Practice Problems and Solutions for Exam 2
The Normal Distribution
As a student in a statistics class, you want to know how easy or hard it will be to pass the class. You know that the mean grade in the
class is 70 and the standard deviation is 9 (from previous semesters). Use this information to answer the following questions:
1. Assuming that grades are normally distributed, what proportion of students do you expect to pass the class (e.g., by scoring higher
than 59)? What proportion of students do you expect to fail the class? If there are 50 people in your class, how many people do you
expect to pass? How many people do you expect to fail?
What you know: Y=59, Y =70,
SY =9, N=50
Y  Y 59  70

 1.22
SY
9
Lower than the mean: p=.3888, p=.1112; higher than the mean: p=.5
You should expect that a proportion of .8888 students will pass the class (a proportion of .8888 cases have a raw score higher than a z
score of -1.22). You should expect that a proportion of .1112 students will fail the class. If there are 50 people in the class, you
should expect that 44 or 45 students will pass (.8888*50) and 5 or 6 will fail (.1112*50).
Z
2. What percentage of the class do you expect to receive a grade of between 60 and 70?
What you know: Y=60, Y =70,
SY =9
Y  Y 60  70

 1.11
SY
9
p=.3665, p=.1335
A proportion of .3665 students or 36.65% should receive a grade of between 60 and 70.
Z
3. What percentage of the class do you expect to receive a grade of between 70 and 80?
What you know: Y=80, Y =70,
SY =9
Y  Y 80  70

 1.11
SY
9
p=.3665, p=.1335
A proportion of .3665 students or 36.65% should receive a grade of between 70 and 80.
Z
4. What percentage of the class do you expect to receive a grade of between 50 and 59?
What you know: Y=50, Y=59, Y =70,
SY =9
Y  Y 50  70
Y  Y 59  70

 2.22 ; Z 

 1.22
SY
9
SY
9
p=.4868, p=.0132; p=.3888, p=.1112
.4868-.3888=.098
A proportion of .098 students or 9.8% should receive a grade of between 50 and 59.
Z
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5. What percentage of the class do you expect to receive a grade of between 40 and 69?
What you know: Y=40, Y=69, Y =70,
SY =9
Y  Y 40  70
Y  Y 69  70

 3.33 ; Z 

 0.11
SY
9
SY
9
p=.4996, p=.0004; p=.0438, p=.4562
.4996-.0438=.4558
A proportion of .4558 students or 45.58% should receive a grade of between 40 and 69.
Z
6. What score do you have to get to be in the top 10% of the class?
What you know: Y =70, SY =9
10%→p=.10
Using Appendix B, column C, I know that the z score associated with a p=.10 is 1.28.
Y  Y  ( Z * S Y )  70  (1.28 * 9)  81.52
You need to score at least 81.52 to be in the top 10% of the class.
Estimation
7. As a researcher for the US Government, you are interested in predicting the mean age in the US. You generate a random sample
(N=1,422) and discover that the mean for the sample is 44.54 years and the standard deviation is 17.03. Estimate the mean age for the
US population. Specifically, what are the point estimate and confidence interval (99%)? Interpret the point estimate and confidence
interval.
The Point estimate is 44.54 years of age
SY 
SY
N

17.03
1422
 .452
The confidence interval is:
CI  Y  ( Z * SY )  44.54  (2.58 * .452)  44.54  1.17 or (43.37, 45.71)
We expect that the mean age for the US population is 44.54 years of age (although we know that this is probably incorrect). We can
be 99% certain that the true population mean for age is between 43.37 and 45.71 years of age. In 99 out of 100 samples, our
confidence interval would contain the true population mean.
8. As an avid fan of Survivor (the TV show, not the band), you want to know what percent of the US population watched last season’s
final episode. You generate a random sample (N=1,000) and discover that 19.3% of the sample watched the final episode. Estimate
the percent for the US population. Specifically, what are the point estimate and confidence interval (95%)? Interpret the point
estimate and confidence interval.
The point estimate is 19.3%
Sp 
( p)(1  p)
(.193)(1  .193)

 .012
N
1000
The confidence interval is: CI  p  ( Z * S p )  .193  (1.96 * .012)  .193  .024 or (.169, .217)
This is the same as 19.3%±2.4% or (16.9%, 21.7%)
We expect that 19.3% of the US population watched the final episode (although we know this point estimate is probably wrong). We
can be 95% certain that the true population percent is between 16.9% and 21.7%. In 95 out of 100 samples, our confidence interval
would contain the true population percent.
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Testing Hypotheses
9. As a researcher, you are interested in studying racial differences in family size - particularly in the number of children. You want
to know if White or Black respondents have larger families. You think that there is a difference, but you don't know which group, on
average, has more children. State your research and null hypotheses (using symbols) and select alpha (=.01). Use the following
information to calculate the t statistic (assume that the variances are equal). Along the way, you’ll need to find the critical t value and
the standard error for the difference in means. Should you reject or fail to reject the null hypothesis? What should you conclude?
race RACE OF
RESPONDENT
N
Mean
Std. Deviation
Variance
Std. Error
Mean
1 WHITE
1111
1.71
1.561
2.437
.047
2 BLACK
202
2.09
1.859
3.456
.131
childs NUMBER OF
CHILDREN
Null hypothesis:
Y  Y
W
Research hypothesis:
B
Y  Y
W
B
Alpha=.01
df = (N1 + N2) – 2 = 1111 + 202 – 2 = 1311
Critical t value = 2.576 (2 tailed hypothesis, alpha=.01, df=∞)
SY 1 Y 2 
t
( N1  1) SY21  ( N 2  1) SY22
( N1  N 2 )  2
N1  N 2
(1111  1)2.437  (202  1)3.456 1111  202

 .124
N1 N 2
(1111  202)  2
1111* 202
Y 1  Y 2 1.71  2.09

 3.06
SY 1 Y 2
.124
We reject the null hypothesis because the observed t value is greater than the critical t value. In other words, the difference in family
size in our sample is larger than we would expect if there really is no difference in the population. We wouldn’t expect to get this
sample difference very often if there is no difference in the population. Thus, we can conclude that there probably is a difference
between White and Black respondents in family size in the US.
10. As a CEO and titan of industry, you are interested in predicting which presidential candidate will win the 2012 election (so that
you can contribute a lot of money to their campaign and win their good graces). You hypothesize that more eligible voters will vote
for Barack Obama (over Sarah Palin) and select an alpha of .05. You generate a random sample (N=500) and discover that 48%
(N=240) of the sample plans to vote for Obama and 42% (210) of the sample plans to vote for Palin (the rest are not sure). What do
the results suggest (i.e., who should get your money)? (HINT: Perform the z test to see if the difference in proportions is significant).
Null hypothesis:  O   M
Research hypothesis:
O   M
Alpha=.05
Critical z value = 1.645 (1 tailed hypothesis, alpha=.05)
S p1  p2 
Z
p1 (1  p1 ) p 2 (1  p 2 )
.48(1  .48) .42(1  .42)



 .047
N1
N2
240
210
p1  p 2
.06

 1.28
S p1  p2
.047
We fail to reject the null hypothesis because the observed z value is not larger than the critical z value. In other words, the difference
in proportions (percents) in the sample is not larger than we would expect if there really is no difference in the population. We would
expect to get this sample difference pretty often if there is no difference in the population. Thus, we should contribute money to both
candidates.
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