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CHAPTER 9
ESTIMATION AND CONFIDENCE INTERVALS
1.
See definition on page 362 of the text.
3.
See definition on page 363 of the text.
5.
(a)
Standard error of the mean = 
8 ; (1 - ) = (1 – 0.1) = 0.9. Since the population
distribution is approximately normal, the distribution of X is approximately normal.

Hence, margin of error =  z ( / n )   z0.05 ( / n )  1.645 
2

confidence interval = X  1.645 
(b)
Standard error of the mean = 
8


8 ; 90%
50 ; (1 - ) = (1 – 0.025) = 0.975. Since sample size is
large enough (n = 50 > 30), the distribution of X is approximately normal. Hence, margin

of error =  z 2 ( / n )   z0.0125 ( / n )  2.24 

interval = X  2.24 
(c)
50

Standard error of the mean = 

50 ; 97.5% confidence
60 ; (1 - ) = (1 – 0.01) = 0.99. Since sample size is
large enough (n = 60), the distribution of X is approximately normal. Hence, margin of

error =  z 2 ( / n )   z0.005 ( / n )  2.575 

X  2.575 
60

Since the population is approximately normally distributed, the distribution of X is
approximately normal. Hence, a 90 percent confidence interval estimate for  is
7.
= X  z

2
n

 X  z0.05
= 26  1.645 6
9.

60 ; 99% confidence interval =


n
16   23.5325, 28.4675
(a)
It is the number in the table of Student’s t Distribution in row corresponding to “18” and
2
column corresponding to “  0.010
” . It equals 34.8053.
(b)
Since the required left tail probability is 0.025, the right tail probability is 1 – 0.025 =
0.955. It is the number in the table of Student’s t Distribution in row corresponding to “30”
2
and column corresponding to “  0.975
”. So, L = 16.7908.
(c)
2
2
Total area to the right of  0.990
is 0.99. We want area between  0.990
and U to be 0.98.
2
Hence, area to the right of U should be (0.99 – 0.98) = 0.01. Thus, U   0.01
. It is the
9-1
2
number in the table of Student’s t Distribution in row “5” and column “  0.01
” and equals
15.0863.
(d)
2
2
We want area between L and  0.05
to be 0.925. Area to the left of  0.05
is 0.05. Hence,
2
total area to the right of L should be 0.925 + 0.05 = 0.975. Thus, L   0.975
. It is the
2
number in the table of Student’s t Distribution in row “15” and column “  0.975
” and equals
6.26214.
11.
(a)
Since the population is normal,
 n  1 S 2
2
follows chi-square distribution with df = (n – 1)
9.
A 99 percent confidence interval estimate for 2 is
2
2


  (n  1) s 2 , (n  1) s 2 


0.005
0.995 

 9(5)
, 9(5)
 (1.908, 25.938)
23.5893
1.734926

(b)

An 80 percent confidence interval estimate for 2 is
2
2


  (n  1) s 2 , (n  1) s 2 


0.1
0.9 

 9(5)
, 9(5)
 (3.065, 10.8)
14.6837
4.16816

13.
15.

(a)
(1 - ) = 0.8, so  = (1 – 0.8) = 0.2.
Hence, t = t/2 = t0.1 (for df = 30) = 1.31 (from the t-table)
(b)
From the t-table, we find that t0.005 (for df = 6) = 3.707.
(c)
By symmetry of t-distribution it follows that t = -t0.1.
From the t-table, we get t0.1 (for df = 24) = 1.318. Hence, t = -1.318.
(d)
The right tail area corresponding to t = 1 – 0.9 = 0.1. Thus, t = t0.1 (for df = 12) = 1.356
(from the t-table).
(e)
The left tail area corresponding to t0.995 is (1 – 0.995) = 0.005. Thus, t0.995 = -t0.005.
From the t-table, we get t0.005 (for df = 14) = 2.977.
Hence, t0.995 = -2.977.
(a)
Since the population distribution is approximately normal, the distribution of
X 
is
S n
approximately t-distribution with df = n – 1.
Calculate x and s. (1 - ) = 0.95. So,  = (1 – 0.95) = 0.05. For df, t/2 = t0.025 = 2.201. 95%
confidence interval estimate for  is x 2.201( s / 12 ) .
9-2
(b)
Calculate x and s. (1 - ) = 0.9. So,  = (1 – 0.9) = 0.1. For df = 19, t/2 = t0.05 = 1.729.

