AP Statistics Second Semester Exam Review Answers
1) Suppose we want a 90% confidence interval for the average amount spent on books by freshmen
in their first year at a major university. The interval is to have a margin of error of $2, and the
amount spent has a normal distribution with a standard deviation   $30 . The number of
observations required is
margin of error = t*
n = 608.86

2  1.645
n
number of observations = 609
30
n
2) Scores on the SAT Mathematics test (SAT-M) are believed to be normally distributed. The
scores of a random sample of three students who recently took the exam are 550, 620, and 480.
A 95% confidence interval for  based on these data is
s
Confidence interval is:
x t*
n
70
95% confidence interval is: 550 173.89 or (376.11, 723.89)
550  4.303
3
3) The following represents some computer output that relates the number of Manatee deaths to the
number of powerboats registered in Florida.
Predictor
Constant
Boats
Coef
-41.430
0.12486
StDev
7.412
0.01290
t ratio
-5.59
9.68
P
.000
.000
a) Write the least squares regression line for predicting the number of manatee deaths from the
number of powerboat registrations.
ˆ  41.430  0.12486( Boats)
Deaths
b) What is the intercept of the least-squares regression line?
The intercept is -41.430. In the context of the problem it would mean that if there were no
boats registered, then we would have an increase of 41.430 manatees. It doesn’t make sense
in this problem.
c) Interpret the slope of the line in the context of the problem.
The slope of 0.12486 means that as the number of powerboats registered in Florida increases
by 100, the number of manatee deaths increases by 12.5.
4) Birth weights at a local hospital have a normal distribution with a mean of 110 oz. and a
standard deviation of 15 oz. Find the proportion of infants with birth weights between 125 oz.
and 140 oz.
Since there is a normal distribution of weights, we can use the z-calculations to find the answer.
x   125  110
z

 1.0

15
Look in the z-table for the proportion with a z-score less than 1.0 P( z  1.0)  .8413
z
x 


140  110
 2.0
15
P( z  1.0)  .9772
P(125  x  140)  .1359
5) The heights (in inches) of males in the United States are believed to be normally distributed with
mean  . The average height of a random sample of 25 American adult males is found to be 69.72
inches and the standard deviation of the 25 heights is found to be sx  4.15 . The standard error of
the mean is
standard error =
s
4.15

 .83
n
25
6) Referring to the information in question 5, if the t-test statistic is found to be 1.75. At what level
is this finding significant?
Look up a z-score of 1.75 in the z-table. A z-score of 1.75 corresponds to a p-value of
.0401. We can say that the finding would be significant at the 5% level of   .05 .
7) Bags of a certain brand of tortilla chips claim to have a net weight of 14 ounces. Net weights
actually vary slightly from bag to bag and are normally distributed with mean   14 and standard
deviation .2. In a random sample of 10 bags of tortilla chips, what is the probability that the mean
net weight is between 13.7 and 14.3?
Since we have a sample of size 10, we need to use the sample size as part of our calculation.
x   13.7  14
z

 .47

2
n
10
Look in the z-table for the proportion with a z-score less than -.47 P( z  .47)  .3192
x   14.3  14
P( z  .47)  .6808
z

