The document provides examples of calculating confidence intervals from sample data. It includes steps for finding 95%, 99%, and 90% confidence intervals using the t-distribution and z-distribution. Sample sizes, means, standard deviations and confidence levels are given for multiple data sets, and confidence intervals are calculated and interpreted for each example.

1. Interval Estimation (Page No. 289)
Answer the question: 1
Given that, 45.  x  ; 25  n ; 90. 2  x
(a) 95% confidence interval is: 100(1 -  ) = .95  = .05  .025
2


So, we know that,
Z x
  
2 2    
n
x
Z x
n
x


=
.45
2.90 .025 .025 
  
25
2.90
.45

25

Z Z
 [ 96. 1 .025  Z ]
= 2.72< <3.08
So, 95% confidence interval range is from 2.7 to 3.08
(b) The probability content associated the interval from 2.81 to 2.99 is:
z x
n
w
  / 2 2
 [w = 2.99 – 2.81 = .18]

18. / 2 
  z
45. 2
25
 2 /  Z =
18.
18.
  / 2 Z = 1 = .8413
So, 1 -

2
= .8413 

2
= .1587   = .3174
Now 1 -  = 1 - .3174 = .6826 or 68.26%
The probability content associated the interval from 2.81 to 2.99 is 68.26%
Answer the question: 2
2. Given that, 12. x  16n 07 . 4  x
(a) 99% confidence interval is: 100(1 -  ) = .999  = .01  005 .
2


So, we know that,
Z x
  
2 2    
n
x
Z x
n
x


=
4.07 .005 .005 
  
16
.16
4.07
16
.12


Z Z
 [ Z.005  2.575 ]
= 3.99< <4.15
So, 99% confidence interval range is from 3.97 to 4.17
(b) Narrower
(c) Narrower
(d) Wider
Answer the question: 3

150.9
   Given that, x  3.8 n  9 16.76
9
n
xi
x
90% confidence interval is: 100(1 -  ) = .90  = .10  .05
2



2. So, we know that,
Z x
  
2 2    
n
x
Z x
n
x


=
1.645 3.8
9
16.76
1.645 3.8
9
16.76

  

  [ Z / 2 1.645 ]
= 14.67< <18.85
So, 90% confidence interval range is from 14.67 to18.85
b) Wider
Answer the question: 4
Here, 4. 32  x  n=9 9. 187  X
(a) 80% confidence interval is: 100(1 -  ) = 80   = .20  10.
2


Z x
  
2 2    
n
x
Z x
n
x


=
1.285 32.4
9
187.9
1.285 32.4
9
187.9

  

  [ 285. 1 2 /   Z ]
= 174.02< <201.78
So, 80% confidence interval range is from 174.02 to 201.78.
(b) The probability content associated the interval from 185.8 to 210 is:
z x
n
w
2  / 2
 [w = 210 – 185.8 = 44.2]
44.2=
2z / 232.4
9

2  Z =4.09
2
 Z =2.046Fz(
2
 Z )=Fz(2.05)
2
1-

2
=.9798

2
=.0202  =.04041-
 =.9596
The probability content associated the interval from 185.8 to 210 is 95.96%
Answer the question: 5
Given that, n=1562 x =3.92 Sx=1.57
Confidence interval, 100(1-∞) =95  α =.05  α∕2=.025  Z.025=1.96
The 95% Confidence interval for the population mean
Z Sx
n
x
Z Sx
n
x
/ 2  / 2


    [ Z / 2 1.91]
= 3.92-
(1.96)(1.57)
1562
<μ< 3.92+
(1.96)(1.57)
1562
= 3.92- .0779 <μ< 3.92 + .079
= 3.84 <μ< 3.99
So, 95% confidence interval range is from 3.84 to 3.99
Answer the question: 6
Given that, n = 541 X =3.81 Sx =1.34

3. The 90% Confidence Interval for the population mean100 %( 1-α) =90%  α = .10
  /2=.05  Z /2=1.65
(a) So, we know that,
Z Sx
n
x
Z Sx
n
x
/ 2  / 2


    [ 645. 12/ Z ]
= 3.81-
(1.65)(1.34)
541
<μ<3.81+
(1.65)(1.34)
541
=3.71<μ<3.90
The 90% Confidence Interval for the population mean 3.71 to 3.91
(b) It will be narrower.
Answer the question: 7
We know,
n
Z
W
  2 / 2
 Given that, W=.2 Sx =1.045
So, .2= 2Z α/2(1.045)
√457
=> Z α/2=2.04=.9793
= >1- α/2=.9793
= > α/2=1-.9793
 α=.0414
Or, 1- α=.9586 or, 95.86%
So, The Confidence Interval is 95.86 %.
Answer the question: 8
Here, n=352 Sx=11.28 X =60.41
Z Sx
 
x 2 2
n
x
Z Sx
    
n
=
1.645 11.28
352
60.41
1.645 11.28
352
60.41

  