90% confidence interval estimate for  is x  1.729 s
(c)

20 .
Calculate x and s. (1 - ) = 0.99. So,  = (1 – 0.99) = 0.01. For df = 7, t/2 = t0.005 = 3.499.

99% confidence interval estimate for  is x  3.499 s

8 .
17.
x
48.16  42.22 
20
 61.46
 49.353
(48.16  49.353) 2  (42.22  49.353) 2 
19
 9.013
s
 (61.46  49.353) 2
Since the population distribution is approximately normal, the distribution of
X 
is
S n
approximately t-distribution with df = n – 1 = 19.
(1 - ) = 0.99. So,  = (1 – 0.99) = 0.01. For df = 19, t/2 = t0.005 = 2.861
99% confidence interval estimate for  is


 x   2.861 s

20  49.353   2.861 9.013
20

  43.587,55.119 
The value 50 lies in the 99% confidence interval estimate. Hence, the analysis does not
provide reason to rule out the probability that the value of  = 50. The value 60 does not lie
in the interval estimate. Hence it will be unreasonable to assume that  = 60.
19.
(a)
The best point estimate of  is the sample mean x = 21.9 eggs/month.
(b)
(1 - ) = 0.98. So,  = (1 – 0.98) = 0.02. For df = 19, t/2 = t0.01 = 2.539
Hence, a 98% confidence interval estimate for  is
= x  t0.01

s
n

= 21.9   2.539 2.1
(c)
The population is normally distributed, but population standard deviation is unknown. In
this case
(d)

20 = (20.708,23.092)
X 
has t-distribution with (n - 1) degrees of freedom.
S n
Value 22 lies in the 98% confidence interval estimate. Hence, the analysis does not provide
evidence against the probability that the value of  = 22. Value 24 does not lies within
confidence interval estimate. Hence it will be reasonable to assume that the value of  is
not 24.
9-3
21.
(a)
The best estimate of the value of the population proportion is pˆ 
(b)
An estimate of the standard error of the proportion is
pˆ 1  pˆ 
n
(c)

300
 0.75 .
400
 0.75 0.25   0.022
400
(400)(0.75) > 5 and (400)(0.25) > 5. Hence, we can assume normal approximation. An
approximate 99% confidence interval estimate of population proportion is
= pˆ  z0.005
pˆ 1  pˆ 
n
= 0.75   2.575 0.022   0.693,0.807 .
23.
(d)
The estimator used in (c) for finding the confidence interval estimate produces an interval
containing the population proportion, p, 99 percent times in the long run. Given the high
probability, it would be reasonable to expect the interval currently obtained to contain the
value of p.
(a)
The best point estimate of the population proportion is pˆ 
(b)
(300)(0.05) > 5 and (300)(0.95) > 5. Hence, an approximate 95% confidence interval
estimate of p is
= pˆ  z0.025
pˆ 1  pˆ 
= 0.05  1.96
(c)
25.
15
 0.05 .
300

n
 0.05 0.95

300   0.025,0.075
The entire 95% confidence interval estimate obtained in (b) lies to the left of 0.1. Hence, it
is reasonable to infer that p is less than 0.1. He should not return the lot.
Let us hope that the sample size, n, will be large enough for the distribution of X to be
approximately normal. Then, the sample size n should be at least
 z 2    2.57515 

