PHCO 0504 – Introduction to Biostatistics
Homework #4
This homework will not be collected. However, please do it conscientiously as the
material in this homework will be included on the midterm. Homework solutions will be
posted on Monday. If you do not get to the homework before Monday, I would strongly
encourage you to do the homework before looking at the solutions. You will not learn the
material as well if you look at the solutions without trying to complete the homework
problems first!
1. Nancy Stearns Burgess (Journal of the Diabetic Association, 1991, pp.430-434)
conducted a study to determine weight loss in obese subjects before and after 12
weeks of treatment with a very-low-calorie diet. 9 women participated in the
study. The average weight of these women before the treatment was 98.53 kg.
(Actual values are 117.3, 111.4, 98.6, 104.3, 105.4, 100.4, 81.7, 89.5, 78.2.)
Below are the changes in the weights (kilograms, negative implies weight loss,
positive would imply a gain).
-34.0, -25.5, -22.8, -21.4, -23.1, -22.7, -19.0, -20.5, -14.3
a. Calculate the mean and standard deviation of these changes.
The mean is –22.589 kg, and the standard deviation is 5.3194.
b. Calculate a 95% confidence interval for the mean loss that results from this
weight loss program. (Think about which interval to use: a z-interval or a tinterval.)
Use t-interval (1 pt) because the sample size is small.
With SEM = 5.3194 / sqrt (9) = 1.7731, d.f. = 9 – 1 = 8, and t = 2.3060, we
have the 95% confidence interval as (-26.6777, -18.5001).
c. Interpret the confidence interval calculated above.
I am 95% confident that the mean weight loss for women in this weight loss
program falls between -26.6777 and -18.5001.
d. Is it plausible that the population distribution from which values given above
were sampled is normal? Explain using the actual values (and not intuition
about weight loss).
A histogram of the weight loss values is as follows.
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WGHTLOSS
5
4
3
Frequency
2
1
Std. Dev = 5.32
Mean = -22.6
N = 9.00
0
-35.0 -32.5 -30.0 -27.5 -25.0 -22.5 -20.0 -17.5 -15.0
WGHTLOSS
This is a uni-modal distribution (there is one peak to the data). The general
trend to the data might be approximately bell shaped. With only nine data
points, there are two outliers (which one can calculate using the formula for
outliers or by looking at the box plot). Hence, the distribution is perhaps more
spread out than a typical normal distribution. So, it is plausible that the
distribution of the population (best represented by this distribution of the
sample) is normal. We really don’t have enough evidence to say that it is not.
e. Using the 5-step format presented in class (ASSUMPTIONS, HYPOTHESES,
TEST STATISTIC, P-VALUE/CRITICAL VALUE, CONCLUSION),
perform a hypothesis test that the there would be, on average in the population
of obese women, a significant decrease in weight with the very-low-calorie
diet. Use a significance level of 0.05. Be sure to pay attention to whether this
should be a one- or two-sided test. If a one-sided test, make sure to get the
direction correct! Also, justify the distribution that you use to calculate the pvalue/critical value (use either approach).
1. Assumptions:
Independent, random sample (we don’t have enough information to
evaluate this assumption.)
Population distribution is approximately normal. Based on the histogram
above, I would not be completely comfortable with saying that this
assumption about the data is satisfied. Therefore, our conclusions
based on this test may not be accurate because this last assumption is
violated. We do, however, know that the t-test is “robust,” i.e., it is
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2.
3.
4.
5.
OK even if the assumption of normality is somewhat violated. Thus, I
would expect the conclusions of this test to hold at least
approximately.
Hypotheses: we are interested in showing that, on average in the
population of obese women, a significant decrease in weight with the
very-low-calorie diet.
H0: μ = 0, HA: μ < 0.
Test Statistic: t = (-22.589 – 0) / (5.3194/sqrt(9)) = -12.740.
For a t-distribution with 8 degrees of freedom and allowing a 0.05 Type I
error rate, we would require the test statistic to be less than –1.8585. That
is, our critical value is –1.8585.
(If we were comparing to a normal distribution which we are not
doing because of the sample size, this critical value would be
somewhere between -1.60 and -1.65. The critical value based on
the t is more stringent criteria, because the t-distribution tries to
account for errors that might be made by random error in the
estimation of the standard deviation/ standard error.)
Because the observed value is less than –1.8585, there is enough evidence
to reject the null hypothesis H0: μ = 0. With 95% confidence, we would
reject the null hypothesis and that the average change in weight before and
after the treatment with the very-low-calorie diet is negative.
f. In the test above, suppose that you used a two-sided alternative hypothesis
(not equal). What would the p-value/critical value be?
For a two-sided alternative, the critical value would be -2.3060.
