**Exam 1 Review Problems Solutions**

**1) **a. Blood type - categorical

b. Waiting time – quantitative (if recorded as number of
minutes between arrival and seen by doctor).

c. Mode of arrival (ambulance, personal car, on foot, other) -
categorical

d. Whether or not men have to wait longer than women – this is
a research question/comparison, not a variable posed to individual visitors
(the variables asked of the visitors would be length of wait and gender)

e. Number of patients who arrive before noon – this is a
summary of the data, not a variable posed to individual visitors, the variable
would be whether or not the patient arrived before noon

f. Whether or not the patient is insured - categorical

g. Number of stitches required – quantitative

h. Whether or not stitches are required - categorical

i. Which patients require stitches – a subgroup of the sample,
not a variable posed to individual visitors (see g and h)

j. Number of patients who are insured – this is a summary of
the data, not a variable posed to individual visitors (see f)

k. Assigned room
number – categorical (numerical but doesn’t make sense to talk about “average”
room number)

**2)** When a tennis racquet is spun, is it equally likely to
land with its label facing up or down? (This technique is often used to decide
who should serve first.) Or does the spinning process favor one outcome more
than the other? A statistics professor once investigated this question by spinning
his tennis racquet many times. For each spin he recorded whether the racquet
landed with the label up or down.

(a) Describe
(in words) the relevant parameter whose value is being investigated with this
study.

The
parameter is the probability that my spun tennis racquet lands with the label
up, denoted by . This is equivalent to the long-run proportion
of times that the racquet would land up if it were spun by this professor
indefinitely.

(b) Write the
appropriate null and alternative hypotheses (in symbols).

H_{0}:
= .5 (the racquet is equally likely to land
the two ways)

vs

H_{a}:
≠ .5 (there is a tendency to land one
way more than the other and this would not be a fair way to determine who
services first)

He spun his
racquet 100 times, finding that it landed with the label up in 46 of those
spins.

(c) Would you
consider these 100 spins to be a sample from a process or from a population?
Explain briefly.

This
would be considered sampling from a process. There is not a large number of
spins out there from which we have selected a subset. Instead, we believe the
probability of landing label up to be constant across the spins and we are
obtaining a representative set of those spins.
We will be cautious not to generalize these results beyond this racquet
or this spinner.

(d) Describe
how you could use a coin to conduct a simulation analysis of whether this
result constitutes strong evidence that his racquet spinning process is not
equally likely to land with its label facing up or down. Provide enough detail
that someone else could implement the simulation and draw the appropriate
conclusion.

Toss
a fair coin 100 times, counting the number of tosses that land heads,
representing a racquet spin that lands up.
Repeat this process of 100 coin flips a large number of times (say,
1000), each time counting the number of heads.
Then determine the proportion of those 1000 repetitions (of 100 spins
each) that produced either 46 or fewer heads or 54 or more heads (to match our
two-sided alternative). If this
proportion is very small (say, less than .05), that indicates that a result as
extreme as the one observed would rarely happen with a 50/50 process, and so in
that case we would conclude that the racquet really does favor one side or the
other. But if this proportion is not
very small (say, greater than .05), that indicates that a result as extreme as
the one observed is fairly consistent with a 50/50 process, and so in that case
we would not reject the hypothesis that the racquet spinning is a 50/50
process.

*Notice how this response
includes a discussion of how decisions will be made based on the simulation
results.*

(e)
Based on the output below, what is the simulation-based p-value?

0.4890
is the simulation-based p-value in the above output.

(f)
What are the parameters of the binomial distribution: *n* = and * *= and we found P(X ? )?

The
parameter (or “inputs”) of the binomial distribution would be *n* = 100
and = 0.50, so we are finding P(X __<__ 46
or X __>__ 54) for the two-sided p-value.

(g) Check
whether the normal approximation (Central Limit Theorem) is valid here.

Yes,
because *n*× = 100(.5) = 50 is larger than
10, as is *n*×(1 - ) = 100(.5) = 50.

(h) Describe
what the CLT says about the (approximate) sampling distribution of the sample
proportion , assuming that the null hypothesis is true. Be sure
to describe all of shape, mean, and standard deviation and include a rough
sketch (but well labeled) of the distribution.

The CLT
says that the sample proportion will vary according to
an approximate normal distribution. We
also find the mean .5 and standard deviation = .05. A sketch of this distribution is shown here:

Using
the distribution in (h), we would expect 95% of sample proportions to fall
within 2 standard deviations of , so within 2(.05) of 0.50, so
between 0.40 and 0.60. This is all
assuming we know to equal 0.50.

(j) Calculate (by
hand) and interpret the test statistic by finding the z-score for the observed
sample proportion .

*z* = = -0.8

This
calculation tells us that the observed sample proportion (.46) fell .8 standard
deviations below the conjectured probability (.5).

(k) Which of the following graphs would be correct for finding
the one-proportion z-test p-value?

The
two-sided p-value would be the blue area below 0.46 and above 0.54, which looks
fairly large to me.

(l) What test
decision would you make at the .05 significance level?

Fail
to reject H_{0}, because the p-value is larger than .05 (and the |*z*|
value is less than 2). We have little/no
evidence to doubt that the racquet lands up 50% of the time in the long run.

(m) Do the
conditions for the (Wald) normal-based confidence interval hold here?

Yes,
because *n*× = 100(.46) = 46 is
larger than 10, as is *n*×(1-) = 100(.54) = 54. This is also an infinite process and we
are assuming under identical conditions for each spin.

