Download Sampling Theory Exercises

Sampling Theory Exercises
Use a standard normal table or the JMP calculator PROBABILITY function NORMAL
DISTRIBUTION or NORMDIST to "compute" the standard normal probabilities. In
fact, the entire exercise can be done on JMP.
1. Dwarf Apple Trees (This builds on the earlier exercise). The yearly growth of
dwarf-apple-tree seedlings can be measured by the increase in the length of the
central leader. Suppose that the second-year growth of such trees is normally
distributed with a mean of 20 cm and a standard deviation of 6 cm.
a. Compute the fraction of such dwarf-apple-tree seedlings that would be
expected to have a second-year growth of between 18 and 22 cm.
22  20 
18  20
P 18  Y  22  P 
Z

6 
 6
 P 0.33  Z  0.33
 P Z  0.33  P Z  0.33
 0.6306  0.3694
 0.2612
b. Now consider the sampling distribution of the sample mean for samples of
size 4. Compute the expected value of the sampling distribution.
 
E Y  E Y   20
c. Compute the standard deviation of the sampling distribution of the mean
for samples of size 4.
 
SD Y   Y 
Y
n

6
 3.0
4
d. For samples of size 5, compute the probability that the sample mean is
between 18 and 22 cm.
22  20 
18  20
P 18  Y  22  P 
Z

6 5 
 6 5


 P 0.745  Z  0.745
 P Z  0.745  P Z  0.745
 0.7720  0.2280
 0.5439
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e. For samples of size 25, compute the probability that the sample mean is
between 18 and 22 cm.
18  20
22  20 
P 18  Y  22  P 
Z

6 25 
 6 25
 P 1.67  Z  1.67


 P Z  1.67  P Z  1.67
 09522  0.0478
 0.9044
f. For samples of size 100, compute the probability that the sample mean is
between 18 and 22 cm.
 18  20
22  20 
P 18  Y  22  P 
Z

6 100 
 6 100


 P 3.33  Z  3.33
 P Z  3.33  P Z  3.33
 099957  0.000429
 0.99914
g. Graph the probability that the mean is between 18 and 22 cm as a function
of the sample size for sample sizes of 1 to 100.
2. A veterinarian found that the average time it takes residents to perform a certain
procedure is 12 minutes. Assume that the time it takes residents to perform the
procedure is normally distributed with a mean of 12 minutes and a standard
deviation of 2 minutes.
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a. Compute the probability that a randomly selected resident would take
between 11 and 13 minutes to perform the procedure, i.e., within 1.0
minute of the mean.
13  12 
11  12
P 11  Y  13  P 
Z

2 
 2
 P 0.50  Z  0.50
 P Z  0.50  P Z  0.50
 0.6915  0.3085
 0.3829
b. Graph the probability that the sample mean would be between 11 and 13
minutes, for samples of size 1 to 100.
1
P(11<Time<13)
0.9
0.8
0.7
0.6
0.5
0.4
-10
0
10
20
30
40
50
60
70
80
90
100
n
c. If you wanted the probability of being within 1.0 minute of the mean to be
95%, what is the minimum sample size that would be required? You can
read this off the graph, or solve the appropriate formula for sample size.
1
P(11<Time<13)
0.9
0.8
0.7
0.6
0.5
0.4
-10
0
10
20
30
40
50
60
70
80
90
100
n
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

P 11  Y  13  0.95
11  12
13  12 
 P
Z

2 n 
 2 n
 1
1 
 P
Z
  0.95
2 n 
 2 n
 P 1.96  Z  1.96  0.95
implies
1
 z0.975  1.96
2 n
where z0.975 denotes the 0.975 quantile of the standard normal distribution.
Solving the equation for n yields
 1.96  2  
n
  15.4
1


2
indicating a sample of size 16 would be required. (notice that we always
round up.)
In general, the formula for prescribing the sample size is
 z1 2 
n

 MOE 
2
where z1 2 is the (1 – /2) quantile of the of the standard normal
distribution,  is the population standard deviation, and MOE is the
desired margin of error.
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