For both the one-sided and the two-sided alternatives, the p-value is <0.0001.
2. The CDC reported in an article that of 224 staff workers at a single high school
within five miles of the World Trade Center (WTC), 49 were considered to have
experienced Post-traumatic stress disorder (PTSD, defined as a person who
responded affirmatively to questions according to the Diagnostic Statistical
Manual of Mental Disorders-IV criteria) soon after the September 11, 2001 attack
on the WTC.
a. Calculate a 95% confidence interval for the proportion of high school
students near the WTC that experienced PTSD. Interpret the confidence
interval formally and informally.
With n = 224, p = 49/224, and SE = 0.0276206, a 95% confidence interval
is calculated as (0.1646, 0.2729).
I am 95% confident that the proportion of staff workers at a single high
school within 5 miles of the WTC that have experienced PTSD is between
0.1646 and 0.2729.
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b. A similar sample of staff workers at single a high school more than five
miles from the WTC were surveyed. Only 9 out of 155 workers reported
symptoms of PTSD. Calculate and interpret a 95% confidence interval.
Use n = 155, p = 9/155, and SE = 0.0187845, the 95% confidence interval
is then calculated as (0.02125, 0.09488).
I am 95% confident that the proportion of staff workers at a single high
school more than 5 miles from the WTC that have experienced PTSD is
between 0.02125 and 0.09488.
c. Critique the sampling method used in this study. (What populations are
being studied? Is this a simple random sample? etc.) What assumptions
for confidence intervals do these violate? Despite the somewhat flawed
sampling method, what might you conclude from the two confidence
intervals you calculated? (I am pointing out that sometimes despite
flawed studies, we sometimes tend to believe results anyways. If the
differences between the two groups were small, we might be less willing
to accept that the differences are “real.”)
The study population (and the sample in this case) are staff workers at
single a high school within five miles of the WTC and the staff workers at
single a high school more than five miles from the WTC. (2 pts)
Presumably for the target population, researchers were interested in
comparing (at least) all high school workers within 5 miles and outside of
a 5 mile radius of the WTC (if not all employees in these areas). The fact
that all employees from each of two high schools were sampled indicates
that this is not a simple random sample. In addition, the assumption of
independent sampling is severely violated: the probability that one
particular worker was selected for the sample is certainly not independent
of whether a worker in the same school was selected or whether a worker
in another school was selected. (If a particular worker was selected for the
sample, then we know that everyone else in his/her same school was also
selected!) In summary, the assumptions for confidence intervals that this
study violates are the random sample assumption and the independent
observations.
d. For each interval, evaluate whether we have a large enough sample to use
the large sample confidence interval for proportions. (Give a “yes/no”
answer as well as an explanation.)
(Also, observe this is one study where it would perhaps not be reasonable
to do a hypothesis test. We might not necessarily have a pre-conceived
notion of what this proportion should be. Presumably the study is
conducted to find out what this proportion is.)
For proportions with large sample sizes, large is defined as “at least
subjects – 5 in each category.” Therefore, we have a large enough sample
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for both samples (49/224 and 9/155) to use the large sample confidence
interval for proportions. However, it would be better to increase the
sample size for the sample (9/155) to increase the 9 subjects to more than
9 subjects.
3. For the following hypotheses, identify the experimental unit, the variable of
interest (measured on each experimental unit) and the parameter of interest. In
addition write down the null and alternative hypotheses (in mathematical
notation), being careful to correctly specify and one-sided or two-sided alternative
hypothesis. Identify the null values as specifically as possible, including an actual
numerical value where possible.
a. More people in New Jersey prefer Coke to Pepsi. (Hint: Compare one
proportion to a number that would indicate an even proportion.)
If you assumed that the question meant “The majority of New Jersey
people prefer Coke,” your answer should have been:
The experimental unit is a single person in New Jersey
The variable of interest (measured on each experimental unit) is the
whether a person prefers Coke rather than Pepsi.
The parameter of interest is the proportion of people who prefer Coke
rather than Pepsi (π).
The null hypothesis is: H0: π = .5, the alternative hypothesis is HA: π > .5.
Note that alternatively, if you assumed that some people preferred neither,
one might have answered:
The experimental unit is a single person in New Jersey
The variable of interest (measured on each experimental unit) is the
whether a person prefers Coke or Pepsi or some other beverage.
The parameters of interest are the proportion of people who prefer Coke
(π1) and those who prefer Pepsi (π 0).
The null hypothesis is: H0: π 0 = π 1, the alternative hypothesis is HA: π 0 <
π 1.
Note that in trying to use the mathematical notation to our advantage, we
generally use Greek letters to represent population parameters and English
letters to represent the corresponding sample statistics. Thus, π represents
a population proportion whereas p represents the corresponding sample
proportion.
b. The number of doctor’s visits is larger for those who carry private
insurance relative to the national average.
