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Review 2
92.1.9
Chapter 7:
 Compute the standard error and the probability of the sample
mean x or the sample proportion p within some range.
 Different sampling methods.
Example 1:
MNM Corporation gives each of its employees an aptitude test. The
scores on the test are normally distributed with a mean of 75 and a
standard deviation of 15. A simple random sample of 25 is taken from a
population of 500.
(a) What is the probability that the average aptitude test score in the
sample will be between 70.14 and 82.14?
(b) What is the probability that the average aptitude test score in the
sample will be greater than 82.68?
(c) Find a value, C, such that P( x  C) = .015.
[solution:]
(a)
Since  X 

n

15
25
3,
P70.14  X  82.14 
 70.14  75 X   X  75 82.14  75 

 P




3

3
3
X


 P 1.62  Z  2.38  0.9387
(b)
 82.68  75 X   X  75 

P82.68  X   P


3
X
3 

 P2.56  Z   0.0052
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(c)
 X   X  75 C  75 
C  75 
  P Z 
PX  C   P


  0.015


3
3
3


X


C  75

 2.17  C  75  2.17  3  81.51
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Example 2:
A new soft drink is being market tested. It is estimated that 60% of
consumers will like the new drink. A sample of 96 taste-tested the new
drink.
(a) Determine the standard error of the proportion
(b)What is the probability that more than 70.4% of consumers will
indicate they like the drink?
(c) What is the probability that more than 30% of consumers will indicate
they do not like the drink?
[solution:]
(a)
p  0.6, n  96,  P 
0.61  0.6
 0.05 (since
96
np  57.6  5, n1  p  38.4  5
(b)
 0.704  0.6 P  p P  0.6 

P0.704  P   P



0
.
05

0
.
05
P


 P2.08  Z   0.0188
(c)
We need to compute the probability that less than or equal to 70% of
consumers will indicate they like the drink?
 P  p P  0.6 0.7  0.6 

PP  0.7   P




0
.
05
0
.
05
 P

 PZ  2  0.9772
2
Example 3:
Suppose we have a population of 40 elements
148 148 149 149 153 154 155
157 157 158 158 158 158 158
160 160 161 162 162 162 163
164 164 164 164 165 165 165
155
159
163
165
156
159
163
165
156
160
163
166
Suppose the first row of the table of random number is
63271 59986 71744 51102 15141 80714 58683 93108 13554 79945
Please use systematic sampling to obtain
(a) a sample of 5 elements.
(b) the sample mean and sample variance based on (a).
[solution:]
(a)
40/5=8. Thus, we need to divide the original data into 5 subsets and select
1 element from these subsets. The subsets are
Subset 148
148
149
149
153
155
155
154
1
Subset 156
156
157
157
158
158
158
158
2
Subset 158
159
159
160
160
161
162
160
3
Subset 162
162
163
163
163
164
164
163
4
Subset 164
164
165
165
165
165
166
165
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The first random number between 1 and 8 are 6. Therefore, the sample
we select are 154, 158, 160, 163, and 165.
(b)
The sample mean is
3
x
154  158  160  163  165
 160
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and the sample variance is
s2 
154  1602  158  1602  160  1602  163  1602  165  1602
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 18.5
Chapter 8:
 Construct a 1   100% confidence intervals in large and
small sample cases.
 Determine sample size based on the desired margin of error
Example:
1. A random sample of 81 workers at a company showed that they work
an average of 100 hours per month with a standard deviation of 27
hours.
(a) Compute a 95% confidence interval for the mean hours per month all
workers at the company work.
(b) At 95% confidence, how many more workers need to be included in
the sample to provide a confidence interval with length 4 (i.e., the
margin of error being 2)?
[solution:]
(a)
As   0.05 ,
x  z
2

27
27
 100  z0.025
 100  1.96   94.12, 105.88
9
n
81
is a 95% confidence interval estimate of the population mean
(b)
Since E  2 ,
4
.
n
z02.05  2
2
2
E
1.962  27 2

 700.13  n  701 .
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Thus, we need 701-81=620 more workers.
2. For a t distribution with 16 degrees of freedom, find the area of
probability.
(a) To the left of -1.746.
(b) Between -1.337 and 2.120.
[solution:]
(a)
P(T (16)  1.746)  P(T (16)  1.746)  0.05 .
(b)
P(1.337  T (16)  2.120)  1  P(T (16)  2.12)  P(T (16)  1.337)
 1  0.025  P(T (16)  1.337)
 1  0.025  0.1
 0.875
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