 =
  59.676 or 60.
E
5


 
2
2
This value of n is large enough for assumption of normality. So the bound is valid.
27.
Let us hope that the sample size, n, will be large enough for normal approximation.
We want (1 - ) = 0.99, so  = 0.01. z = z0.005 = 2.575.
9-4
An approximate lower bound on the required sample size is
2
  2.575   0.45  0.55  

  164.108 or 165.
0.1




This value of n is large enough for assumption of normality. So the bound is valid.
29.
We want the margin of error to be no more than + 0.05. Let us hope that the sample size
will be large enough for distribution of X to be approximately normal.
An approximate lower bound on the required sample size is
 z0.025 s   1.96  0.2  
  61.466 or 62.
 E  = 
0.05 

 
2
2
This value of n is large enough for assumption of normality. So the bound is valid.
31.
(a)
Let us hope that the sample size will be large enough for applying normal approximation.
An approximate lower bound on the required sample size is
2
2
 z0.05 p 1  p    1.645   0.3 0.7  

 =
  5682.65 or 5683.

 
E
0.01


 

This value of n is large enough for assumption of normality. So the bound is valid.
(b)
One way is to first obtain a better estimate of p by first choosing a pilot sample of
moderate size, say 100. If the sample size based on new estimate is still large, then we shall
have to increase the allowable margin of error and/or decrease the confidence level. For
example, if the allowable margin of error is changed to 0.05, then an approximate lower
2
 1.645   0.3 0.7  
 = 227.3 or 228.
bound on sample size n will be 
0.05




33.
The ratio (n/N) = (49/500) = 0.098 is greater than 0.05, but not too large.
The population size is moderately large, and the population distribution is approximately
normal. Hence, we shall use Formula 9-18 in the textbook.
(1 - ) = 0.99. So  = (1 – 0.99) = 0.01. t/2 = t0.005 (for df = 48) is approximately 2.68.
An approximate 99% confidence interval estimate for  is:
= x  t0.005
s  N n 
9  451 

 = 40  2.68


n  N 1 
49  499 
= (36.724, 43.276).
35.
The ratio (n/N) = (30/300) = 0.1 is greater than 0.05, but not too large.
pˆ 
18
 0.6 .
30
9-5
npˆ =30 (0.6) > 5 and n 1  pˆ  = 30 (40) > 5.
Hence n is large enough for normality assumption.
An approximate 95% confidence interval estimate for p is:

 ( pˆ )(1  pˆ )   N  n 


0.6

1.96




  N 1 

n




= pˆ  z0.025 
 0.6  0.4   
30
270 


  299 

= (0.433, 0.767).
37.
Let us hope that the sample size will be large enough for assumption of normality.
An approximate lower bound for the sample size is
2
2
 z0.05 p 1  p  
 1.96   0.35  0.65  

 = 
  2184.91 , or 2185.


E
0.02






This value of n is large enough for normality assumption. So the bound is valid.
39.
(a)
The best point estimate 54. That is, the best choice of a single value as an estimate of  is
54.
(b)
n = 49 is large enough for assumption of normality of distribution of X . For df = 48, t0.025
= approximately 2.01. A 95% confidence interval estimate
 10 
   51.13,56.87  .
 49 
is 54  2.01
41.
Let us hope that the sample size will be large enough for assumption of normality.
2
 z s
An approximate lower bound on the required sample size is  0.025  =
 E 
 1.96  500  

