AP Statistics
Hypothesis Testing: Mixed Practice
For each problem:
a. State the Null and Alternative Hypotheses.
b. Choose an appropriate test after listing and checking appropriate conditions.
c. Perform the computations.
d. Determine the P-value and make a decision regarding H0.
e. State your conclusion in context.
f. Construct a 90% confidence interval for the value of the parameter in question.
1. A coffee vending machine dispenses coffee into a paper cup. The sign on the machine
indicates that I’m supposed to get 10 ounces of coffee, but the amount varies slightly from cup to
cup. Even if I assume that the amount of coffee is normally distributed, I’m still wondering if there
is evidence at the level  = .05 that the machine is shortchanging me. Here are the amounts
measured in a random sample of 20 cups.
9.9
9.6
9.7
9.8
9.7
9.8
10.1
10.0
10.0
9.8
9.9
9.9
10.1
10.0
9.6
9.5
9.9
9.5
10.2
9.9
2. During the 1990’s about 4.5% of high school seniors enlisted in the military after graduating
from school. There is some speculation that the percentage of high school graduates enlisting has
since gone down. To investigate, pollsters randomly selected 5 cities in Upstate New York and
then randomly selected one high school in each city. They obtained the following compilation of
responses from 100 randomly chosen seniors at each of the 5 schools. Do these data suggest
that the percentage who enlist in the military has changed significantly? ( = .05)
Plans
College
Work
Military
Other (travel, parenting, etc.)
Undecided
Count
289
112
26
51
22
3. An admissions officer at a small liberal arts college suspects that male applicants to its honors
engineering program outscore female applicants on a specially designed placement test. A
random sample of applicants for admissions in 2011 resulted in the following data:
Gender
Male
Female
Sample
size
32
36
Mean
Score
75.5
71.2
Standard
Deviation
10.3
11.4
Do these data indicate there is a significant difference ( = .05) among these two groups of
applicants?
4. A bio-engineering firm claims that no more than 5% of its artificial joints will fail within 4 years
of implantation. A simple random sample of 192 recipients found that 17 had artificial joints that
had failed within 4 years. Test at the level  = .01 whether or not these sample data suggest that
the company’s claim is too low.
5. A recent article in a local magazine stated that high school seniors contributed an average of
87.5 hours of community service. In a random sample of 52 seniors, the mean number of
community service hours worked was 90 with a standard deviation of 12.5 hours. Test whether or
not these data cast doubt on the magazine’s claim at the level  = .05.
Mixed Hypothesis Test Answers
1. Coffee Machine.
I think the machine is dispensing less than the advertized 10 ounces of coffee.
H0:  = 10
HA:  < 10
Random? stated
Normal? assumed
Independent? Presumably, the machine dispenses more than 200 cups of coffee.
x 
9.845  10

 3.49
Use a one-sample t-test. t 
 s x   .1986 


 n


20


P-value: P  t  3.49  =.0012 < .05 so we reject H0.
There is statistically strong evidence that the coffee machine is dispensing less than 10 ounces of coffee.
Based on my sample evidence, I am 90% confident that the average cup of coffee from the machine contains
between 9.77 and 9.92 ounces.
Notice: Even the high end of the confidence interval was lower than the advertized figure of 10 ounces. It’s no
surprise, then that we rejected the null hypothesis.
2. Military Enlistment
We wonder if the proportion of seniors enlisting has gone down from the advertized 4.5% figure.
H0:  = .045
HA:  < .045
Random? stated
Normal? np > 10, n(1 – p) > 10? Yes, 26 > 10 and 474 > 10
Independent? Yes, there are more than 5000 seniors in Upstate New York schools.
p̂  
.052  .045

 .755
Use a one-sample z-test. z 
(1  )
(.045)(.955)
n
500
P-value: P  z  .755  =.774 > .05 so we cannot reject H0.
The evidence does not indicate that the percentage of seniors enlisting in the military has gone down. On the
contrary, the data suggest the exact opposite, that the percentage may have gone up slightly, (since .052 > .045),
although, there too, the amount is not statistically significant.
Based on this recent tally, we can be 90% confident that the actual percentage of seniors who will enlist in the military
is between 3.6% and 6.8%, so it may not have changed from the previous mark.
3. Engineering Placement Test
We wonder if the mean score for male applicants is higher than the mean score for female applicants.
H0: Male – Female = 0
HA: Male – Female > 0
Random? stated
Normal? n1 = 32 > 30, n2 = 36 > 30 so the distribution will be roughly normal.
Independent? Applicants’ scores are presumably independent.
( x1  x2 )  (1   2 )
(75.5  71.2)  (0)

 1.634
Use a two-sample t-test. t 
2
2
2
2
s1
s2
(10.3)
(11.4)


32
36
n1
n2
P-value: P  t  1.634  =.0535 > .05 so we cannot reject H0. (close, but not quite.)
Although the males in the sample had a higher mean score than the females, the evidence is not strong enough to
assert that men, in general, score higher on the test than women do.
Based on our evidence, we 90% confident that the actual difference in the average scores of the two groups is
between = -.09 and 8.69.
4. Artificial Joints
We wonder if the proportion of artificial joints that fail is more than the advertized figure of 5%.
H0:  = .05
HA:  > .05
Random? stated
Normal? np > 10, n(1 – p) > 10? Yes, 17 > 10 and 165 > 10
Independent? Yes, we presume there were a large number of joint replacements.
p̂  
.0885  .05

 2.448
Use a one-sample z-test. z 
(1  )
(.05)(.95)
n
192
P-value: P  z  2.448  =.007 < .05 so we reject H0.
We have strong evidence that the percentage of artificial joints that fail within the first 4 years is higher than the
company’s advertized figure of 5%.
Based on our evidence, we are 90% confident that the actual percentage of artificial joints that fail within the first 4
years is between 5.48% and 12.23%.
Notice: The lower end of the confidence interval was higher than the advertized figure. It should come as no
surprise, then, that we rejected the null hypothesis.
5. Community Service
We wonder if the number of hours high school seniors devote to community service is higher than 87.5, the figure
published by the local magazine.
H0:  = 87.5
HA:  > 87.5
Random? stated
Normal? n > 30? Yes, our sample had 52
Independent? Yes, 52 is presumably a small fraction of the total number of seniors attending local high schools..
Use a one-sample t-test. t 
x 
 sx 


 n

90  87.5
 12.5 


 52 
 1.442
P-value: P  t  1.442  =.078 > .05 so we cannot reject H0.
Although the mean of 90 hours community service posted by our sample was higher than the advertized 87.5 hours, it
was not strong enough evidence to refute the magazine’s claim.
Based on our evidence, we are 90% confident that the mean number of hours local high school seniors devote to
community service is between 87.1 and 92.9.
The calculator does a lot of this work for us, but to receive full credit on the AP, we need to write
enough of to convince the examiners that we know what the calculator is doing. That includes
showing the computations and stating the conclusion in context.