Stat 301 – Review 2
Problems Solutions
1) Weights of 30 (funsize) Mounds candy bars and 20 (funsize) PayDay candy bars, in grams, are shown in the dotplots below.
(a) Which distribution
would you consider skewed to the right?
The
Mounds distribution is a bit skewed to the right and the PayDay distribution is
strongly skewed to the left.
(b) Which
distribution do you expect has a larger mean?
The
PayDay distribution is clearly centered around larger values than the Mounds
distribution.
(c) Which
distribution do you expect has a larger standard deviation?
The
PayDay values are more spread out/less consistent than the Mounds distribution.
In
other words, the Mounds weights are more consistent, but occasionally a few
weigh more. The PayDay distribution is
less predictable and often has weights that are much lower than typical,
perhaps the difference of one or two peanuts?
(d) Which
distribution would you suspect will have its mean larger than its median?
Mounds
because it is skewed to the right
2) The highway miles per gallon rating of the 1999 Volkswagen
Passat was 31 mpg (Consumer Reports, 1999). The fuel efficiency that a driver
obtains on an individual tank of gasoline naturally varies from tankful to
tankful. Suppose the mpg calculations per tank of gas have a mean of = 31 mpg and a standard deviation of = 3 mpg.
(a) Would it be
surprising to obtain 30.4 mpg on one tank of gas? Explain.
Not
really, 30.4 is well within one standard deviation of the “population” mean of
31.
z =
(30.4 – 31)/3 = 0.20
(b) Would it be surprising for a sample of 30
tanks of gas to produce a sample mean of 30.4 mpg or less? Explain, referring
to the CLT and to a sketch that you draw of the sampling distribution.
First,
does the CLT apply here? We don’t know
much about the shape of the population distribution, though it’s reasonable to
assume the mileage from different tanks will by symmetric and roughly normal. But we also don’t care too much because our
sample size of 30 is considered large.
We are also assuming these observations are taken under identical
conditions.
So we
will model the distribution of the averages of 30 tanks for be normally distributed
with mean equal to 31 mpg and standard deviation equal to 3/sqrt(30) = 0.5477
mpg.
So a
sample mean of 30.4 mpg would be (30.4 – 31)/.5477 = 1.095 standard deviations
below the mean. This is still not larger
than 2.
Using
the normal distribution, P(< 30.4)
About
13.7% of random samples of 30 tanks will have an average mileage of 30.4 mpg by
random chance alone. We would probably not
consider this a surprising outcome (happens more than 10% of the time).
(c) Assess the
validity of your calculations in (a) and (b)
It’s always
reasonable to calculate a “standard score” as I did in (a). If I wanted to convert this zvalue to a
probability, then I would need to know that the tank MPGs follow a normal
distribution. We aren’t told that here though it seems a reasonable assumption. As stated in (b), we can use the CLT if we
continue to have this belief in the normality of the MPG values in general or
if it’s not too crazy behaving because then the sample of size 30 tells us that
the distribution of sample means should still be approximately normal.
If
you go to the Sampling from a Finite Population applet and
check the box for Population Model, you can simulate drawing random samples
from a probability distribution rather than a finite population. When the
probability distribution is a normal distribution, everything works very well:
If
the theoretical probability distribution is not normal but symmetric, things
still work pretty well.
If
the theoretical probability distribution is not normal to begin with, things
still work pretty well due to the “large” sample size
3)
The file AgeGuesses.txt
contains students’ guesses of my age on the first day of class a few years ago.
(a) Estimate and interpret a 95%
confidence interval for the population mean.
Confidence interval for the
population mean:
+ t*
(s /) = 48.43 + t* (10.89/sqrt(30))
For 95% confidence, we
estimate t* to be around 2.
48.42 + 2(1.99) =
(44.4, 52.4) years
I’m 95% confident that the
average guess of my age in the population of all Cal Poly students on such an
activity would be between 44.4 and 52.4 years
(More precisely, t* = 2.045)
+ t*
(s /) = 48.43 + 2.045 (10.89/sqrt(30)) = (44.36,
52.50). A little bit wider.
On an exam without the computer, for 95% confidence you can use 2
for either z* or t*. That’s why I said “estimate”
(b) Estimate and interpret a 95%
confidence interval for the next student’s guess of my age.
+ t*
(s ) = 48.43 + 2 (10.89 × sqrt(1+1/30)) = (26.29, 70.6)
I’m 95% confident that any one
Cal Poly student would guess my age between 26.3 and 70.6 years.