  96.04 or 97.
100


2
This value of n is large enough for assumption of normality. Hence the bound is valid.
43.
(a)
A point estimate of population mean is 1.01 kg.
(b)
The sample size (n = 36) is large enough for us to assume that distribution of
approximately student’s t-distribution
For df = 35, t0.025 = approximately 2.031. A 95% confidence interval estimate
 0.02 
  1.003,1.017 .
 36 
is 1.01  2.031
9-6
X 
is
S/ n
45.
(a)
The sample size (n = 50) is large enough for assumption of normality.
For df = 49, t0.005 = approximately 2.68.
A 99% confidence interval estimate is x  t0.005
s
=
n
 0.01 
0.64  2.68 
   0.636, 0.644 .
 50 
47.
(b)
The result, thus provides significant evidence that value of  is not 0.63. It will therefore be
reasonable to conclude that the value of  is not 0.63.
(a)
The value of sample proportion is p̂ = 0.63.
n p̂ = (1000) (0.63) > 5 and n (1 - p̂ ) = (1000) (0.37) > 5.
Hence, the sample size is large enough for normality assumption.
A 95% confidence interval estimate is
pˆ  z0.025
(b)
49.
 0.63 0.37   
1000


 0.6, 0.66 .
The entire interval in (a) lies above 0.6. The analysis thus provides sufficient evidence that
the population proportion p is greater than or equal 0.6. It is therefore reasonable to
conclude that the claim, that p is less than 0.6 is false.
The sample size is large enough to assume normality.
For df = 35, t0.05 = approximately 1.69. A 90% confidence interval estimate of  is
x  t0.05
51.

pˆ (1  pˆ )
= 0.63  1.96 

n

(a)
s
 8200 
= 32000  1.69 
   29690.33,34309.67  .
n
 36 
Let us hope that the sample size will be large enough for assumption of normality. p =
0.21.
2
 z0.025 p 1  p  
 =
An approximate lower bound for n is 


E


2
 1.96   0.21 0.79  

  708.13 or 709.
0.03




This is large enough for assumption of normality. Hence the bound is valid.
 z 2  0.5 
An approximate lower bound on the required sample size n is 
 =

E


2
(b)
 1.96  0.5 

  1067.11 or 1068.
 0.03 
2
9-7
Hence, 1068 accountants should be contacted.
Let us hope that the sample size will be large enough for assumption of normality. p =
5/50 =0.1.
53.
2
 z0.025 p 1  p  
 =
An approximate lower bound for n is 


E


2
 1.96   0.1 0.9  
  864.36 or 865. This is
An approximate lower bound for n is 
0.02




large enough for assumption of normality. Hence the bound is valid.
55.
(a)
x
94  78  83 
15
 84
 89.4667 .
Thus the best point estimate of  is 89.4667.
(b)
The sample mean is
s
(94  89.4667) 2  (78  89.4667) 2 
14
 (84  89.4667) 2
 8.08
Since the population distribution is approximately normal, the distribution of
X 
is
S n
approximately t-distribution with df = n – 1 = 14.
For df = 14, t0.025 = 2.145. Hence, 95% confidence interval estimate is
 8.08 
89.4667  2.145 
   84.99,93.99  .
 15 
(c)
57.
The confidence interval estimate in (b) lies entirely above 80. Hence it is reasonable to
conclude that the mean stress level is in the dangerous level.
The sample size (n = 60) is large enough for assumption of normality.
 = (1 – 0.98) = 0.02. For df = 59, t/2 = t0.01 = approximately 2.39. Hence, an approximate
 0.75 
   2.53, 2.99  .
 60 
98% confidence interval estimate of  is 2.76  2.39 
59.
(a)
The best point estimate of the population mean, , we can obtain is
x
(b)
2698  2028 
10
 2379
 2408.8 .
Since the population distribution is approximately normal, the distribution of
approximately t-distribution with df = n – 1 = 9.
The sample standard deviation is
9-8
X 
is
S n
s
(2698  2408.8) 2  (2474  2408.8) 2 
9
 (2379  2408.8) 2
= 304.4276.
For df = 9, t0.025 = 2.262.
A 95% confidence interval estimate for  is
 304.4276 
2408.8  2.262 
   2191.04, 2626.56  .
10 

