Download Duffy Template

Chapter 6: Sampling Distributions
CCHAPTER 6 – Sampling Distributions
6.1
Each unit is the mean of four population measurements.
6.2
a.
Normal
b.
 x   ; the mean of the population of sample means is the same as the mean of the
population.
x 

; the standard deviation of the population of the sample means is smaller than
n
that of the population.
6.3
They are the same.
6.4
The variability of the sample means is smaller because the distribution is for means, where
high and low values are averaged out.
6.5
The sampling distribution of the sampling mean will be approximately normally distributed
when the sample size is large, no matter the shape of the population.
6.6
a.
An unbiased point estimate is a sample statistic where the mean of all possible sample
statistic values would equal the population parameter.
b.
The unbiased point estimate that has the smallest variance of the population of all
possible values.
85
Chapter 6: Sampling Distributions
6.7
a.
 x  10
2
22
4

.16
n
25 25

2
2
x 

 .4
n
25 5
 x2 
b.

 x  500
2
(.5) 2 .25

.0025
n
100 100

.5
.5
x 

 .05
n
100 10
 x2 
c.

x  3
2
(.1) 2 .01

.0025
n
4
4

.1 .1
x 

 .05
n
4 2
 x2 

d.  x  100
2
(1) 2
1

.000625
n 1600 1600

1
1
x 


.025
n
1600 40
 x2 
6.8

a.  x  3 x  [10  3(.4)]  [8.8,11.2] ; assume normal population
b.  x  3 x  [500  3(.05)]  [499 .85, 500 .15]
c.  x  3 x  [3  3(.05)]  [2.85, 3.15] ; assume normal population
d.  x  3 x  [100  3(.025)]  [99.925,100 .075]
86
Chapter 6: Sampling Distributions
6.9
a.
Normally distributed; no, because the sample size is large ( 30)
b.
 x  20,  x 
c.
21  20 

P( x  21)  P z 
  P( z  2)  .5  .4772  .0228
.5 

d.
19.385  20 

P( x  19.385)  P z 
  P( z  1.23)  .5  .3907  .1093
.5



4

n
64

4
 .5
8
6.10
 x  12.4,  x  9.21
 x  2 x  [12.4  2(9.21)]  [6.02, 30.82]
6.11
a.
Normal because the sample is large (n  30)
b.
 x  6, x 
6.12
s

2.475
n
 .2475
100
c.
5.46  6 

P( x  5.46)  P z 
  P( z  2.18)  .5  .4854  .0146
.2475 

d.
1.46%; conclude that  is less than 6.
a.
  42, s  2.6424
 x  42, x 
2.6424
 .32775
65
42.95  42 

P( x  42.95)  P z 
  P( z  2.90)  .5  .4981  .0019
.32775 

b.
6.13
a.
b.
.19%; conclude  > 42.
 x    1.4

1.3
1.3
x 


.13
n
100 10
1.5  1.4 

P( x  1.5)  P z 
  P( z  .77)  .5  .2794  .2206
.13 

 x    1.0

1.8
1.8
x 


.18
n
100 10
87
Chapter 6: Sampling Distributions
1.5  1.0 

P( x  1.5)  P z 
  P( z  2.78)  .5  .4973  .0027
.18 

6.14
c.
Yes, the probability of observing the sample is very small if the mean is actually 1.0.
a.
(1) Normally distributed, because population normal
(2)  x    45

6

 2.6833
(3)  x 
n
5
(4) 45 ± 3(2.6833) = [45 ± 8.0499] = [36.9501, 53.0499]
b.
55 is higher than upper end at interval. Delivery process is not operating effectively.
6.15
Population of all possible sample proportions ( p̂ values)
6.16
a.
np  5
n(1 – p)  5
b.
 pˆ  p
 pˆ 
p(1  p)
n
6.17
Decreases the spread of all possible pˆ values.
6.18
a.
np = (100)(.4) = 40 : n(1-p) = (100)(.6) = 60: Yes, sample is large enough
b.
np = (10)(.1) = 1 : n(1-p) = (10)(.9) = 9: No, sample is not large enough
c.
np = (50)(.1) = 5 : n(1-p) = (50)(.9) = 45: Yes, sample is large enough
d.
np = (400)(.8) = 320 : n(1-p) = (400)(.2) = 80: Yes, sample is large enough
e.
np = (1000)(.98) = 980 : n(1-p) = (1000)(.02) = 20: Yes, sample is large enough
f.
np = (400)(.99) = 396 : n(1-p) = (400)(.01) = 4: No, sample is not large enough
6.19
a.
 pˆ  p  .5
p(1  p) .5(1  .5) .25