 .47

2
n
10
P(13.7  x  14.3)  .3616
8) Dentists are increasingly concerned about the growing trend of local school districts to grant soft
drink companies exclusive rights to install soda pop machines in schools in return for money that
goes directly into school coffers. According to a recent study by the National Soft Drink
Association, 62% of schools nationally already have such contracts. This comes at a time when
dentists are seeing an alarming increase in horribly decayed teeth and eroded enamel in the mouths
of teenagers and young adults. With ready access to soft drinks, children tend to drink them all day.
That, combined with no opportunity to brush, leads to disaster, dentists say. Suppose that 20
schools around the county are randomly selected and asked if they have a soft drink contract. What
is the probability that at most 8 of these schools have contracts?
This is a binomial setting for probability.
n
P( X  8)    p k (1  p) n k 
k 
 20 
 20 
 20 
 20 
0
20
1
19
2
18
8
12
  .62  .38     .62  .38     .62  .38   ...    .62  .38   .0381
0
1
2
8
Enter it in the calculator as Binomcdf (20,.62,8).
9) In a process for manufacturing glassware, glass stems are sealed by heating them in a flame.
The temperature of the flame varies a bit. Here is the distribution of the temperature X measured in
degrees Celsius:
Temperature
540
545
550
555
560
Probability
0.1
0.25
0.3
0.25
0.1
Find the mean temperature  x and the standard deviation  x .
This is a probability distribution, so we can use the probability formulas for expected value:
x   xi pi   540.1  545.25  550 .3  555.25  560 .1  550
 x2    xi   x  pi 
2
 540  550  .10    545  550  .25  550  550  .30   555  550 
2
2
2
To find the standard deviation, take the square root of the variance:
2
(.25)  560  550  .10   32.5
2
32.5  5.70
You can do the calculations in the calculator by putting the temperature data in L1 and the
probabilities in L2. You can find the mean and standard deviation by doing 1-variable statistics
L1,L2. (Stat >>>CALC>>>1-Var Stats L1, L2)
10) Consider a screening test for prostate cancer that its maker claims will detect the cancer in 85%
of the men that actually have the disease. One hundred seventy-five men who have been
previously diagnosed with prostate cancer are given the screening test, and 141 of the men are
identified as having the disease. Does this finding provide evidence that the screening test
detects the cancer at a rate different from the 85% claimed by its manufacturer?
We need to do a hypothesis test to test the sample results against the claim. Follow the
PHANTOMS procedure.
P: Proportion of prostate cancer detected in men using a screening test.
H: H o : p  .85 Proportion of prostate cancer detected is .85
H a : p  .85 Proportion of prostate cancer detected is not equal to .85
A: Assume SRS of men with prostate cancer. Our sample is less than 10% of the population.
There are more than 1750 men in the US with prostate cancer.
We have a possibility for more than 10 successes and 10 failures in the sample:
np  10
175*.85  148.75
n(1  p)  10 175*.15  26.25
N: 1-proportion z-test
T: z 
pˆ  p
.806  .85

 1.64
p (1  p )
.85*.15
n
175
O: p-value = .10
  .05
M: Retain the null hypothesis.
S: We do not have enough statistical evidence, p-value of .10 at   .05 , to reject the null
hypothesis. We must retain the null hypothesis that the screening test for prostate cancer
successfully detects cancer is 85% of patients with prostate cancer.
11) A geneticist hypothesizes that half of the given population will have brown eyes and the
remaining half will be evenly split between blue and green-eyed people. In a random sample of
60 people from this population, the individuals are distributed as shown in the table below.
Perform a goodness of fit test of the geneticist’s hypotheses about eye color.
Brown Eyes
34
30
Observed
Expected
Green Eyes
15
15
Blue Eyes
11
15
P: Population will have the proportion of eye color that a geneticist believes is correct.
H:
H o : Proportions of eye color is the same as what a geneticist believes is correct
H a : Proportions of eye color is not the same as what a geneticist believes is correct
A: All expected counts are greater than 5. There was a random sample of people from the
population. See table for expected counts.
N: Chi-Square Goodness of Fit test
T:  2  
 Observed  Expected 
O: p-value = .449
Expected
2
 34  30

30
2
15  15

15
2
11  15

15
2
 1.6
df=2
  .05
M: Retain the null hypothesis.
S: We do not have enough statistical evidence, p-value of .449 at   .05 , to reject the null
hypothesis. We must retain the null hypothesis that the proportion of eye color for the population is
the same as what the geneticist believes is correct. One half the population will have brown eyes,
25% will have blue eyes and 25% will have green eyes.
12) A researcher wished to compare the average amount of time spent in extracurricular activities
by high school students in a suburban school district with that in a school district of a large city.
The researcher obtained an SRS of 60 high school students in a large suburban school district and
found the mean time spent in extracurricular activities per week to be x1  6 hours with a standard
deviation of s1  3 hours. The researcher also obtained an independent SRS of 40 high school
students in a large city school district and found the mean time spent in extracurricular activities per
week to be x2  4 with a standard deviation s2  2 . Construct a 95% confidence interval for the
difference in the mean time spent in extracurricular activities of the two groups.
P: Difference in the mean time spent in extracurricular activities for students in a suburban school
district and those students in a large city school district.
A: SRS of both student groups
Large sample size (60 and 40) so by the Central limit theorem we can assume normality of the
sample distributions.
N: 2-sample confidence interval
I:
 xs  xc   t *
2  .9922
ss2 sc2
32 22

 =  6  4  t *
60 40
ns nc
df=98
(1.0078, 2.9922)
C: We are 95% confident that the true difference in mean time spent in extracurricular activities
between suburban district students and city students is 1.0078 to 2.9922. Since 0 is not within the
interval, we are confident that there is a difference in the mean time spent by the 2 groups of
students. Suburban students spend more time on extracurricular activities than do city students.