different in
NYC and Boston; it is about .25 less in Boston. Because it is an observational study we
can’t be sure that it is due to the difference in laws. One possible remaining confounder
(remember it can’t be gender or race) is the age distribution of people in the cities.