(More precisely)
+ t*
(s ) = 48.43 + 2.045 (10.89 × sqrt(1+1/30)) = (25.79,
71.07)
(c) Which interval do you feel is more
meaningful in this context?
Opinions will vary, the prediction
interval is quite wide due to the huge amount of variation in the responses
given to this question. Typically a
prediction interval is more meaningful (what will happen next, vs. what is the
longrun mean), but because it’s so wide this one is not very informative,
basically saying I went to graduate school but I’m still alive!
(d) What information would you need to
know to decide whether students’ are “biased” in how they guess my age in this
activity? If you did a test of
significance, would this be a onesided or a twosided test?
You would need to know my
actual age, then we could see whether the sample mean fell above that
(overestimating my age on average) or below that (underestimating my age on
average).
(e) Evaluate the validity of your
calculations in (a) and (b).
The distribution is pretty
symmetric and the sample size is 30 so the confidence interval in (a) is
probably ok (achieves the stated 95% confidence in the long run), but with the
outliers on both sides, the sample distribution of age guesses has heavier
tails than we might expect for a normally distributed population. If we believe
these long tails exist in the population, then this would cast some doubt as to
the validity of the prediction interval (though again, at least the
distribution is symmetric, but there may be less than 95% of the population
distribution falling within 2 standard deviations of the mean, or more if the
population standard deviation is inflated by such outliers). The nonlinear nature of the normal probably
plot suggests these data are not coming from a normally distributed population.
(f) Column 2 indicates whether the data
were collected in Section 1 or Section 2.
I changed something about my appearance between the two sections.
Suppose I find a statistically significant difference in the average guess of
my age between the two classes, flipping a coin in advance to decide which
appearance I would use in each section. Would you be willing to attribute the
change in the ages to the change I made in my appearance? Explain why or why
not.
While I did randomly assign
the two treatments in a sense, I did so at the class level rather than at the
individual student level. So there could
still be a confounding variable between the two sections (e.g., I looked more
tired later in the day) and we should not draw any causeandeffect conclusion
here. (Actually the average guess was 10 years larger in section 1!)
4)
In a recent study (Klein, Thomas, and Sutter, 2007), researchers found that
current smokers were more likely to have used candy cigarettes as children than
current nonsmokers were.
(a) Identify and classify the
explanatory and response variables.
EV = whether used candy cigarettes as child
RV = whether or not current smoker
(b) When first hearing of this study,
someone responded by saying, “Isn’t the smoking status of the parents a
confounding variable here?”
Explain
what “confounding variable” means in this context, and describe how parents’
smoking status could be confounding (i.e., describe what would need to be
true).
It would be a confounding variable if it provides an alternative
explanation for the observed association. To do this, it must differ between
the explanatory variable groups and potentially impact the response
variable. So if those with smoking
parents are more likely to be allowed to play with candy cigarettes as children
but also more likely to smoke due to the environment they were raised and/or
genetics, then the smoking habits of the parents might better predict who is a
later smoker, but would also explain why current smokers are more likely to
have played with candy cigarettes.
5) Newspaper headlines proclaimed that
chocolate lovers live longer, following the publication of a study titled “Life
is Sweet: Candy Consumption and Longevity” in the British Medical Journal (Lee
and Paffenbarger, 1998). In 1988, researchers sent a health questionnaire to
men who entered Harvard University as undergraduates between 1916 and 1950. The
study included 7841 men, free of cardiovascular disease and cancer. From the
questionnaire they determined whether the respondents consumed candy “almost
never” (3312 men) or “sometimes or often” (4529 men), and then they tracked the
participants to determine whether or not they had died by 1993.
(a) Identify
the observational units.
men
(b) Identify
the response variable.
Whether
or not the person had died by 1993.
(c) Identify
the explanatory variable.
Whether
the person was classified a candy consumer (sometimes or often) or not a candy
consumer (almost never)
(d) Was this an
experiment or an observational study? If an experiment, was it a randomized,
comparative experiment? If observational, was if a casecontrol study? This was an observational study because the candyconsumption
levels were not imposed on the men in
the study, the men in the study chose for themselves. This is probably best classified as a cohort
study because they were identified, their candy consumption determined, and
then followed for 5 years to determine the outcome for the response variable.
This means its legitimate for us to use this data to estimate the probability
of still being alive.