  [ Z / 2 1.645 ]
=59.42 <μ< 61.40
Comment: Here, we see that, if 57% to more mark than they are adequate understanding
the material. So we can say that students are adequate understanding of the written
material.
Answer the question: 9
Here, n=174 Sx=1.43 X =6.06 W=.2
(a) We know,
n
Z
W
2  / 2

So, .2=2Z α/2(1.43)
√174
=> Z α/2=.92=.8212
= >1- α/2=.8212
= > α/2=1-.8212

4.  α=.3576
Or, 1- α=.6424 or, 64.24%
So, The Confidence Interval is 64.24 %.
(b) Comment: Here confidence is decrease that different factor exiting. Such as
sample size and sample standard deviation are difference compare than exercise
7.
Answer the question: 10
Here, n=9 Sx=38.89 X =157.82 V= (n-1)=(9-1)=8
tv Sx
n
x
tv Sx
n
x
/ 2  / 2


   
=
8,.025 38.89
9
157.82
8,.025 38.89
9
157.82

  


t t
 [ 025 .
2


]
=
2.306 38.89
3
157.82
2.306 38.89
3
157.82

  

 
=127.93 <μ< 187.71
Answer the question: 11
(a) nx / 1   i X 1/7(523) =74.7143
1 2 2 x nx
n i
 2 Sx  

{ }
( 1)
= 1/6{39321-(7) (74.7143) 2 }
=40.90
 Sx 90. 40 =6.3953
(b) 95% confidence interval is: 100(1 -
i i  2
i 
1 79 6241
2 73 5329
3 68 4624
4 77 5929
5 86 7396
6 71 5041
7 69 4761
 i X 523   i X 2 39321
 ) = .95  = .05 025 .
2


So, we know that, x -
t Sx (n1) / 2
n
< < x +
t Sx (n1) / 2
n
[ n 1= 6] [ n1 t  / 2 =2.447]
= 74.7143-
(2.447 )(6.3953 )
7
< < 74.7143+
(2.447 )(6.3953 )
7
= 74.7143-5.9149 < < 74.7143+5.9149
=68.7994 < < 80.6292
So, 95% confidence interval range is from 68.7994 to 80.6292

5. Answer the question: 12
(a) n x / 1   i X 1/10(163.7) =16.37
1 2 2 x nx
n i
 2 Sx  

{ }
( 1)
= 1/9{2939.85 - (10) (16.37) 2 }
= 28.8979
Sx 8979 . 28 = 5.3757
Here,  Sx 5.3757 X = 16.37 n=10
99% confidence interval is: 100(1 -
 ) = .99  = .01  /2=.005
i i  2
1 18.2 331.24
2 25.9 670.81
3 6.3 39.69
4 11.8 139.24
5 15.4 237.16
6 20.3 412.09
7 16.8 282.24
8 19.5 380.25
9 12.3 151.29
10 17.2 295.84
  i X 163.7  i X 2 2939.85
So, we know that, x -
Sx t (n1) / 2
n
i 
<< x +
Sx t (n1) / 2
n
 [ 1  n = 9] [ 1n t 2 /
=3.250]
=16.37-
(3.250)(5.3757 )
10
<< 16.37+
(3.250)(5.3757 )
10
=16.37- 5.5248<<16.37+5.5248
= 10.8452<<21.8948
So, 99% confidence interval range is from 10.8452 to 21.8948
(b) narrower range.
Answer the question: 13
(a) n x / 1 
1   i X 1/25(1508) =60.32
1 2 2 x nx
n i
 2 Sx  

{ }
( 1)
= 1/24{95628 - (25) (60.32) 2 }
= 194.39
Sx  194.39 = 13.94
(b) So, we know that, x -
t Sx (n1) / 2
n
<< x +
t Sx (n1) / 2
n
[ n 1= 24] [ n1 t  / 2
=1.711]
 60.32-
1.71113.94
25
< <60.32+
1.71113.94
25
55.55< <65.09
Answer the question: 14
Based on the material of section 8.4 .I have to follow the t distribution. But in t distribution I
have to calculate the degree of freedom. The degree of freedom is n-1 but here n = 1 so

6. degree of freedom is 0. Because of the degree of freedom is 0 it’s not possible to find
confidence interval for the population mean.
Answer the question: 15
Given that, 4780  Sx 25  n
90% confidence interval is: 100(1 -  ) = .90   = .10  05.
2