61.
(a)
(b)
x
64  66 
12
 66
 62.583
Since the population distribution is approximately normal, the distribution of
X 
is
S n
approximately t-distribution with df = n – 1 = 11.
The sample standard deviation is
s
(64  62.583) 2  (66  62.583) 2 
11
 (66  62.583) 2
 3.942
For df = 11, t0.05 =1.796. Hence, 90% confidence interval estimate
 3.942 
   60.539, 64.627 .
 12 
is 62.583  1.796 
(c)
63.
The value 60 lies in the interval estimate obtained in part (b). Thus, our analysis does not
provide significant evidence against the claim. We have no reason to doubt the claim.
A point estimate of proportion, p, of the entire population (Ontario) who blame the
provincial government for the tragedy is p̂ = 0.22.
n p̂ = (1001) (0.22) > 5 and n (1 - p̂ ) = (1001) (0.78) > 5.
Hence, the sample size is large enough for normality assumption.
An approximate 95% confidence interval estimate for p is
pˆ  z0.025
65.
(a)

pˆ (1  pˆ )
= 0.22  1.96 

n

 0.22 0.78  = (0.194, 0.246).
1001


A point estimate of proportion, p, of Canadians who favour user fees and charges is p̂ =
0.54.
n p̂ = (1400) (0.54) > 5 and n (1 - p̂ ) = (1400) (0.46) > 5.
Hence, the sample size is large enough for normality assumption.
9-9
An approximate 95% confidence interval estimate is pˆ  z0.025

