NAME Mrs. Jenson
AP STATISTICS
WORKSHEET 10.3
David, a starting player for a major college basketball team, made only 40% of his free throws last
season. During the summer he worked on developing a softer shot in the hope of improving his freethrow accuracy. In the first eight games of this season, David made 25 free throws in 40 attempts. Let p
be his probability of making each free throw he shoots this season.
1. What is
, David’s sample proportion of successes?
25
p̂ =
≈ 0.625 n = 40
40
2. Construct and interpret a 90% confidence interval for David’s free-throw success.
We want to estimate p, the true proportion of free-throws that David will make during this basketball
season. We will use p̂ to estimate the population parameter p with a 90% level of confidence. The
population in this case is David’s total free throw attempts this season.
Conditions:
SRS – We are using all of David’s free throw attempts in his first 8 games of the season, so this is not an
SRS. However it is representative of the basketball season (the population of all free throw attempts).
We will proceed with caution, but feel confident that our sample is representative of the population of all
free throw attempts.
Normal? – Because np̂ ≥ 10 →(40)(.625) ≥ 10 and n(1 − p̂) ≥ 10→ (40)(.375) ≥ 10 , we can assume that
the sampling distribution for p̂ is approximately normal and can use normal calculations to calculate
our confidence interval.
Independent? It is safe to assume that one free throw does not
impact another. We can also use the rule that
10n ≤ N → 10(40) ≤ 400 and safely assume that the 40 free throw
attempts we are using in our sample is less than 10% of all free throw
attempts David will have this season.
Since our conditions are met, it is OK TO PROCEED with a Z-Interval for a population proportion.
p̂ ± z *
p̂(1 − p̂)
where p̂ = 0.625, n = 40, z* = 1.645 → (0.4991 , 0.7509)
n
Based on our sample of David’s free throws this season, we are 90% confident that the interval (0.4991 ,
0.7509) captures the true population proportion of all free throws to be made this season. Because the
value of 0.40 (40%), his free throw percentage from last year, does NOT fall within this interval, it is
likely that David has actually improved his free throw shot this year.
3. If we want to decrease the margin of error for this confidence interval so that we are within 5%
on either side, how many shots will David have to take?
Since we do not know p, we should use p* = 0.50 to estimate n:
(.5)(.5)
1.645
≤ 0.05 → n ≥ 278 David should take 278 shots for our sample.
n
NAME Mrs. Jenson
AP STATISTICS
WORKSHEET 10.3
Sample surveys usually contact large samples, so that we can use the large-sample confidence interval if
the sample design is close to an SRS. Scientific studies often use small samples that require the plus
four method. For example, the small round holes you often see in seashells were drilled by other sea
creatures, who ate the former owner of the shell. Whelks often drill into mussels, but this behavior
appears to be more or less common in different locations. Investigators collected whelk eggs from the
coast of Oregon, raised the whelks in the laboratory, then put each whelk in a container with some
delicious mussels. Only 9 of 98 whelks drilled into a mussel. We want to find the proportion of whelks
that drill holes into a mussel.
4. Why can’t we use the large-sample confidence interval?
The sample is not large enough given the extreme value of p-hat, so we cannot assume that the sampling
distribution of p-hat is approximately normal. (it fails the test:
9
np̂ ≥ 10 →(98)( ) = 9 and p̂(1 − p̂) ≥ 10→ (98)(.375) ≥ 10
98
5. Construct and interpret the plus four 90% confidence interval for the proportion of Oregon
whelks that will spontaneously drill mussels.
We want to estimate p, the true proportion of whelks that drill holes into mussels. We will use p̂ to
estimate the population parameter p with a 90% level of confidence. Because the sample size is small
relative to the very low value of p̂ , we will use the “plus four” method to construct a confidence
p̂ + 2
9+2
=
≈ 0.1078 to estimate p.
interval for p. We will use p =
p̂ + 4 98 + 4
Conditions:
SRS – Our sample is representative but is not an SRS. We are using the 98 whelks collected as eggs
from the coast of Oregon and raised in a laboratory. We will use the plus – four method and proceed
with caution.
9
91
) = 9 and p̂(1 − p̂) ≥ 10→ (98)( ) ≥ 10 , we are using the plus
98
98
four method to be more conservative in our estimation. Using this
method, we can assume that the sampling distribution for p̂ is
approximately normal and can use normal calculations to calculate
our confidence interval.
Normal? – Because np̂ ≥ 10 →(98)(
Independent? It is safe to assume that one whelk does not impact
another. We can also use the rule that 10n ≤ N → 10(98) ≤ 980 and
safely assume that there are many more than 980 whelks in the world
(more than 10 times our sample size).
Since our conditions are met, it is OK TO PROCEED with a Z-Interval for a population proportion.
p ± z *
p (1 − p )
where p = 0.0.1078, n = 102, z* = 1.645 → (0.0573, 0.1584)
n
We are 90% confident that the interval (0.0573,0.1584) captures the true population proportion of all
whelks that bore holes into mussels.
NAME Mrs. Jenson
AP STATISTICS
WORKSHEET 10.3
6. [From the 2010 AP Exam]. A humane society wanted to estimate with 95 percent confidence the
proportion of households in its county that own at least one dog.
a. Interpret the 95 percent confidence level in this context.
Careful! It is asking for you to interpret the confidence LEVEL: The 95% confidence level means that
if we were to take many, many samples of the same size from the population and construct a 95%
confidence interval from each sample, in the long run, 95% of the intervals would contain the true
population of the households that own at least one dog (the population parameter we are trying to find).
The human society selected a random sample of households in its county and used the sample to
estimate the proportion of all households that own at least one dog. The conditions for calculating a
95% confidence interval for the proportion of households in this county that own at least one dog were
checked and verified, and the resulting confidence interval was 0.417+/- 0.119.
b. A national pet products association claimed that 39% of all American households owned
at least one dog. Does the humane society’s interval estimate provide evidence that the
proportion of dog owners in its county is different from the claimed national proportion?
Explain.
No! The 95% CI is (0.298 , 0.536) which includes the value of 0.39 as a possible value for the
population proportion of households in this county that own at least one dog. Therefore, the CI does
NOT provide evidence that the proportion of dog owners in this county is different from the claimed
national proportion.
c. How many households were selected in the humane society’s sample? Show how you
obtained your answer.
The 95% CI has a margin of error = +/- 0.119 as given in the problem. Using the equation
p̂(1 − p̂)
p̂ ± z *
= 0.119 → n = 65.95 so we can conclude that the humane society must have selected
n
66 households for its sample.