 .001
n
250
250
p(1  p)
 pˆ 
 .001  .0316
n
 2pˆ 
88
Chapter 6: Sampling Distributions
b.
 pˆ  p  .1
p(1  p) .1(.9)

 .0009
n
100
p(1  p)
 pˆ 
 .03
n
 2pˆ 
c.
 pˆ  p  .8
p(1  p) .8(.2)

 .0004
n
400
p(1  p)
 pˆ 
 .02
n
 2pˆ 
d.
 pˆ  p  .98
p(1  p) .98(1  .98)

 .0000196
n
1000
p(1  p)
 pˆ 
 .004427
n
 2pˆ 
6.20
6.21
a.
 pˆ  2 pˆ  .5  2(.0316 )  [.4368, .5632 ]
b.
 pˆ  2 pˆ  .1  2(.03)  [.04, .16]
c.
 pˆ  2 pˆ  .8  2(.02)  [.76, .84]
d.
 pˆ  2 pˆ  .98  2(.004427 )  [.971146 , .988854 ]
a.
np = (100)(.9) = 90 : n(1-p) = (100)(.1) = 10: The sample is large enough.
The distribution of pˆ is normally distributed.
b.
 pˆ  p  .9
 pˆ 
c.
p(1  p)
.9(1  .9)

 .03
n
100
.96  .9 

  P( z  2)  .5  .4772  .0228
(1) P( pˆ  .96)  P z 
.03 

.945  .9 
 .855  .9
 pˆ 
  P(1.5  z  1.5) = 2(.4332) =
(2) P(.855  pˆ  .945)  P
.03 
 .03
.8664
.915  .9 

  P( z  .5)  .5  .1915  .6915
(3) P( pˆ  .915)  P z 
.03 

89
Chapter 6: Sampling Distributions
6.22
a.
 pˆ  p  .20
(.20)(1  .20)
 .01265
1000
.17  .20
P ( pˆ  .17 )  P ( z 
)  P ( z  2.37 )  .5  .4911  .0089
.01265
 pˆ 
b.
6.23
a.
Yes, the probability of observing this sample is very small if p=.20. The true p is more
likely less than .20.
 pˆ  p  .30
(.30)(1  .30)
 .0144
1011
.32  .30
P ( pˆ  .32 )  P ( z 
)  P ( z  1.39 )  .5  .4177  .0823
.0144
 pˆ 
6.24
b.
Perhaps, evidence not very strong.
a.
 pˆ  p  .40
(.40)(.60)
 .0154
1014
.36  .40
P ( pˆ  .36)  P ( z 
)  P ( z  2.60)  .5  .4953  .0047
.0154
 pˆ 
b.
Yes, the probability of observing this sample is very small if p=.40. The true p is more
likely less than .40.
90
Chapter 6: Sampling Distributions
6.25
 pˆ .4, pˆ 
a.
.4(1 .4)
.0154
1014
.43  .4 
 .37  .4
z
  P(1.95  z  1.95)  2(.4744 )  .9488
(1) P(.37  pˆ  .43)  P
.0154 
 .0154
.41  .4 
 .39  .4
z
  P(.65  z  .65)  2(.2422 )  .4844
(2) P(.39  pˆ  .41)  P
.0154 
 .0154
b.
6.26
a.
b.
6.27
a.
Only 48.44% probability of being within 1%.
.5(1 .5)
.0349
205
.5415 .5 
111



