Math 1070-003
Exam 3 Review
15 April 2013
The following questions are designed to help you prepare for Exam 3. You
should also study the relevant vocabulary, concepts, and formulas. You will
be allowed one 8.5 × 5.5 sheet of paper (half of a letter-size sheet of paper)
in the exam. The relevant tables will be provided for you.
For each question, determine which procedures to use, and then carry out the hypothesis
test or find the confidence interval.
1. A researcher believes that the weights (in grams) of adult males in two species of beetles
follow the Normal distribution. He takes an SRS of each species and calculates the
following information:
Group n x
s
A
31 15 4.61
B
17 21 7.34
Is there strong evidence (at the level α = 0.10) that species B is larger than species A,
on average?
Solution: Based on the fact that there are two SRS’s from different populations,
we should use a two-sample t-test. Let µ1 be the mean weight of species B and µ2
be the mean weight of species A.
Step 1: State Hypotheses
H0 : µ1 = µ2
Ha : µ1 > µ2
. Step 2: Calculate t-statistic
x1 − x2
t= q 2
s1
s2
+ n22
n1
=q
21 − 15
7.342
17
+
4.612
31
6
3.169 + 0.686
6
=
1.963
= 3.057
=√
Step 3: Find the P -value.
We should use the t distribution with 16 degrees of freedom (the minimum of n1 -1
and n2 - 1). Based on the alternative hypothesis, we want to find the area under the
density curve to the right of 3.057:
2.921 < t-score < 3.252 so 0.005 > P > 0.0025
.
Step 4: State Conclusion
Since the P -value is smaller than the significance level α = 0.10, we conclude that
the data is significant at the level α = 0.10. We reject the null hypothesis in favor
of the alternative, that species B is in fact larger than species A.
2. Many years of data support the assumption that the lifetimes of bald eagles are Normally
Distributed with mean 37 years and standard deviation 6 years. Rachel Carson finds
that an SRS of 9 bald eagles in agricultural areas with long-term heavy DDT use have
an average lifespan of 15 years. Is there sufficient evidence, at the level α = 0.10 that
DDT causes bald eagles to die prematurely? Find a 99% confidence interval for the
lifespan of bald eagles in the presence of DDT.
Solution: We should use z-procedures, because we are given the population standard deviation.
Step 1: State Hypotheses
Let µ be the mean lifespan of a bald eagle in an area treated by DDT.
H0 : µ = 37
Ha : µ < 37
Step 2: Calculate z-Statistic
x − µ0
15 − 37
=
√
σ/ n
6/√9
−22
=
= −11.00
2
z=
Step 3: Calculate P -value
Since the alternative hypothesis is that µ is less than 37 years, the P -value is the
area under the standard Normal density curve to the left of the z-score:
P = P (Z ≤ −11.00) ≈ 0
(Certainly, we know that the P -value is less than 0.002, which is smaller than
α = 0.10)
Step 4: State Conclusion
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Since the P -value is less than the significance level (P < α), we conclude that there is
strong evidence that DDT causes bald eagles to die earlier than they would normally
live. We reject H0 in favor of the alternative, that the mean lifespan of bald eagles
in DDT treated areas is less than 37 years.
A 99% confidence interval for the lifespan of bald eagles in the presence of DDT
is
σ
6
x ± z ∗ √ = 15 ± √
n
9
15 ± 2
We are 99% confident that the mean lifespan of bald eagles in areas treated with
DDT is between 13 and 17 years.
3. Some people claim that eating meat causes higher cancer rates. In the general population, 8.70% of adults ages 40-60 are diagnosed with some form of cancer. Let’s say
that the cancer rate in a simple random sample of 300 adults ages 40-60 who eat meat
is 8.92%. Is this difference significant at the level α = 0.05? (In other words, is there
sufficient evidence to conclude that the proportion of meat-eaters who are diagnosed
with cancer is higher than the proportion in the general population?)
Solution: Since this question concerns a population proportion Let p be the proportion of meat eaters who are diagnosed with cancer.
Step 1: State Hypotheses
H0 : p = 0.087
Ha : p > 0.0870
Step 2: Calculate z-statistic.
We know that the sample proportion is p̂ = 0.0892.
p̂ − p0
z=q
p̂(1−p̂)
n
0.0892 − 0.0870
= q
0.0892(1−0.0892)
300
0.0022
0.01646
= 0.1337
=
Step 3: Calculate P -value.
