**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) Use
technology to simulate repetitions of 100 spins each. Use the simulation result
to produce an approximate p-value. Be very clear how you are carrying out this
simulation and how you are finding the approximate p-value.

Using
the One Proportion Inference applet, my approximate p-value is (253+204)/1000 =
.457, the proportion of the 1000 simulated samples (of 100 spins each) that
resulted in 46 or fewer heads or 54 or more heads.

(f) Use the
binomial distribution to calculate the p-value exactly. (Be sure to indicate
how you calculate this probability: what values you use for *n* and , and what region you find the
probability of.)

This
p-value is P(X ≤ 46) + P(X ≥ 54), where X has a binomial
distribution with *n* = 100 and = 0.5.
JMP reveals this probability to be 0.484, as shown below (note, specify
Qb before Qa):

(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:

95%
of sample proportions should fall within 2 standard deviations of the
mean. But now we are taking the value to be 0.50 and so they will fall within
2 x .05 = 0.10 or 0.50, so between 0.40 and 0.60.

*Note
how this is different from a confidence interval….*

(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?

Because
the p-value is two-sided, we want to look for the portion of the null
distribution that is “at least as extreme as” 0.46 in both directions. Only the
second graph is two-sided.

We
can also run this as a one proportion z-test in JMP or Theory-Based Inference
applet

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

“Test
decision” is referring to whether you reject or fail to reject the null hypothesis. Because the standardized statistic is only
0.80 (less than 2), we would fail to reject the null hypothesis.

*Note:
it is considered incorrect to say we “accept” the null hypothesis or we have
evidence “for” the null hypothesis.*

(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 **and
interpret **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.

No,
above we assumed we know the value of and we created an interval for sample
proportions. Here, we only know one
sample proportion and created an interval around that value to predict the
value of the underlying process probability.

(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 (be sure
to comment on significance, confidence, and generalizability).

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. 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.

d. 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.

e. 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.

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

g. 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.

h. 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.

i. 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.

j. 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.