(n) Produce a
95% confidence interval for the parameter, using the Wald procedure if the
conditions are met but using the adjusted Wald procedure if they are not met.

A 95%
confidence interval for is: .46 ± 1.96 = .46 ± 1.96(.049) = .46 ± .098, which is
(.362, .558).

We
can be 95% confident that in the long run between 36.2% and 55.8% of all spins
with this racquet would land with the label up.

We
don’t have to use the Plus Four Method here, but we certain could:

48/104 = 0.4615

0.4615 __+__ 1.96 sqrt(.4615(.5385)/104) = (0.366, 0.557),
not much different

No,
these “intervals” are telling us two different things. In (i), we found the likely range of sample
proportions in a distribution with a known value of . Here, we are using the (single) observed
sample proportion to find the range of plausible values of .

(p) Is the
confidence interval consistent with the test decision? Explain.

Yes,
we did not reject the hypothesis that = .5 at the = .05 level, and .5 appears within the 95%
confidence interval for .

(q) Summarize
your conclusion about the original question that motivated this study.

The
sample data provide no reason to doubt that this racquet lands “up” 50% of the
time. We are 95% confident that the
probability of landing label up is between .362 and .558. These results may only apply to this racquet
and this spinning technique however.

(r) Summarize
how your calculations and conclusions would change if you instead examined the
54 spins that landed label down.

The
z-score would now be positive but the (two-sided) p-value would remain the
same. We would conclude that we are 95%
confident that the probability of landing label down falls between .442 and
.638.

(s) Use the
normal approximation to determine how large the sample size *n* needs to be in order for the 95% confidence
interval to have margin-of-error __<__ .08.

Using
the observed sample proportion .46 as an estimate for , we need to solve for n. Solving gives ≈ 149.1, so 150
spins would be needed.

**3)** a. The population is all Cal Poly students and the sample
is the 97 who responded.

b. The
parameter is the proportion of all Cal Poly students who eat breakfast 6 or 7
times a week. We could call this (unknown) value . The statistic is the proportion of the
sampled students who report eating breakfast 6 or 7 times a week. We know the
second number to be = 35/97 = .361.

Note: .21 is
the population proportion at James Madison, not or .

c. The
population of students at Cal Poly is on the order of 20,000, which is more
than 20 times larger than the sample size of 97 in this study. So yes, we can use the binomial distribution
here as an approximation.

d. We can
consider two approaches, simulation and theory-based.

For the
simulation approach we can use the Simulation-Based One Proportion Inference
applet (note we are assuming the population here is much larger than the
sample, more than 20 times as large). We want to set the sample size to be 97
and we are going to conduct the simulation assuming the proportion of *all Cal Poly students* is also .21 as it
was at James Madison University. Then we see how often random samples from such
a population yield a sample proportion of .361 or higher (the direction
conjectured by the researchers).

Here, we see it
is very surprising for a random sample of 97 students from a population with =
.21 to have a sample proportion of .361 or higher. With a p-value of
approximately zero, this is a small p-value (e.g., less than .05) and we reject
the hypothesis that =
.21 for all Cal Poly students. Instead
we will conclude that , the population proportion *of all Cal Poly students* who eat
breakfast 6 or 7 times a week, is larger than .21. (We could construct a
confidence interval to estimate a range of plausible values for but we don’t think .21 is one of them.)

Alternatively
we could consider using the normal approximation because there were 35
“successes” and 62 “failures,” both at least 10, so the approximation should be
valid. Using the Theory-Based Inference applet

We again find a
very small p-value. In addition, this applet tells us that our observed
statistic (_{observed} = .361) is 3.65 standard
deviations above the hypothesized value for of .21.

e. The p-value
is clearly small (less than .001), so we reject the null hypothesis that , the proportion of all Cal Poly students
who eat breakfast 6 or 7 times as week, equal .21, in favor of the alternative
the hypothesis that > .21.
Thus, we conclude that there is convincing evidence that more than 21%
of all Cal Poly students eat breakfast 6 or 7 times a week. So at least on this
measure of “healthiness” we have strong evidence that the Cal Poly population
is healthier.

f. The p-value
says that if we were to repeatedly take simple random samples of 97 students
from a population with =
.21, then in less than .1% (depending what you find for the p-value) of those
samples would we find our sample proportion who eat
breakfast at least 6 times a week to be .361 or higher.

g. The
statistic () will probably be different but not the parameter ().
The population parameter is a fixed (but unknown to us) value.

h. This
indicates a potential source of sampling bias. Those who chose to respond to
the survey could be different from those who did not. Perhaps they are more aware of their eating
habits and more interested in health overall and that’s why they were more
likely to reply, leading to an overestimate of the population proportion in
this study.

i. We need a
list of all Cal Poly students (the sampling frame) probably from the
registrar’s office. Assign everyone on
the list a unique 5 digit number. Then
use a random number table or computer or calculator to randomly select 1590
unique ID numbers. Find and survey those
students.

j. If there is
still a high nonresponse rate, even with the larger sample size we would still
be concerned about how representative the sample is. It would be much, much better to recontact
the originally selected people (several times if necessary) to get their
responses than to simply sample more people.

k. With the
larger sample size and the same value for the statistic we would expect the
p-value to be even smaller. The sampling
distribution of the sample proportions (’s) would cluster even closer around .21 (less
sampling variability) and it would be even more shocking to randomly obtain a
sample proportion as extreme as .361 when =
.21.