STAT 302 Midterm: STAT 302 UW Madison Solutions 06
Solutions to Homework 6
Statistics 302 Professor Larget
Textbook Exercises
5.29 (Graded for Completeness) What Proportion Have College Degrees? According
to the US Census Bureau, about 27.5% of US adults over the age of 25 have a bachelor’s level (or
higher) college degree. For random samples of n= 500 US adults over the age of 25, the sample
proportions, ˆp, with at least a bachelor’s degree follow a normal distribution with mean 0.275 and
standard deviation 0.02. Draw a sketch of this normal distribution and label at least three points
on the horizontal axis.
Solution
AN(0.275,0.02) curve is centered at its mean, 0.275. The values two standard deviations away,
0.235 and 0.315, are labeled so that approximately 95% of the area falls between these two values.
They should be out in the tails, with only about 2.5% of the distribution beyond them on each
side. See the figure.
5.32 (Graded for Completeness) Critical Reading on the SAT Exam In the table from
Exercise 5.30, we see that scores on the Critical Reading portion of the SAT (Scholastic Aptitude
Test) exam are normally distributed with mean 501 and standard deviation 112. Use the normal
distribution to answer the following questions:
(a) What is the estimated percentile for a student who scores 690 on Critical Reading?
(b) What is the approximate score for a student who is at the 30th percentile for Critical Reading?
Solution
The plots below show the required endpoint(s) and/or probabilities for the given normal distribu-
tions. Note that a percentile always means the area to the left. Using technology, we can find the
endpoints and areas directly, and we obtain the answers below. (Alternately, we could convert to
a standard normal and use the standard normal to find the equivalent area.)
(a) The area below 690 is 0.95, so that point is the 95th percentile of a N(501,112) distribu-
tion.
(b) The point where 30% of the scores are below it is a score of 442.
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5.36 (Graded for Accurateness) Commuting TImes in St. Louis A bootstrap distribution
of mean commute times (in minutes) based on a sample of 500 St. Louis workers stored in Com-
muteStLouis is shown in the book. The pattern in this dot plot is reasonably bell-shaped so we
use a normal curve to model this distribution of bootstrap means. The mean for this distribution
is 21.97 minutes and the standard deviation is 0.65 minutes. Based on this normal distribution,
what proportion of bootstrap means should be in each of the following regions?
(a) More than 23 mintues
(b) Less than 20 minutes
(c) Between 21.5 and 22.5 minutes
Solution
The plots below show the three required regions as areas in a N(21.97,0.65) distribution. We see
that the areas are 0.0565, 0.0012, and 0.5578, respectively.
If converting to a standard normal, the relevant z-scores and areas are shown below.
(a) z=23−21.97
0.65 = 1.585. The area above 1.585 for N(0,1) is 0.0565.
(b) z=20−21.97
0.65 =−3.031. The area below -3.031 for N(0,1) is 0.0012.
(c) z=21.5−21.97
0.65 =−0.7231 and z=22.5−21.97
0.65 = 0.8154. The area between −0.7231 and 0.8154
for N(0,1) is 0.5578.
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5.62 (Graded for Accurateness) To Study Effectively, Test Yourself! Cognitive science
consistently shows that one of the most effective studying tools is to self-test. A recent study
reinforced this finding. In the study, 118 college student studied 48 pairs of Swahili and English
words. All students had an initial study time and then three blocks of practice time. During the
practice time, half the students studied the words by reading them side by side, while the other
half gave themselves quizzes in which they were shown one word and had to recall it partner. Stu-
dents were randomly assigned to the two groups, and total practice times was the same for both
groups. On the final test one week later, the proportion of items correctly recalled was 15% for
the reading-study group and 42% for the self-quiz group. The standard error for the difference in
proportions is about 0.07. Test whether giving self-quizzes is more effective and show all details of
the test. The sample size is large enough to use the normal distribution.
Solution
The relevant hypotheses are H0:pQ=pRvs Ha:pQ> pR, where pQand pRare the proportions
of words recalled correctly after quiz studying or studying by reading alone, respectively. Based on
the sample information the statistic of interest is
ˆpQ−ˆpR= 0.42 −0.15 = 0.27
The standard error of this statistic is given as SE = 0.07 and the null hypothesis is that the
difference in the proportions for the two group is zero. We compute the standardized test statistic
with
z=SampleStatistic −N ullP arameter
SE =0.27 −0
0.07 = 3.86
Using technology, the area under a N(0,1) curve beyond z= 3.86 is only 0.000056. This very small
p-value provides very strong evidence that the proportion of words recalled using self-quizzes is
more than the proportion recalled with reading study alone.
5.64 (Graded for Completeness) Penalty Shots in World Cup Soccer A study of 138
penalty shots in World Cup Finals games between 1982 and 1994 found that the goalkeeper cor-
rectly guessed the direction of the kick only 41% of the time. The article notes that this is “slightly
worse than random chance”. We use these data as a sample of all World Cup penalty shots ever.
Test at a 5% significance level to see whether there is evidence that the percent guessed correctly
is less than 50%. The sample size is large enough to use the normal distribution. The standard
error from a randomization distribution under the null hypothesis is SE=0.043.
Solution
We test H0:p= 0.5 vs Ha:p < 0.5 where pis the proportion of all World Cup penalty shots
for which the goalkeeper guesses the correct direction. The statistic from the original sample is
ˆp= 0.41 and the null parameter is p= 0.5. The standard error is SE = 0.043. We use this
information to find the standardized test statistic:
z=Samplestatistic −Nullparameter
SE
=0.41 −0.5
0.043
=−2.09.
This is a lower tail test, so we find the area below z=−2.09 in the lower tail of the standard
normal distribution. We see in the figure below that this area is 0.0183. The p-value for this test
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Document Summary
According to the us census bureau, about 27. 5% of us adults over the age of 25 have a bachelor"s level (or higher) college degree. For random samples of n = 500 us adults over the age of 25, the sample proportions, p, with at least a bachelor"s degree follow a normal distribution with mean 0. 275 and standard deviation 0. 02. Draw a sketch of this normal distribution and label at least three points on the horizontal axis. A n (0. 275, 0. 02) curve is centered at its mean, 0. 275. 0. 235 and 0. 315, are labeled so that approximately 95% of the area falls between these two values. They should be out in the tails, with only about 2. 5% of the distribution beyond them on each side. 5. 32 (graded for completeness) critical reading on the sat exam in the table from. Exercise 5. 30, we see that scores on the critical reading portion of the sat (scholastic aptitude.