0.54  1.96 


(b)
 0.54 0.46  
1400


pˆ (1  pˆ )
=
n
 0.514, 0.566 .
The entire interval lies above 0.5. Hence, it is reasonable to conclude that p is greater than
0.5, that is, majority of Canadians favour moderate user fees and charges.
Select a sample of size 10; calculate x and s. For df = 9, t0.025 = 2.262. A 95% confidence
67.
 s 
 . The answer will vary according to the sample
 10 
interval estimate is x  2.262 
chosen.
69.
The Minitab and Excel outputs are given below. (See the Minitab and Excel instructions
given in the chapter.)
It should be noted that Minitab output directly gives the confidence interval
estimates, while the Excel output gives in the row labeled “Confidence level”, the margin
of error.
(a)
Using the Excel output, we get for selling price of the homes:
95 percent confidence interval estimate for population mean
= ( sample mean  margin of error)
= (221.1029  9.116045) = (211.9868122, 230.2189021).
Minitab Output
One-Sample T: Selling price
Variable
Price
N
105
Mean
221.10
StDev
47.11
SE Mean
4.60
EXCEL OUTPUT
Selling price
Mean
Standard Error
Median
Mode
Standard Deviation
Sample Variance
Kurtosis
Skewness
Range
Minimum
Maximum
Sum
Count
Confidence
Level(95.0%) 9-10
221.1029
4.597017
213.6
188.3
47.1054
2218.919
-0.2768
0.474013
220.3
125
345.3
23215.8
105
9.116045
95.0% CI
(211.99, 230.22)
(b)
A 90 percent confidence interval estimate for the fraction of home that were sold more than
$240000 = ( pˆ  z
( pˆ )(1  pˆ )
)  (0.249,0.399) .
n
2
MINITAB OUTPUT
Test and CI for One Proportion: Sold more than $240000
Variable
Prices
X
34
N
105
Sample p
0.323810
90.0% CI
(0.248697, 0.398922)
Z-Value
-3.61
P-Value
0.000
MEGASTAT OUTPUT
Confidence interval - proportion
90% confidence level
0.32381 proportion
105 n
1.645 z
0.324 half-width
0.249 upper confidence limit
0.399 lower confidence limit
(c)
For number of bedrooms: 90 percent confidence interval estimate for the population mean
= (mean  margin of error)
= (3.8  0.346460433) = (3.45354, 4.14646).
Minitab Output
One-Sample T: Bedrooms
Variable
Bedrooms
N
105
Mean
3.800
9-11
StDev
1.503
SE Mean
0.147
98.0% CI
(3.454, 4.146)
Excel Output
# of bedrooms
Mean
Standard Error
Median
Mode
Standard Deviation
Sample Variance
Kurtosis
Skewness
Range
Minimum
Maximum
Sum
Count
Confidence
Level(98.0%)
(d)
3.8
0.146635028
4
4
1.502561915
2.257692308
-0.199838427
0.660884776
6
2
8
399
105
0.346460433
A 90 percent confidence interval estimate for the proportion of homes sold that have a
garage = = ( pˆ  z
2
( pˆ )(1  pˆ )
)  (0.601,0.751) .
n
Minitab Output
Test and CI for One Proportion: Garage
Variable
Garage
X
71
N
105
Sample p
0.676190
90.0% CI
(0.601078, 0.751303)
MEGASTAT OUTPUT
Confidence interval - proportion
90% confidence level
0.67619 proportion
105 n
1.645 z
0.676 half-width
0.601 upper confidence limit
0.751 lower confidence limit
9-12
Z-Value
3.61
P-Value
0.000
(e)
A 95 percent confidence interval estimate for the proportion of homes sold that have two
or more bathrooms = (0.779, 0.916)
Minitab Output
Test and CI for One Proportion: Baths
Variable
Baths
X
89
N
105
Sample p
0.847619
95.0% CI
(0.778878, 0.916361)
Z-Value
7.12
P-Value
0.000
MEGASTAT OUTPUT
Confidence interval - proportion
95% confidence level
0.847619 proportion
105 n
1.960 z
0.848 half-width
0.779 upper confidence limit
0.916 lower confidence limit
71.
The Minitab and Excel outputs are given below. (See the Minitab and Excel instructions
given in the chapter.)
It should be noted that Minitab output directly gives the confidence interval
estimates, while the Excel output gives in the row labeled “Confidence level”, the margin
of error.
Thus, using Excel, we get :
For TSE 300 data : 92 percent confidence interval =
(mean  margin of error)  (9776.126  272.6599)  (9503.4661,10048.7859)
For BCE data : 95 percent confidence interval =
(mean  margin of error)  (38.199  2.117026)  (36.081974, 40.316026) .
For Air Canada data : 90 percent confidence interval =
(mean  margin of error)  (17.3875  0.960303)  (16.427197,18.347803).
These are the same (except for rounding) as the ones obtained using Minitab.
Minitab Output:
One-Sample T: TSE 300
VARIABLE
TSE 300
N
20
MEAN
9776
STDEV SE MEAN
659
147
92.0% CI
(9503, 10049)
One-Sample T: BCE
VARIABLE
BCE
N
20
MEAN
38.20
One-Sample T: Air Canada
9-13
STDEV SE MEAN
4.52
1.01
95.0% CI
(36.08, 40.32)
EXCEL OUTPUT
EXCEL OUTPUT
BCE
TSE 300
Mean
Standard Error
Median
Mode
Standard Deviation
Sample Variance
Kurtosis
Skewness
Range
Minimum
Maximum
Sum
Count
Confidence
Level(92.0%)
Mean
Standard Error
Median
Mode
Standard Deviation
Sample Variance
Kurtosis
Skewness
Range
Minimum
Maximum
Sum
Count
Confidence
Level(95.0%)
9776.126
147.4212
9517.935
10778.8
659.2875
434660
-1.15449
0.608497
1821.2
9020.88
10842.08
195522.5
20
272.6599
EXCEL OUTPUT
Air Canada
Mean
Standard Error
Median
Mode
Standard Deviation
Sample Variance
Kurtosis
Skewness
Range
Minimum
Maximum
Sum
Count
Confidence
Level(90.0%)
9-14
17.3875
0.555367
17.475
19.4
2.483677
6.168651
-0.24516
-0.53549
9.25
11.5
20.75
347.75
20
0.960303
38.199
1.011467
37.475
37
4.523419
20.46132
-0.94464
0.308152
15.92
31.75
47.67
763.98
20
2.117026