So, we know that, x -
t Sx (n1) / 2
n
< < x +
t Sx (n1) / 2
n
[ 1  n =24]
[ 1n t 2 / =1.711]
= 42740 -
) 4780 )( 711 . 1(
25
<< 42740 +
) 4780 )( 711 . 1(
25
= 42740-(1635.716) < <42740+ (1635.716)
= 41104.284 < < 44375.716
So, 90% confidence interval range is from 41104.284 to 44375.716
Confidence interval for Proportion and variance (Page no. 299)
Answer the question: 19
Here, n=189 X=132 Px=
x
=
n
132
189
=.698
(a) 90% confidence interval for population proportion is;
Ṗ-Z 2 / 
Px(1 Px)
n
< P < Ṗ+Z 2 / 
Px(1 Px)
n
[ Z 2 /  =1.645]
.698-1.645
.698(1.698)
189
< P < .698+1.645
.698(1.698)
189
.6430<P< .7529
(b) Here 95% confidence so we can say that range will be wider than previous (a).
Answer the question: 20
Here, n=323 X= 155 Px=
x
=
n
155
323
=.4799 W= [.5-.458] = .042
We know,
Px Px
n
W Z
(1 )
2 / 2

 

.4799(1 .4799)
323
.042 2 / 2

 Z
Z / 2  .755 =.7764
1 / 2 =.7764

7. =.44721-=.5528
So, The Confidence Interval is 55.28 %.
Answer the question: 21
Here, n=134 X=82 Px=
x
=
n
82
134
=.612
95% confidence interval for population proportion is;
Ṗ-Z 2 /
Px(1 Px)
n
< P < Ṗ+Z 2 /
Px(1 Px)
n
[ Z 2 /  =1.955]
.612-1.955
) 612. 1( 612. 
134
< P < .612+1.955
) 612. 1( 612. 
134
.53 <P< .694
Answer the question: 22
Here, n=95 X=29 Px=
x
=
n
29
95
=.3053
(a) 99% confidence interval for population proportion is;
Ṗ-Z 2 /
Px Px ) 1( 
n
< P < Ṗ+Z 2 /
Px Px ) 1( 
n
[ Z 2 /  =2.575]
.3053-2.575
.3053(1.3053)
95
< P < .3053+2.575
.3053(1.3053)
95
.1836 <P< .4270
(b) If confidence is decreases than rang will be narrower.
Answer the question: 23
Here, n=96 X=32 Px=
x
=
n
32
96
=.333
80% confidence interval for population proportion is;
Ṗ-Z 2 / 
Px(1 Px)
n
< P < Ṗ+Z 2 / 
Px(1 Px)
n
[ Z 2 /  =1.285]
.333-1.285
.333(1.3053)
95
< P < .333+1.285
.333(1.333)
96
.2712<P< .3948
Answer the question: 24
Here, n=198 X= 98 Px=
x
=
n
98
198
=.495 W= [.545-.445] = .10
We know,
Px Px
n
W Z
(1 )
2 / 2

 

.495(1 .495)
198
.10 2 / 2

 Z
Z / 2 1.41=.9207

8.  2 / 1  =.9207
=.15861-=.8414
So, The Confidence Interval is 84.14 %.
Answer the question: 25
Given that, 15 n 880. x s so, 7744. 2 x s
100(1 -  ) = .95   = .05  025 .
2


We know,
n s x
( 
1)
v
,1 / 2
n s x
( 1)
v
, / 2
2
2
2

2
2
 
x
 

  
=
(15 
1).7744
14,1 .025
(15 
1).7744
14,.025
2
2
 
2 



x
=
(15 1).7744 2 
(15 1).7744
5.63
26.12
 

 x
= .4151 < x 2 <1.9257
Answer the question: 26
7  n
nx / 1 i X 1/7(523) =74.7143
1 2 2 x nx
n i
 2 Sx  

{ }
( 1)
= 1/6{39321-(7) (74.7143) 2 }
=40.90
100(1 -  ) = .80  = .20
i i  2
1 79 6241
2 73 5329
3 68 4624
4 77 5929
5 86 7396
6 71 5041
7 69 4761
 1.
2