The experimental unit is a single person who carries private insurance
The variable of interest (measured on each experimental unit) is the
number of doctors’ visits in a specified period of time.
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The parameter of interest, μ, is the average number of doctor’s visit for
those who carry private insurance. In the single sample cases that we are
currently studying, the “national average,” call it μ0, would be known.
The null hypothesis is: H0: μ = μ0, the alternative hypothesis is HA: μ > μ0.
c. A standard drug treatment for lowering blood cholesterol is known to be
effective in 40% of adults in the US. A generic version of the drug has
been developed. Researchers want to test whether the effectiveness of the
generic drug is different from the effectiveness of the standard drug
treatment.
The experimental unit is a single adult (possibly with high blood
cholesterol) in the US.
The variable of interest (measured on each experimental unit) is whether
the generic drug is effective for a person.
The parameter of interest (π) is the proportion of the population in which
the generic drug treatment is effective.
The null hypothesis is: H0: π = 0.4, the alternative hypothesis is HA: π is
not equal to 0.4 (HA: π ≠ 0.4). (Note the null hypothesized value of the
proportion is π0 = 0.4.)
d. Last year, the average score on a standardized test for high school students
in the state of New Jersey was 79. Developers of this year’s new
curriculum, claim that the curriculum will result in better scores on the
standardized test.
The experimental unit is a single high school student in NJ.
The variable of interest (measured on each experimental unit) is the score
on a standardized test for a high school student this year.
The parameter of interest (μ) is the average score of NJ high school
students from this year (after the new curriculum).
The null hypothesis is: H0: μ = 79, the alternative hypothesis is HA: μ >
79, where the null value μ0 is the average score from last year, 79.
Alternatively, you could very reasonably argue that curriculums are
applied at the level of the classroom, since there is one teacher
implementing a curriculum for an entire class. (This is a very important
point for understanding what we mean by “experimental unit!”) In this
case, the experimental listed previously is incorrect and you should change
the answers to the following:
The experimental unit is a single high school classroom in NJ.
The variable of interest (measured on each experimental unit) is the
average score on a standardized test for a single high school class this
year.
The parameter of interest (μ) is the average score of the average scores for
the NJ high school classes from this year (after the new curriculum).
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The null hypothesis is: H0: μ = 79, the alternative hypothesis is HA: μ >
79, where the null value μ0 is the average score from last year, 79.
e. You hypothesize that a newly designed computer software program will
take shorter than thirty minutes to learn.
The experimental unit is a single person who is learning the computer
software.
The variable of interest (measured on each experimental unit) is the time
taken by a single person to learn the newly designed computer software.
The parameter of interest is the mean time to learn the newly designed
computer software.
The null hypothesis is: H0: μ = 30, the alternative hypothesis is HA: μ < 30.
(The null hypothesized value is μ0=30.)
4. A study examined a sample of 100 records of patients seen at a chronic disease
hospital on an outpatient basis. The mean number of outpatient visits per patient
was 4.8, and the sample standard deviation was 2.
a. Can it be concluded from these data that the population mean is greater
than four visits per patient? To answer this question, conduct an
appropriate hypothesis test (choose between one- or two-sided based on
the question) with a significance level of 0.05. Use the 5-steps form of the
hypothesis test that was presented in class.





Assumptions:
Independent, random sample (This assumption cannot be evaluated
using the information provided.)
Large sample size (N=100, which means that the sampling distribution
is approximately normal)
Hypotheses: we are interested in showing the mean number of
outpatient visits per patient (m) is greater than 4. Therefore,
H0: μ = 4, HA: μ > 4.
Test Statistic: t = (4.8 - 4) / (2/sqrt(100)) = 4.
p-value is 0.00003167 < 0.05.
There is strong evidence to reject the null hypothesis in favor of the
alternative that says that the mean number of visits is greater than 4.
b. What theorem allows us to say that the sampling distribution of the test
statistic used above is approximately normal? What conditions must be
true in order to apply this theorem?
Central Limit Theorem allows us to say that the sampling distribution of
the test statistic used above is approximately normal. Sample size should
be large enough (usually N > 30). We assume random and independent
sample in order to apply this theorem.
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5. What does it mean to say that the t-statistic, whether used to form a confidence
interval or to conduct a hypothesis test is “robust?”
For means (not proportions), the t-interval is robust to violations of the Normality
assumption. If from the sample distribution it appears that the distribution of the
population is not normal, the results based on a t-interval or hypothesis test based
on the t-distribution will be more or less correct, as long as the distribution of the
population is not too drastically far from normal.
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