P pˆ 
.5415   P z 
  P( z  1.19) .5 .3830 .1170
205
.0349 



 pˆ .5, pˆ 
Probably not, evidence not strong enough.
 pˆ  p  .20
(.20)(.80)
 .0126
1000
.15  .20
P ( pˆ  .15)  P ( z 
)  P ( z  3.97 )
.0126
 pˆ 
= less than .001
6.28
b.
Yes, the probability of observing this sample is very small if p=.20. The true p is
probably less than .20.
a.
 pˆ  p  .70
(.70)(.30)
 .0265
300
.75  .70
P ( pˆ  .75)  P ( z 
)  P ( z  1.89 )  .5  .4706  .0294
.0265
 pˆ 
b.
6.29
Yes, the probability of this sample is small if p=.70. The true p is probably greater
than .70.
  2000,  300
a.
2150  2000 

P( x  2150 )  P z 
  P( z  .5)  .5  .1915  .3085
300


91
Chapter 6: Sampling Distributions
b.
6.30
 x  2000 ,  x 

300
 50
36
2150  2000 

P( x  2150 )  P z 
  P( z  3)  .5  .4987  .0013
50


n

c.
More difficult to achieve an average that exceeds $2150; yes
d.
The distribution of the individuals is wider than the distribution of the sample mean.
The highs and lows are averaged out.
a.
m = .2(7) + .4(6) + (.2)(5) + .1(4) + .05(3) + .05(2) = 5.45
 2  .2(7) 2  .4(6) 2  .2(5) 2  .1(4) 2  .05(3) 2  .05(2) 2  (5.45) 2
 31.45  29.7025  1.7475
  1.32
b.
 x    5.45

1.32
x 

 .22
n
36
c.
X
f (x)
X
5.45
d.
z
5  5.45
1.32
36

mean rating, X
5  5.45
 2.05
.22
P( x  5)  P( z  2.05) .5 .4798  0.0202
6.31
e.
Probably not very informative.
a.
 pˆ  p  .20
(.20)(.80)
 .0125
1031
.23  .20
P ( pˆ  .23)  P ( z 
)  P ( z  2.4)  .5  .4918  .0082
.0125
 pˆ 
b.
Yes, the probability that p=.20 is very small. Based on a sample of p̂ =.23, the true p
is probably larger than .20.
92
Chapter 6: Sampling Distributions
6.32
a.
b.
6.33
a.
 x    50,  x 

1
 .1
n
100
50.2  50 

P( x  50.2)  P z 
  P( z  2) .5 .4772 .0228
.1


No, because n  30.
 x    50,  x 

n


1

1
15
225
50.2  50 

P( x  50.2)  P z 
  P( z  3)  .5  .4987  .0013
.06667 

Smaller because sample means are clustered more closely together.
 x    50

.6
.6
x 


 .06
n
100 10
50.12  50 
 49.88  50
P(49.88  x  50.12)  P
z
  P(2  z  2)
.06
.06


= 2(.4772) = .9544
b.
49.85  50 

P( x  49.85)  P z 
  P( z  2.5)  .5  .4938  .0062
.06


93
Chapter 6: Sampling Distributions
6.34
6.35
6.36
6.37
 pˆ .5, pˆ 
a.
.54  .5 
 .46  .5
P(.46  pˆ  .54)  P
z

.01414 
 .01414
P  (2.83  z  2.83)  2(.4977 )  .9954
b.
.52  .5 
 .48  .5
P(.48  pˆ  .52)  P
z
  P(1.41  z  1.41)  2(.4207 )  .8414
.01414 
 .01414
c.
.51  .5 
 .49  .5
P(.49  p  .51)  P
z
  P(.71  z  .71)  2(.2611)  .5222
.01414 
 .01414
d.
No; No; the probabilities are too small.
a.
.41  .5 

P( pˆ  .41)  P z 
  P( z  6.36); less than .001.
.01414 

b.
Yes, conclude p < .5.
a.
 x    84

8
8
x 

 2
n
16 4
89  84 

P( x  89)  P z 
  P( z  2.5)  .5  .4938  .0062
2 

b.
Explanations will vary.
a.
 x  50.6,  x 
 .7245
n
5
[50.6 ± 2(.7245)] = [49.151, 52.049]
b.
 x  50.6, x 
c.
6.38
.5(1 .5)
.01414
1250
a.