(e) Researchers
found that of respondents who admitted to consuming candy regularly, 267 had
died by the end of 1993, compared to 247 of the nonconsumers of candy. Set up
the calculation for Fisher’s Exact Test for deciding whether candy consumers are
significantly less likely to have died than nonconsumers by completing the
following:
Note:
The conditional proportions of death are 267/4529 = .05895 and 247/3312 =
.07458
Best
bet is to set up the twoway table:

candy
consumer 
nonconsumer 
Total 
still
alive 
4262 
3065 
7327 
Died 
267 
247 
514 
Total 
4529 
3312 
7841 
If we
let X represent the number still alive in the candy consumer group, then we want
to find above X (even more survivors in candy consumer group)
pvalue = P(X > 4262 ) where X
follows a hypergeometric distribution with
parameters
N
= 7841 M = 7327 n = 4529
We
can also look at the number deaths in the candy consumer group, which we expect
(in the long run) to be less than the number of deaths in the nonconsumer
group. In this case, pvalue = P(X < 267) where X follows a
hypergeometric distribution with parameters N
= 7841, M = 514, and n = 4529.
(There
are other correct set ups as well.)
(f) Suppose you
wanted to carry out a simulation to determine how surprising it is for two
random samples from the same population to give a difference in sample
proportions at least this large.
Describe the simulation process (if describing an applet, name the
applet and the input information you would use).
We
would randomly sample 4529 men and 3312 men, each from a population with a
probability of success of 7327/7841 = 0.934
Then we
would count how many samples have a difference as extreme (onesided
alternative hypothesis) as 0.05895 – 0.07458 = 0.0156.
For
fun:
With such
large sample sizes, this is extremely statistically significant.
Also
notice that even though random sampling is probably a better model here, the
FET pvalue is quite similar as well.
(g) The study
reported: Between 1988 and 1993, 514 men died: 7.5%
of nonconsumers, but only 5.9% of consumers (age adjusted relative risk 0.83;
95% confidence interval 0.70 to 0.98). Interpret
this statement as if to someone who has never taken a statistics class. In particular, what do you think is meant by
“age adjusted relative risk”?
This
interval provides an assessment for how much less likely a candy consumer is to
die in this time frame than a nonconsumer. The values in the interval are all
less than one, so if we knew the death rate of nonconsumers, we would multiply
by .70 to .98 to find the death rate for those who eat candy.
“Age
adjusted relative risk” essentially looks at the relative risks in different
ages groups (so only comparing men of similar ages) and then roughly averages
across those values to get an ageadjusted relative risk. This helps ensure we
have “controlled” for age since we couldn’t do random assignment.
(h) Based on
this interval, I would consider the comparison statistically significant. Why?
Yes,
because 1 is not inside this 95% confidence interval, we know the twosided
pvalue is less than .05.
(i) This does
not appear to be a large difference (7.5% vs. 5.9%), are you surprised that
this result is statistically significant? Explain.
1. No
because the relative risk takes the magnitudes of the values into account: 1.6
percentage points may not be a lot but it’s a decent fraction of 5.9%.
2.
The sample sizes are pretty large so even a weak association will probably end
up being “statistically significant.”
(j) The study
also reports: We then examined different levels of candy
intake. Compared with nonconsumers, the relative risks of mortality among men
who consumed candy 13 times a month (1704 men), 12 times a week (1589 men),
and 3 or more times a week (1236 men) were 0.64 (0.48 to 0.86), 0.73 (0.55 to
0.96), and 0.84 (0.64 to 1.11),
Does this
result provide evidence of a “doseresponse”? Explain.
Yes,
the relative “risk” of surviving that long is increasing with increasing
amounts of candy!
(k) And then: Finally, using life table analysis
truncated at age 95, we estimated that (after adjustment for age and cigarette
smoking) candy consumers enjoyed, on average, 0.92 (0.04 to 1.80) added years
of life, up to age 95, compared with nonconsumers.
Based on these
results, are you willing to conclude that eat candy leads to a longer life?
No,
this was not a randomized comparative experiment, so we can’t draw any
causeandeffect conclusions.
A
possible confounding variable is “happiness” – those who are happy and relaxed
and not worried about what they eat are more likely to consume candy than those
who are stressed and worried and watching their diet closely. But that happier lifestyle may also be
responsible for longer lives.
(l) What
population are you willing to generalize these results to? Explain.
At
most welloff males (graduates from Harvard), but even that is risky as this study
did not involve random sampling. It’s possible the access to medical care and
longlife span for such individuals is not representative of all adults
(certainly not women).
