Download AP Statistics Name_____KEY___________________________

AP Statistics
Wkst Chapters 18-20
Name_____KEY___________________________
Date________Period______Score____________
Distribution of Sample Proportion
1. A random sample of 188 books purchased at a local bookstore showed that 66 of the books were murder
mysteries. Suppose in fact that 40% of all books sold by this store in the past have been murder mysteries.
ˆ  66 / 188  .3511
a) What is the value of p̂ , the sample proportion? p
b) Can we safely apply the Normal model? Show the checks.
Yes, because the sample is stated to be random, each book’s style is independent of the next in the
sample, 188 books < 10% of all books sold by the store, and 66 successes and 122 failures are both at
least 10, thus the sample size is large enough.
c) For the distribution of the sample proportions, it will be approximated by a Normal model with the
mean of p̂ , µ( p̂ )= .4 and the standard deviation of p̂ , σ( p̂ ) =
(0.4)(0.6) / 188  .0357 .
d) Find probability that less than 70 murder mysteries are sold. (Remember to convert to a %)
P( pˆ  70 / 188)  P( z  (.3511  0.40) / 0.0357)  P( z  .775)  .2192
e) Find probability that between 60 and 75 murder mysteries are sold. (Remember to convert to a %)
P(60 / 188  pˆ  75 / 188)  P(2.2647  z  0.0298)  0.4763
2. Suppose that 80% of those receiving a flu shot and then are exposed to the flu never get the flu.
Suppose that 800 students at a university were selected at random from a student body of 20,000 and
given shots to prevent a certain type of flu. All 800 students were exposed to the flu, and 600 of them
did not get the flu.
ˆ  60 / 800  .75
a) What is the value of p̂ , the sample proportion? p
b) Can we safely apply the Normal model? Check Conditions.
Yes, because the sample was selected at random, 800 students <10% of the student body (20,000),
the likelihood of getting the flue for each student in the sample is independent, and there were at
least 10 successes (600 no flu) and 10 failures (200 flu), thus the sample size is large enough.
f) For the distribution of the sample proportions, it will be approximated by a Normal model with the
mean of p̂ , µ( p̂ )=.8 and the standard deviation of p̂ , σ( p̂ ) =
(0.8)(0.2) / 800  .0141.
c) Find probability that at least 680 students do not get the flu. (Remember to convert to a %)
P( pˆ  680 / 800)  P( z  .
.85  .8
)  P( z  3.5461)  0
0.0141
d) Find probability that between 500 and 650 students do not get the flu..(Remember to convert to %)
P(500 / 800  pˆ  650 / 800)  P(12.41  z  .8865)  .8123
1-Proportion Confidence Interval
3. Retailers report that the use of cents-off coupons is increasing. The Scripps Howard News Service (July
9, 1991) reported that the proportion of all households using coupons is 0.77. Suppose that this estimate
was based on a random sample of 800 households (that is, n = 800 and p̂ =.77).
a) Construct a 95% confidence interval for the true proportion of all households that use coupons.
CI  0.77  1.96 9.77)(.23) / 800  0.77  0.029  .741,.779
b) Explain what this interval means.
We are 95% confident that between 74.1% and 77.9% of all households use cents-off coupons.
c) Explain what “95% confidence “means.
Approximately 95% of all samples will produce intervals that capture the true proportion.
Note: This is similar to saying that there is approximately a 95% likelihood that this interval
contains the true proportion.
d)
What sample size should be selected to estimate this proportion within .05 and 95% confidence?
.05  1.96( (.77)(.23) / n
n  273
4. The Gallup Organization conducted a telephone survey on attitudes towards AIDS ( Gallup Monthly, 1991).
A total of 1014 individuals were contacted. Each was asked whether they agreed with the following
statement: “Landlords should have the right to evict a tenant from an apartment because that person has
AIDS.” One hundred one individuals in the sample agreed with this statement.
a) Use this information to construct a 90% confidence interval for the portion of who are in agreement
with this statement.
pˆ  101/ 1014  0.0996
CI  .0996  1.645
(.0996)(.9004)
 0.0996  .0155  (0.0841,.01151)
1014
b) Explain what this interval means.