1.62

1.62
 .256
n
40
[50.6 ± 2(.256)] = [50.088, 51.112]
n = 40, A sample of 40 results in a more accurate estimate of 
 x = .65(0) + .2(1) + .1(2) + .05(3) = .55
 x2  .65(0)2  (.2)(1)2  .1(2)2  .05(3)2  (.55)2  0  .2  .4  .45  .3025  .7475
 x  .8646
94
Chapter 6: Sampling Distributions
b.
 x    .55

.8646
x 

 .08646
n
100
Normally distributed
c.
X
f (x)
X
.55
6.39
number of flaws, X
d.
.75  .55 

P( x  .75)  P z 
  P( z  2.31)  .5  .4896  .0104
.08646 

a.
 pˆ  p  .50
(.50)(.50 )
 .020
622
.59  .50
P ( pˆ  .59 )  P ( z 
)  P ( z  4.5)  less than .001
.020
 pˆ 
b.
6.40
6.41
a.
Yes, the probability of observing this sample is extremely small if p=.50. More likely
the true p is greater than .50.
.4(.6)
.01526
1031
.42  .40 

P( pˆ .42)  P z 
  P( z  1.31) .5 .4049 .0951
.01526 

 pˆ  p .40, pˆ 
b.
Perhaps, but the evidence is not terribly strong.
a.
 x  500
x 
41
 6.93
35
P ( x  538 )  P( z 
b.
538  500
)  P( z  5.48)  less than .001
6.93
Yes, the probability of this sample is extremely small if μ=$500. The mean is more
likely greater than $500.
95
Chapter 6: Sampling Distributions
6.42
a.
 pˆ  p  .60
(.60)(.40)
 .0155
1000
.64  .60
P ( pˆ  .64 )  P ( z 
)  P ( z  2.58)  .5  .4951  .0049
.0155
 pˆ 
b.
Yes, the probability of this sample is very small if p=.60. The true p is more likely
greater than .60.
Internet Exercises
6.43
Screen from CLT module:
96
Chapter 6: Sampling Distributions
Exercises:
Sampling a Normal Population
Press the Show Notebook button, select the Scenarios tab, click on Normal Distribution, and select Width
of Car Hood. Read the scenario and click on OK.
1. What is the mean and standard deviation of the population being sampled? What is the mean and
standard error of the sampling distribution? What is the relationship between the standard deviation
of the population being sampled and the sampling distribution? If you aren’t sure use the Help
option on the menu bar.
Population Being Sampled: Mean
48
Standard Deviation
Sampling Distribution:
48
Standard Error
Mean
0.1
0.033
Standard Error = Standard Deviation / Sample Size
2. Push the One Sample button on the Sample panel. Answer the two exercises in the scenario. You
can look at the scenario by clicking on the Show Notebook button. After reading the scenario press
the Cancel button to return to the main screen.
No, it does not meet the company’s criterion. Since the width of car hoods is normally
distributed with a mean of 48 inches and a standard deviation of 0.1 inches, the width of
individual car hoods will range between 47.75 inches and 48.25 inches (2.5 standard deviations
from the mean). It is extremely likely (p = 0.942) that the sample of 9 will have at least one
sample point less than 47.9 or greater than 48.1 inches.
3. Was the salesman wrong when he said that his company’s machine has a standard error of 0.033?
Why was his statement misleading?
No, the salesman was not wrong. However, it was misleading because what was relevant was
the standard deviation of the distribution of car hoods produced by the machine, not the
standard error of the sampling distribution.
4.
Increase the sample size to 25 by using the n spin button or typing the number directly into the box
and pressing the Enter key on your computer. What is the standard deviation of the population
being sampled? What is the standard error of the sampling distribution? Which of these statistics
changed and why? What does the standard deviation measure? What does the standard error
measure?
Standard Deviation
0.1
Standard Error
0.02
The standard error changes since the standard error equals  n and n (Sample Size)
changed from 9 to 25. The standard deviation is a measure of dispersion in the population
being sampled while the standard error is a measure of dispersion in the sampling distribution.
97