6) A study of whether AZT
helps to reduce transmission of AIDS from mother to baby (Connor et al., 1994):
Of the 180 babies whose mothers had been randomly assigned to receive AZT, 13
babies were HIVinfected, compared to 40 of the 183 babies in the placebo
group.
(a) Create a segmented bar graph to display
these results. Comment on what the graph reveals.
This bar
graph (and the conditional proportions of 13/180 vs. 40/183) indicates that
mothers given the placebo were about 3 times as more likely to have babies that
were HIV positive than were the mothers given AZT.
(b) Check the validity conditions for whether a
twosample ztest can be applied to these data. Be sure to mention
whether the study involves random sampling from populations or random
assignment to treatment groups.
The number of
successes and failures in each group should be at least 5. The four values
are 13, 18013 = 167, 40, 18340=143. This condition is met.
(c) If you were to carry out a simulation to
obtain a pvalue, would you simulate random sampling or random assignment? Explain.
The data are
from randomly assigning subjects to two treatment groups. So our pvalue
will want to reflect the random variation from random assignment (e.g.,
shuffling the 363 cards (53 successes and 310 failures) to groups of 180 and
183).
(d) Conduct an appropriate test of significance
to determine whether the data provide convincing evidence that AZT is more
effective than a placebo for reducing mothertoinfant transmission of AIDS.
Report the hypotheses, test statistic, and pvalue. Also indicate the test
decision using .01 as the level of significance.
The null
hypothesis is that AZT and a placebo are equally effective in reducing
mothertoinfant transmission of AIDS. Specifically, the probability of
HIVpositive babies born to mothers who could potentially take AZT is the same
as the probability of HIVpositive babies born to mothers who could potentially
take a placebo. In symbols, the null hypothesis is H_{0}: π_{AZT}
 π_{placebo} = 0.
The
alternative hypothesis is that AZT is more effective than a placebo for
reducing mothertoinfant transmission of AIDS, or that the probability of
HIVpositive babies born to mothers who could potentially take AZT is smaller
than the probability of HIVpositive babies born to mothers who could
potentially take a placebo. In symbols, the alternative hypothesis is H_{a}:
π_{AZT}  π_{placebo} < 0.
Because this
is a randomized experiment and the counts are on the small size, we could carry
out Fisher’s Exact Test.
Or we could
carry out the random assignment simulation
And find the
pvalue by counting how many rerandom assignments have a difference in
proportion with HIV positive babies (_{AZT} – _{placebo}) of .146 or less
Or, because
we said in (b) that the theorybased approach should be valid, we could go
straight to the TheoryBased applet to carry out a ‘twosample ztest’
With such a
small pvalue, reject H_{0} at the .01 level of significance.
We have very
strong statistical evidence that AZT is more effective than a placebo for
reducing mothertoinfant transmission of AIDS. We can say ‘more effective”
because this was a randomized, comparative experiment.
(e) Estimate the relative risk of transmission
with the placebo compared to AZT with a 95% confidence interval. Also be sure
to interpret this interval in context.
Sample
relative risk (with placebo in numerator): (40/183)/ (13/180) = 3.03
For 95%
confidence, z* = 1.96
For relative
risk, we first take the ln rel risk: ln(3.03) = 1.106
SE(ln rel
risk) = sqrt(1/40 – 1/183 + 1/13 – 1/180) = 0.3015
Confidence
interval: 1.106 + 1.96(.3015) = (0.515, 1.697)
Backtransforming:
exp(0.515, 1.697) = (1.67, 5.46)
We are 95%
confident that the probability of transmission is 1.67 to 5.46 times higher
with the placebo than with AZT.
(f) Summarize the conclusion that you could draw
from this study (significance, estimation, causation, and generalizability).
Also explain the reasoning behind each component.
Because
this was a welldesigned experiment with a small pvalue, we can
conclude that AZT caused the observed difference in HIV transmission
rates. If AZT and a placebo were equally effective in reducing
mothertoinfant transmission of AIDS, we virtually never see sample results as
or more extreme as those we saw in this experiment by random assignment alone
(pvalue < .0001). We are 95% confident that using the placebo increases the
probability of transmission by 67% to 546%. We might have some caution in
generalizing these results to a larger population as we don’t know how the
HIVpositive mothers willing to participate in this study were recruited.
7) Consider the question of whether exposure to secondhand smoke is
harmful to the health of children. EV = whether or not
exposed to second hand smoke, RV = health of child
(a) Describe a prospective
cohort observational study that could address this question.
