AP Statistics
Review: Sampling Dist. & Confidence Intervals
Name______________________
Part I. The facts you need to know.
1.
Know the properties of a Sampling Distribution of Sample Mean.
*x  
* X 

n
*Approximately Normal if
the population is normal or

n  30, (Central Limit Theorem)
2. Know the properties of a Sampling Distribution of Sample Proportions
*p  p
pq
n
*Approximately Normal if
* p 
np  5

nq  5
3. Know each variable & whether it's a statistic or a population.
Sample
(Statistic)
Population
(Parameter)
x

p
p
s

mean
proportion
st. Dev
4. What is the Central Limit Theorem?
If n is sufficiently large enough (30 or more), then the sampling distribution is well approximated by a
normal curve.
5. What is an unbiased statistic?
The mean of the sampling distribution is equal to the mean of the population.
6. How does sample size, confidence level, proportion, & standard deviation affect a confidence interval?
Larger sample size gives a smaller interval.
Larger degree of confidence gives a larger interval.
Smaller standard deviation gives a smaller interval.
The closer the proportion gets to 0.5, the larger the deviation – thus the larger the interval.
7. Know the properties of the t-distribution.
*Bell shaped curve with the center at 0.
*More spread out than the z-distribution (normal).
*As the degree of freedom increases, it approaches the z-distribution (normal).
*As the degree of freedom increases, the spread (st. dev) decreases.
8. Be able to construct a sampling distribution of means and find
x
and
x.
This is like the front side of the quiz on Chapter 7. Given population, do size 2. (sample means with
probabilities)
Part II. Computational Problems.
9. The nicotine content in a single cigarette of a particular brand has a distribution with a mean 0.8 mg
and standard deviation 0.1 mg. If 100 of these cigarettes are analyzed, what is the probability that
the resulting sample mean nicotine content will be
a.
less than 0.79?
*Approximately normal since n>30.



.79  .80 

P ( x  0.79)  P  z 
.1




100 

=P ( z  1)
=ncdf ( , 1)
=0.1587
b. between 0.785 to 0.81?
*Approximately normal since n>30.


 .785  .8
.81  .8 

P (0.785  x  0.81)  P 
z
.1
.1 



100 
 100
=P ( 1.5  z  1)
=ncdf ( 1.5,,1)
=0.7745
10. The mean time for taxi and takeoff for commercial jets is 8.5 minutes and standard deviation is 2.5
minutes. Assume it is approximately normal. What is the probability that for 36 jets on a given
runway the mean taxi and takeoff time will be at least 9 minutes?
*Approximately normal since n>30.



9  8.5 

P ( x  9)  P  z 
2.5 



36 

= P ( z  1.2)
=ncdf (1.2,  )
=0.1151
11. It was found that about 73% of American men would favor a law requiring a police permit to buy a gun.
What is the probability that in a sample of 38 American men, at least 66% will support gun permits?
*Approximately normal since
np  38(.73)  28  5

nq  38(.27)  10  5


.66  .73
P ( p  0.66)  P  z 

.73(.27

38

=P (z  0.97)






=ncdf (0.97,  )
=0.8340
12. A physician at the clinic in Grand Canyon Village estimates that 31% of the boating accidents on the
Colorado River in the Grand Canyon National Park occur at Crystal Rapids. Suppose there are 28
recently reported boating accidents in the park, what is the probability that between 25% and 50%
are from Crystal Rapids?
*Approximately normal since
np  28(.31)  9  5

nq  28(.31)  19  5

 .25  .31
.5  .31
P (0.25  p  0.5)  P 
z
 .31(.69)
.31(.69)

28
28

=P (0.686  z  2.17)
=ncdf (0.686, 2.17)
=0.7388






13. Suppose that 20% of the subscribers of a cable television company watch the shopping channel at
least once a week. The cable company is trying to decide whether to replace this channel with a new
local station. A survey of 100 subscribers will be undertaken. The cable company has decided to keep
the shipping channel if the sample proportion is greater than 0.25. What is the approximate
probability that the cable company will keep the shopping channel, even though the true proportion
who watch it is only 0.20?
*Approximately normal since
np  100(.2)  20  5

nq  100(.8)  80  5



.25  .2 
P ( p  0.25)  P  z 

.2(.8) 


100 

=P (z  1.25)
=ncdf (1.25,  )
= 0.1056
14. Retailers report that the use of cents-off coupons is increasing. It is reported the proportion of all
households that use coupons is 0.77. Suppose that this estimate was based on a sample of 800
households. Construct a 95% confidence interval for the true proportion of all households that use
coupons.
p = True population proportion of all households that use coupons
*SRS