We know,
i 
2 Xi 39321
  i X 523  
n s x
( 
1)
v
,1 / 2
n s x
( 1)
v
, / 2
2
2
2

2
2
 
x
 

  
=
(7 1)40.90
6,1 .1
(7 1)40.90
6,.1

2
2
 
2 




x
=
(7 1)40.90 2 
(7 1)40.90
2.20
10.64
 

 x
= 23.064 < x 2  <111.545

9. Answer the question: 27
Here, 10  n
n x / 1  i X 1/10(163.7) =16.37
1 2 2 x nx
n i
 2 Sx  

{ }
( 1)
= 1/9{2939.85 - (10) (16.37) 2 }
= 28.89
100(1 -  ) = .90   = .10
 05.
2


i i  2
We know,
2Xi 2939.85
n s x
( 
1)
v
,1 / 2
n s x
( 1)
v
, / 2
2
2
2

2
2
 
x
 

  
=
(10 
1)28.89
9,1 .05
(10 1)28.89
9,.05
2
2
 

2 



x
=
(10 1)28.89 2 
(10 1)28.89
3.33
16.92
 

 x
= 15.370 < x 2 <78.097
So, the confidence interval for population standard deviation is 15.370 x  78.097
= 3.92 < x < 8.84
Confidence interval for population standard deviation range is from 3.92 to 8.84 of weight
looses for patients of the clinic’s weight reduction program.
Answer the question: 28
Here, n  25
60.32

1508
   25
n
xi
x And  2 Sx
2 2
 
xi nx
1

n
=
95628 25(60.32)2
25 
1

= 194.393
95% confidence interval for population standard deviation is: 100(1 -  ) = .95  =
.05  .025
2


i 
1 18.2 331.24
2 25.9 670.81
3 6.3 39.69
4 11.8 139.24
5 15.4 237.16
6 20.3 412.09
7 16.8 282.24
8 19.5 380.25
9 12.3 151.29
10 17.2 295.84
  i X 163.7  

10. We know,
n s x
( 
1)
v
,1 / 2
n s x
( 1)
v
, / 2
2
2
2

2
2
 
x
 

  
=
(25 
1)194.393
24,1 .025
(25 1)194.393
24,.025
2
2
 

2 



x
=
(25 1)194.393 2 
(25 1)194.393
12.40
39.36
 

 x
= 118.53 < x 2 <376.24
So, the confidence interval for population standard deviation is
118.53 x  376.24 = 10.89 < x < 19.40
Answer the question: 31
Here, n 18 Sx=10.4
(a) 90% confidence interval for population standard deviation is: 100(1 -  ) = .95
  = .05  05.
2


We know,
n s x
( 
1)
v
,1 / 2
n s x
( 1)
v
, / 2
2
2
2

2
2
 
x
 

  
=
(18 1)(10.4)
17,1 .05
(18 1)(10.4)
17,.05
2
2
2
2
2


 




x
=
(17)108.16
8.67
(17)108.16 2  x 
27.59
= 66.64 < x 2 <212.08
Answer the question: 32
Given that, 15n % 36 . 2  x s so, 5.57% 2 s x 
a) So, 95% confidence interval for variance is: 100(1 -  ) = .95  = .05  025 .
2


=
(15 
1)5.57
14,1 .025
(15 
1)5.57
14,.025
2
2
 
2 



x
=
(15 1)5.57 2 
(15 1)5.57
5.63
26.12
 

 x
= 2.99 < x 2  <13.85
Hence, confidence interval for variance range is from 2.99 to 13.85
b) So, 99% confidence interval for variance is: 100(1 -  ) = .99  = .01  005 .
2


=
(15 
1)5.57
14,1 .005
(15 
1)5.57
14,.005
2
2
 
2 



x
=
(15 1)5.57 2 
(15 1)5.57
4.07
31.34
 

 x
= 2.49 < x 2  <19.16
Confidence interval for variance range is from 2.49 to 19.16

11. So, it is wider than a.
Answer the question: 33

182.4
   Given that, n  9 20.27
9
n
xi
x
[xi = 19.8, 21.2, 18.6, 20.4, 21.6, 19.8, 19.9, 20.3, and 20.8]
And  2 Sx
 
xi x
( )
1
2

n
=
6.3001
9 
1
= .788
2
x xi
So, ) (
= .2209 + .8649 + 2.7889 + .0169 + 1.7689 + .2209 + .1369 + .0009 + .2809 = 6.3001
So, 90% confidence interval for population variance is: 100(1 -  ) = .90   = .10
 05.
2


We know,
n s x
( 
1)
v
,1 / 2
n s x
( 1)
v
, / 2
2
2
2

2
2
 
x
 

  
=
(9 
1).788
8,1 .05
(9 1).788
8,.05
2
2
 

2 



x
=
(9 1).788 2 
(9 1).788
2.18
15.51
 

 x
= .406 < x 2 <2.892
Hence, the confidence interval for population variance range is from .406 to 2.892