Find children who will be
exposed to second hand smoke and children who won’t. In a few years,
compare the health of the two groups.
(b) Describe a retrospective
casecontrol observational study that could address this question.
Find healthy children and unhealthy children and then see how much
secondhand smoke they were exposed to growing up.
(c)
Describe
a crossclassified observational study that could address this question.
Find older children and then
determine their health status and whether they were exposed to different amount
of secondhand smoke.
(d) Describe how you could (in
principle) design an experiment to address this question.
Randomly assign some children
to be exposed to secondhand smoke and some children not to be exposed to
secondhand smoke.
(e) Would it be ethical to
conduct an experiment to address this question? Explain.
Because secondhand smoke
is so potentially hazardous, it would not be ethical to willing impose this
treatment on children.
8)
Investigation 3.10
(a) The observational units are drivers;
Explanatory variable is whether or not got a full night’s sleep in previous
week; Response variable is whether or not were involved in a car crash; This is
a casecontrol observational study.
(b)

No full night’s sleep 
Full night’s sleep 
Total 
Crash 
61 
53561 = 474 
535 
No Crash 
44 
588 – 44 = 544 
588 
Total 
105 
1018 
1123 
(c) Because this was a casecontrol
study we should use the odds ratio. If we look at the odds of being in a crash
if they did not get a full night’s sleep compared to the odds of being in a
crash if they did get a full night’s sleep: (61/44)/(474/544) = 1.5911
Note,
this is the same as the odds of not getting a full night’s sleep for the crash
victims vs. the odds of not getting a full night’s sleep for the noncrash
subjects.
The
odds of being in a crash if didn’t get a full night’s sleep were 1.59 times
higher than the odds of being in a crash if did get at least one full night’s
sleep.
(d) We could replicate the sampling
design by sampling independently from two binomial processes with the same
probability of success (this will model H_{0}: = 1). One
process will represent the sampling of the crash victims (n = 535) and the other will represent the sampling of the nocrash
population (n = 588). We can use 105/1123 » 0.581 as the common probability of
success. Once we get the two samples, we
will calculate the simulated odds ratio.
Then we will see how often we get a sample odds ratio of 1.5911 or
larger (H_{a}: > 1,
note, it’s not clear here whether they had a one or two sided alternative in
mind, but it’s reasonable to think that they suspected the lack of sleep would
be associated with an increase in odds of a car crash.).
(e) Let X_{crash} be the number
of successes (no full night’s sleep) in the crash group, so we are modeling X_{crash}
as binomial with n = 535 and =
0.0935.
Let
X_{no crash }be the number of successes in the “no crash” group, and
we are modeling X_{no crash} as binomial with n = 588 and p = 0.0935.
Then
odds ratio = (X_{crash}/X_{no crash}) / [(535  X_{crash})/(588  X_{no crash})]
The
distribution should appear skewed to the right with mean close to 1 and
standard deviation near 0.2.
Example
results:
mean
= 1.108, standard deviation = 0.2129
The
observed ratio of 1.59 is a fair bit out in the tail of the distribution and
appears to have a smallish pvalue. If
we count how many of these observations are 1.59 are larger (rounding down from
1.5911 so that 1.5911 is included), we find about 1% (10 out of 1000) of the
simulated sample odds ratios are at least this extreme. Using Fisher's Exact test, the pvalue is 0.0158 = P(X ≥ 61, for X
hypergeometric with N = 1123, M = 535, n = 105. This gives us
strong evidence, of a relationship in the population.
(f) Theoretical
standard error for the log odds ratio:
_{} »
0.2075
ln(1.59)
+ 1.96(0.2075) = 0.4637 + 0.4067 Þ (0.057, 0.870)
exp(0.057,
0.070) Þ (1.06,
2.39)
We
are 95% confident that the odds of being in a car crash are 1.06 to 2.39 times
larger for those without a full night’s sleep in the previous week compared to
those with at least on full night’s sleep.
This
interval does not capture one, so we have statistically significant evidence of
an increase in odds for those without a full night’s sleep.
R output (R inverts a test (doubling the
onesided pvalue) rather than using the zformula)
JMP output
These
are very close to the confidence interval we calculated.
(g) We have
strong evidence (pvalue = 0.0158) that sleep deprived drivers have higher odds
(4% to 140% higher) of being in a car crash, at least for drivers like these
New Zealand drivers. We cannot draw a causeandeffect conclusion however as
this was an observational study. We
should probably apply these results only to drivers in New Zealand at that
time.