Assumptions: 
np  800(.77)  616  5
* Approx.Normal  nq  800(.23)  184  5



pq 

p  z  
2
n 


 .77(.23) 
0.77  1.96 


800 

0.77  0.029
 0.741, 0.799 
  0.05
  0.025
2
z  invnorm(0.025)  1.96
2
I’m 95% confident that the true population
proportion of all households that use coupons
is between 74.1% and 79.9%.
15. A manufacturer of small appliances purchases plastic handles for coffeepots from an outside vendor.
If a handle is cracked, it is considered defective and must be discarded. A large shipment of plastic
handles is received. The proportion of defective handles is of interest. How many handles from the
shipment should be inspected to estimate p to within 0.1 with 95% confidence?
 pq 
ME  z 

2
 n 
 .5(.5) 
0.1  1.96 


n 

.25
(0.051) 
n
0.25
0.0026 
n
0.0026n  0.25
n  96.04
2
2
If p is not known – use 0.5.
So I need at least 97 in my sample.
16. The Gallup Organization conducted a telephone survey on attitudes toward AIDS. A total of 1014
individuals were 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. Use these data to construct a 90% confidence
interval for the proportion who are in agreement with this statement. Give an interpretation of your
interval.
p = True population proportion of all people who agree with the statement above.
*SRS


 101 

np  1014 
  101  5

Assumptions: 

 1014 
* Approx.Normal  
nq  1014  913   913  5

 1014 




  0.10
  0.05
2
z  invnorm(0.05)  1.645
2
 pq 

p  z 
2
n 


 0.0996(0.9004) 
0.0996  1.645 


1014


0.0996  0.0155
 0.0841, 0.1151
I’m 90% confident that the true population
proportion of all people who agree with the
statement is between 8.41% and 11.51%.
17. The center for Urban Transportation Research released a report stating that the average commuting
distance in the U. S. is 10.9 miles. Suppose that this average is actually the mean of a random sample
of 300 commuters and that the sample standard deviation is 6.2 miles. Estimate the true mean
commuting distance using a 99% confidence interval.
 = True population mean commuting distance for all US citizens
Assumptions:
*SRS

* Approx.Normal  (n  30)
*Use t since σ is unknown
  
x  t 

2
n
 6.2 
10.9  2.59 

 300 
10.9  0.92
 9.97, 11.83 
  0.01
  0.005
2
t  invT (0.005,299)  2.59 or 2.626(chart )
2
I’m 99% confident that the true population
mean commuting distance for all US citizens is
between 9.98 miles and 11.82 miles.
18. A manufacturer of college textbooks is interested in estimating the strength of the bindings
produced by a particular binding machine. Strength can be measured by recording the force required
to pull the pages from the binding. If this force is measured in pounds, how many books should be
tested to estimate with 95% confidence to within 0.1 lb, the average force required to break the
binding? Assume that  is known to be 0.8 lb.
  
ME  z 

2 n 
 .8 
0.1  1.96 

 n
.8
0.051 
n
0.051 n  .8
n  15.68
n  245.8624
So I need at least 246 in my sample.
19. Construct a 95% confidence interval for the population mean. You took a random sample of 12 twoslice toasters and found the mean price was $61.12 and the standard deviation was $24.62.
 = True population mean price of all toasters
Assumptions:
*SRS

*PopulationApprox.Normal  but use t-dist since n<30
*Use t since σ is unknown
 s 
x  t 

2
n
  0.05
  0.025
 24.62 
61.12  2.201

 12 
61.12  15.64
t
2
, df
2
Use t-chart.
 2.201
I’m 95% confident that the true population
mean price of all toasters is between $45.45
and $76.76
 45.45, 76.76 
20. Construct a 90% confidence interval for the ACT scores shown below.
26
17
22
26
23
23
12
24
19
20
25
14
23
21
21
23
25
20
10
22
 = True population mean ACT score for all students
Assumptions:
*SRS

*PopulationApprox.Normal  but use t-dist since n<30
*Use t since σ is unknown
  0.10
  0.05
2
t
, df
2
 1.725
Use t-chart.
 s 
x  t 

2
n
 4.4792 
20.8  1.725 

 20 
20.8  1.728
19.072, 22.528 
I’m 90% confident that the true population
mean ACT score for all students is between
19.072 and 22.528.
21. Explain what a 90% confidence level means.
If I repeat this process over and over, 90% of the intervals formed will contain the true
population mean.