STAT 302 Midterm: STAT 302 UW Madison Solutions 09
Solutions to Homework 9
Statistics 302 Professor Larget
Textbook Exercises
In Exercise 6.160, situations comparing two proportions are described. In each case, determine
whether the situation involves comparing proportions for two groups or comparing two proportions
from the same group. State whether the methods of this section apply to the difference in propor-
tions.
6.160 (a) In a taste test, compare the proportion of tasters who prefer one brand of cola to
the proportion who prefer the other brand.
(b) Compare the proportion of males who voted in the last election to the proportions of females
who voted in the last election.
(c) Compare the graduation rate (proportion to graduate) of students on an athletic scholarship to
the graduation rate of students who are not on an athletic scholarship.
(d) Compare the proportion of voters who vote in favor of a school budget to the proportion
who vote against the budget.
Solution
(a) This compares two proportions (one brand of cola vs the other brand) drawn from the same
group (tasters). The methods of this section do not apply to this type of difference in proportions.
(b) This compares the proportion who voted using two different groups (males vs females). The
methods of this section are appropriate for this type of difference in proportions.
(c) This compares the proportion who graduate using two different groups (athletes and non-
athletes). The methods of this section are appropriate for this type of difference in proportions.
(d) This compares two proportions (proportion in favor vs proportion opposed) drawn from the
same group (voters). The methods of this section do not apply to this type of difference in propor-
tions.
6.164 Is Argentina or Bolivia More Rural? We see in the AllCountries dataset that the
precent of the population living in rural areas is 8.0 in Argentina and 34.3 in Bolivia. Suppose we
take random samples of size 200 from each country, and compute the difference in sample propor-
tions ˆpA−ˆpB, where ˆpArepresents the sample proportion living in rural areas in Argentina and ˆpB
represents the proportion of the sample that lives in rural areas in Bolivia.
(a) Find the mean and standard deviation of the distribution of differences in sample propor-
tions, ˆpA−ˆpB.
(b) If the sample sizes are large enough for the Central Limit Theorem to apply, draw a curve
showing the shape of the sampling distribution. Include at least three values on the horizontal axis.
(c) Using the graph drawing in part (b), are we likely to see a difference in sample proportions as
large in magnitude as −0.4? As large as −0.3? Explain.
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Solution
(a) The differences in sample proportions will be centered at the difference in population propor-
tions, so will have a mean of pA−pB=.080 −.344 = −0.264, and a standard deviation equal to
the standard error SE. We have
SE =spA(1 −pA)
nA
+pB(1 −pB)
nB
=r0.08(0.92)
200 +0.344(0.656)
200 = 0.039
(b) The sample sizes of 200 are large enough for the normal distribution to apply. The distribution
of differences of proportions will be N(−0.264,0.039). We use technology to sketch the graph, or
we can sketch it by hand, noting that the differences in sample proportions will be centered at
−0.264 and roughly 95% of the distribution lies within two standard deviations on either side of
the center. This goes between −0.264 ±2(0.039) or −0.342 to −0.186.
c) We see from the figure that -0.4 is a very unlikely result for the difference in sample proportions
(it’s more than 3 SE’s from the center) whereas -0.3 is plausible (it’s less than one SE from the
center).
6.176 Has Support for Capital Punishment Changed over Time? The General Social
Survey (GSS) has been collecting demographic, behavioral, and attitudinal information since 1972
to monitor changes within the US and to compare the US to other nations. Support for the capital
punishment (the death penalty) in the US is shown in 1974 and in 2006 in the two-way table below.
Find a 95% confidence interval for the change in the proportion supporting capital punishment
between 1974 and 2006. Is it plausible that the proportion supporting capital punishment has not
changed?
Year Favor Oppose Total
1974 937 473 1410
2006 1945 870 2815
Solution
Letting ˆp1and ˆp2represent the proportion supporting capital punishment in 2006 and 1974, re-
spectively, we have
ˆp1= 1945 = 0.691 and ˆp2= 937 = 0.665
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The sample sizes are both large, so it is reasonable to use a normal distribution. For 95% confidence
the standard normal endpoint is z∗= 1.96. This gives
(ˆp1−ˆp2)±z∗·sˆp1(1 −ˆp1)
n1
+ˆp2(1 −ˆp2)
n2
(0.691 −0.665) ±1.96∗·r0.691(1 −0.691)
2815 +0.665(1 −0.665)
1410
0.026 ±0.030
−0.004 to 0.056
We are 95% sure that, between 1974 and 2006, the percent change in support of the death penalty
is between a decrease of 0.4% and an increase of 5.6%. Since a difference of zero (no change) is
within this interval, it is plausible that there have been no change in support or opposition to the
death penalty in this 22-year period.
6.178 Mental Tags on Penguins and Breeding Success Data 1.3 on page 10 discusses a
study designed to test whether applying metal tags is detrimental to penguins. One variable ex-
amined is the survival rate 10 years after tagging. The scientists observed that 10 of the 50 metal
tagged penguins survived, compared to 18 of the 50 electronic tagged penguins. Construct a 90%
confidence interval for the difference in proportion surviving between the metal and electronic
tagged penguins (pM−pE). Interpret the result.
Solution
We have ˆpM= 10/50 = 0.20 and ˆpE= 18/50 = 0.36. The 95% confidence interval for pM?pEis
(ˆpM−ˆpE)±z∗·sˆpM(1 −ˆpM)
nM
+ˆpE(1 −ˆpE)
nE
(0.20 −0.36) ±1.645 ·r0.20(0.80)
50 +0.36(0.64)
50
−0.16 ±0.145
−0.305 to −0.015
We are 90% sure that the survival rate for metal tagged penguins is between 0.305 and 0.015 less
than for electronic tagged penguins. This shows a significant difference at a 10% level.
Comparing Normal and Bootstrap Confidence Intervals In Exercise 6.185, find a 95%
confidence interval for the difference in proportions two ways, using StatKey or other technology
and percentiles from a bootstrap distribution and using the normal distribution and the formula
for standard error. Compare the results.
6.185 Difference in proportion who use text messaging, using ˆpt= 0.87 with n= 800 for teens and
ˆpa= 0.72 with n= 2252 for adults.
Solution
We use StatKey or other technology to create a bootstrap distribution with at least 1000 simulated
differences in proportion. We find the endpoints that contain 95% of the simulated statistics and
see that this 95% confidence interval is 0.120 to 0.179.
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Document Summary
In exercise 6. 160, situations comparing two proportions are described. In each case, determine whether the situation involves comparing proportions for two groups or comparing two proportions from the same group. State whether the methods of this section apply to the di erence in propor- tions. Solution (a) this compares two proportions (one brand of cola vs the other brand) drawn from the same group (tasters). The methods of this section do not apply to this type of di erence in proportions. (b) this compares the proportion who voted using two di erent groups (males vs females). The methods of this section are appropriate for this type of di erence in proportions. (c) this compares the proportion who graduate using two di erent groups (athletes and non- athletes). The methods of this section are appropriate for this type of di erence in proportions. (d) this compares two proportions (proportion in favor vs proportion opposed) drawn from the same group (voters).
