AP Statistics
Quiz Review
1. It is generally believed that nearsightedness affects about 12% of children. A
school district gives vision tests to 133 incoming kindergarten children.
a. Describe the sampling distribution model for the sample proportion by naming
the model and telling its mean and standard deviation. Justify your answer.
b. Sketch and clearly label the model.
c. What is the probability that in this group over 15% of the children will be found
to be nearsighted?
2. Wildlife scientists studying a certain species of frogs know that past records
indicate the adults should weigh an average of 118 grams with a standard
deviation of 14 grams. The researchers collect a random sample of 50 adult frogs
and weigh them. In their sample the mean weight was only 110 grams. One of
the scientists is alarmed, fearing the environmental changes may be adversely
affecting the frogs. Do you think this sample result is unusually low? Explain.
3. A state’s Department of Education reports that 12% of the high school students
in that state attend private high schools. The State University wonders if the
percentage is the same in their applicant pool. Admissions officers plan to check
a random sample of over 10,000 applicants on file to estimate the percentage of
students applying for admission who attend private school.
a. The admissions officers want to estimate the true percentage of private
school applicants to within ±4%, with 90% confidence. How many applicants
should they sample?
b. They actually select a random sample of 450 applicants, and find that 46 of
those students attend private school. Create the confidence interval.
c. Interpret the confidence interval in this context.
d. Explain what 90% confidence means in this context.
e. Should the admissions officers conclude that the percentage of private school
students in their applicant pool is lower than the statewide enrollment rate of
12%? Explain.
Answers:
1. p = 0.12; n = 133
a) Conditions:
1) Randomization: This is not a random sample but we will assume that it is
representative of the population of the school district
2) 10% Condition: 133 children is less than 10% of all children in the school district.
3) Success/Failure: np = 133(0.12) = 15.96 ≥ 10
nq = 133(0.88) = 117.04 ≥ 10
All conditions have been met to use the Normal model for the distribution of sample
proportions.

(0.12)(0.88) 
N  0.12,
 = N(0.12, 0.0282)
133


68%
95%
99.7%
b)
c) pˆ  0.15
0.15  0.12
z
 1.064
0.0282
P( pˆ  0.15 ) = P(z > 1.064) = 0.1437
There is a 14.4% chance that this group would have over 15% nearsighted.
2.   118 g;   14 g n = 50
Conditions:
1) Randomization: Stated as a random sample
2) 10% Condition: 50 adult frogs is less than 10% of all adult frogs of this species
3) Large Enough?: 50 is a large enough sample since the distribution of weights
should be unimodal and symmetric
All conditions have been met to use the Normal model for the distribution of sample
means.
14 

N  118,
 = N(118, 1.980)
50 

x  110
110  118
z
 4.041
1.980
This sample mean is very low. It is over 4 standard deviations below the mean of 118 g.
The scientist should be alarmed.
3.
a) ME = z*SE( p̂ )
0.04  1.645
(0.12)(0.88)
n
n ≈ 179
b) n = 450
46
pˆ 
450
Conditions:
1) Randomization: Stated as a random sample
2) 10% Condition: 450 applicants is less than 10% of all applicants to the school
3) Success/Failure: npˆ = 46 ≥ 10
nqˆ = 404 ≥ 10
All conditions have been met to use the Normal model for a one-proportion z-interval
 46 
CI = pˆ  z * SE ( pˆ )  
  1.645
 450 
CI: (0.079, 0.126)
 46 450 404 450
450
c) We are 90% confident that the true proportion of applicants that attend private school
lies between 7.9% and 12.6%.
d) 90% of all random samples of size 450 will produce a confidence interval that
captures the true proportion.
e) Since 12% falls within the confidence interval the admissions officer can not conclude
that the percentage of private school students in their applicant pool is less than 12%.