We are 90% confident that between 8.4% and 11.5% of people believe that landlords have the right
to evict tenants because of AIDS.
c) Explain what “90% confidence “means.
Approximately 90% of all samples will produce intervals that capture the true proportion.
Note: This is “approximately” for two reasons: 1) We use p-hat instead of the true proportion, thus
we are estimating the SD(p-hat) with SE(p-hat); 2) the exact distribution is a binomial
transformation, not a normal distribution.
5. A consumer group is interested in estimating the p
roportion of packages of ground beef sold at a
particular store that have an actual fat content exceeding the fat content listed on the label.
a) How many packages of ground beef should be tested to estimate this proportion to with .06 with 98%
confidence?
.05  2.326 (.5)(.5) / n
b)
n  542
If we want a narrower Confidence Interval, what two things can be changed to accomplish this?
1) Increase the sample size (which decreases the SE and variability, or
2) Decrease the confidence level (which decreases z*)
1-Proportion Z-Test
6. A number of initiatives of the topic of legalized gambling have appeared on state ballots in recent years.
Suppose that a political candidate has decided to support legalization of casino gambling if he is convinced
that more than 2/3 of US adults approve of casino gambling. USA Today (June 17, 1999) reported the
results of a Gallup poll in which 1523 adults (selected at random from households with telephones) were
asked if they approved of casino gambling. The number in the sample who approved was 1035. Does the
sample provided convincing evidence that more than 2/3 approve>
Hypotheses:
Conditions:
H 0 : p  2 / 3; two- thirds of candidates support legalizati on of gambling
1) Sample was random
2) Each person in sample opinion on issue
H a : p  2 / 3; more than two- thirds support legalizati on of gambling
independent
p is the prop. of residents whosupport the legalizati on of casino gambling.
3) 1523<10% of all residents in population
4) np0=1523(2/3)=1015 and
Test name: 1 proportion z-test
nq0=1523(1/3)=508; thus, the sample size is
Mechanics (with labeled sketch):
large enough because both at least 10.
z
1035 2

1523 3  1.069
(2 / 3)(1 / 3)
1523
P  Value  P( pˆ  .680)  P( z  1.069)  .1423
Conclusion:
Fail to reject the null hypothesis that p=2/3 with such a big P-value of .14. There is insufficient evidence
to suggest that the true proportion of residents who support legalization of casino gambling is greater
than two-thirds.
7. The Associated Press (Feb. 27, 1995) reported that 71% of Americans age 25 and older are overweight, a
substantial increase over 58% figure from 1983. Although this information came from a Harris Poll (a
survey) rather than a census of the population, let’s assume that the nationwide population proportion is
exactly 0.71, Suppose that an investigator wishes to know whether the proportion of such individuals in
her state that are overweight differs from the national proportion. A random sample of size n = 600
results in 450 who are classified as overweight. What can the investigator conclude?
Hypotheses:
Conditions:
H 0 : p  .71; proportion of state overweight same as nat. avg.
1) Sample was random
H a : p  .71; prop. of state overweight not the same as nat. avg. 2) Each person in samples weight independent
3) 600<10% of all states residents
p is the prop. of overwieght people
4) np0=600(.71)=426 and nq0=600(.29)=174;
thus, the sample size is large enough because
in this state
both at least 10.
Test name: 1 proportion z-test
Mechanics (with labeled sketch):
.75  .71
z
 2.159
(.71)(.29) / 600
P  value  2 P( pˆ  .75)  2 P( z  2.159)  2(.0154)  .0308
Conclusion:
Reject the null hypothesis that p=.71 with such a low P-value of .031. There is evidence to suggest that
the true proportion of this states residents overweight is not 71%.
Compute a 95% confidence interval and explain how the CI verifies your conclusion.
CI=(.715,.785) Since the CI does not include .71, this is additional evidence to suggest that the true
proportion is not .71 and thus that claim should be rejected.