To find the P -value, we want to find the area under the standard normal curve to
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the right of the z-score. In this case, that area is
P (Z ≥ 0.1337) = 1 − P (Z ≤ 0.1337) = 1 − .5517 = 0.4483
Step 4: State conclusion
Since the P -value is larger than the significance level α = 0.05, there is not strong
evidence against the null hypothesis and in favor of the alternative. Therefore, we
accept the null hypothesis, that the proportion of meat-eaters who are diagnosed with
cancer is roughly the same as the proportion of the general population diagnosed with
cancer.
4. Researchers want to determine whether children with obese parents are more likely to
be obese. In a matched-pairs study, 10 single-child two-parent families with two obese
parents were matched with similar families with two non-obese parents. The average
difference between the children’s body mass indexes (obese parents minus non-obese
parents) was 1.25 with sample standard deviation 0.97. Is there significant evidence (at
the level α = 0.01) that children with obese parents are more likely to be obese?
Solution: Since the experiment is a matched pairs design, we should use the single
sample t-test. There are 10 pairs of families. Let µ represent the mean difference
in BMI between children with obese parents and children with non-obese parents
(obese minus non-obese).
Step 1: State Hypotheses.
H0 : µ = 0
Ha : µ > 0
Step 2: Calculate t-statistic.
x − µ0
s/√n
1.25 − 0
= 0.97 √
/ 10
1.25
=
0.3067
= 4.076
t=
Step 3: Find the P value.
Since the sample size is 10, we should use the t-distribution with df = 9.
3.690 < P < 4.297
so,
0.0025 > P > 0.001
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Step 4: State conclusion: Since the P value is smaller than α = 0.01, the data is
significant at this level. We reject the null hypothesis in favor of the alternative, that
children with obese parents are more likely to be obese themselves.
5. The National Longitudinal Study of Adolescent Health interviewed several thousand
teens (grades 7 to 12). One question asked was “What do you think are the chances
that you will be married in the next ten years?” Here is a table of the responses by
gender:
Female
Almost no chance
119
150
Some chance, but probably not
A 50-50 chance
447
735
A good chance
Almost certain
1174
Male
103
171
512
710
756
Is there evidence of a difference between male and female teenagers in their distributions
of opinions about marriage?
Solution: See Exercises 22.18-26
6. Give an example of a Type I error.
Solution: A Type I error is when we incorrectly reject the (true) null hypothesis.
For example, a small study on the interaction between vaccines and autism rejected
the null hypothesis that vaccines have no influence on an autism diagnosis in favor of
the alternative, that there is some link between vaccines and autism. In many follow
up studies with very large sample sizes, the medical community has come to the
consensus that there is no link between vaccines and autism. The original conclusion
was a Type I error.
7. Give an example of a Type II error.
Solution: A Type II error is an incorrect acceptance of the (false) null hypothesis.
In this error, there may be some evidence against the null hypothesis, but not enough
to reject it in favor of the alternative. For example, a preliminary study find that
smoking does not increase cancer risk by a statistically significant amount. We now
know that smoking does increase lung cancer risk, so the initial conclusion was a
Type II error.
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8. Assuming that the null hypothesis (H0 : µ = µ0 ) is true, what distribution do each of
the following statistics have? (Assume a random sample of size n, with σ, µ0 , x, s, etc
as usual.)
x − µ0
(a) z = σ √
/ n
x − µ0
(b) t = s √
/ n
p̂ − p0
(c) z = q
p̂(1−p̂)
n
Solution:
(a) Standard Normal distribution (N(0,1) distribution)
(b) t distribution with n − 1 degrees of freedom
(c) Standard Normal distribution
9. What is the relationship between the margin of error of a 99% confidence interval and
the margin of error of a 90% confidence interval, calculated from the same data?
Solution: The margin of error for the 99% confidence interval will be larger than
the margin of error for the 90% confidence interval
10. Suppose that Alice and Bob are researchers studying the same variable in the same
population. They find the same sample mean and sample standard deviation and are
testing the same hypotheses, but Alice’s study has a larger sample size than Bob’s. What
can you say about the P -value of Alice’s hypothesis test in relation to Bob’s?
Solution: Alice’s P -value will be smaller than Bob’s.
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