United International University, MBA Faculty: Rashed Mohammad Saadullah Assistant Professor School of Business and Economics

1. Business Statistics (BUS 505) Assignment 9
Answer the question: 1
Given that, sx =.45; n = 25; x =2.90
a (a) 95% confidence interval is: 100(1 - a ) = .95 Þa = .05 Þ =
.025
2
So, we know that,
Z x
s a a
n
x
Z x
- 2 < < + 2
n
x
s
m
=
Z Z .45
- m [ Z = 1.96 .025 ]
2.90 .025 .025 ´
25
2.90
.45
25
< < +
´
= 2.72<m <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:
w 2 z s x = a / 2 [w = 2.99 – 2.81 = .18]
n
Þ
2 ´
.45
.18 / 2 25
= a z
Þ a / 2 Z =
.18
.18
Þ a / 2 Z = 1 = .8413
So, 1 -
a
2
= .8413 Þ
a
2
= .1587 Þ a = .3174
Now 1 - a = 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, sx =.12 n =16 x =4.07
a (a) 99% confidence interval is: 100(1 - a ) = .999 Þa = .01 Þ =
.005
2
So, we know that,
Z x
s a a
n
x
Z x
- 2 < < + 2
n
x
s
m
=
Z Z .16
- m [ Z.005 = 2.575 ]
4.07 .005 .005 ´
16
4.07
.12
16
< < +
´
= 3.99<m <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 = å
Given that, sx =3.8 n = 9 16.76
9
n
xi
x
a =
90% confidence interval is: 100(1 - a ) = .90 Þa = .10 Þ .05
2
So, we know that,
Z x
s a a
n
x
Z x
- 2 < < + 2
n
x
s
m
Page 1 of 16

2. Business Statistics (BUS 505) Assignment 9
1.645´3.8 1.645 ´ 3.8
=
16.76 -<m < 16.76 + [ Z=1.645 ]
9
9
a / 2 = 14.67<m <18.85
So, 90% confidence interval range is from 14.67 to18.85
b) Wider
Answer the question: 4
Here, s x = 32.4 n=9 X =187.9
a =
(a) 80% confidence interval is: 100(1 - a ) = 80 Þa = .20 Þ .10
2
Z x
s a a
n
x
Z x
- 2 < < + 2
n
x
s
m
1.285´32.4 1.285 ´ 32.4
=
187.9 -<m < 187.9 + [ Z=1.285 ]
9
9
a / 2 = 174.02<m <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 a / 2s
= [w = 210 – 185.8 = 44.2]
Þ44.2=
2za / 232.4
9
2Zs =4.09Þ
Þ 2
Zs
2
=2.046ÞFz(
Zs
2
)=Fz(2.05)
Þ1-
s =.9798Þ
2
s =.0202Þs =.0404Þ1-s =.9596
2
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-∞) =95Þ α =.05Þ α∕2=.025Þ Z.025=1.96
The 95% Confidence interval for the population mean
x - Za / 2Sx < m < x + Z a / 2 Sx
[ Zn
n
a / 2 =1.91]
(1.96)(1.57)
= 3.92-
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
The 90% Confidence Interval for the population mean100 %( 1-α) =90% Þ α = .10
Þa /2=.05 Þ Za /2=1.65
Page 2 of 16

3. Business Statistics (BUS 505) Assignment 9
(a) So, we know that,
x - Za / 2Sx < m < x + Z a / 2 Sx
[ Za / 2 =1.645 ]
n
n
= 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,
W = 2Za / 2d Given that, W=.2 Sx =1.045
n
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
a a
n
x 2 x
2
Z Sx
- <m < +
n
1.645´11.28 1.645 ´ 11.28
=
60.41-<m < 60.41 + [ Z=1.645 ]
352
352
a / 2 =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,
W = 2Za / 2d
n
So, .2=2Z α/2(1.43)
√174
=> Z α/2=.92=.8212
= >1- α/2=.8212
= > α/2=1-.8212
α=.3576
Or, 1- α=.6424 or, 64.24%
Page 3 of 16

4. Business Statistics (BUS 505) Assignment 9
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
x - tva / 2Sx <m < + a / 2
x tv Sx
n
n
=
157.82 -t8,.025´38.89 <m < 157.82 +t 8,.025 ´ 38.89
[ .025
9
9
a = ]
2
=
157.82 - 2.306´38.89 <m < + ´
=127.93 <μ< 187.71
157.82 2.306 38.89
3
3
Answer the question: 11
(a) x =1/ nå i= X 1/7(523) =74.7143
1 x2 nx2
n i
Sx2 = å -
{ }
( -
1)
= 1/6{39321-(7) (74.7143) 2 }
=40.90
Sx = 40.90 =6.3953
a =
(b) 95% confidence interval is: 100(1 - a ) = .95 Þa = .05 Þ .025
2
So, we know that, x -
t Sx (n-1)a/ 2 <m < x +
n
t Sx (n-1)a / 2 [ n -1= 6] [ n-1 t a / 2
n
=2.447]
Page 4 of 16
i i c
2
i c
1 79 6241
2 73 5329
3 68 4624
4 77 5929
5 86 7396
6 71 5041
7 69 4761
å i= X 523 åX 2i= 39321

5. Business Statistics (BUS 505) Assignment 9
= 74.7143-
(2.447)(6.3953)
7
<m < 74.7143+
(2.447)(6.3953)
7
= 74.7143-5.9149 <m < 74.7143+5.9149
=68.7994 <m < 80.6292
So, 95% confidence interval range is from 68.7994 to 80.6292
Answer the question: 12
(a) x =1/ nå i= X 1/10(163.7)
=16.37
1 x2 nx2
n i
Sx2 = å -
{ }
( -
1)
= 1/9{2939.85 - (10) (16.37) 2 }
= 28.8979
Sx = 28.8979 = 5.3757
Here, Sx =5.3757 X = 16.37 n=10
99% confidence interval is: 100(1 - a ) = .99 Þa = .01 Þa /2=.005
So, we know that, x -
t Sx (n-1)a/ 2 <m < x +
n
t Sx (n-1)a / 2 [ n -1= 9] [ n-1 t
n
a / 2 =3.250]
=16.37-
(3.250)(5.3757)
10
<m < 16.37+
(3.250)(5.3757)
10
=16.37- 5.5248<m <16.37+5.5248
Page 5 of 16
i i c
2
i c
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
å i= X 163.
7
åX 2i= 2939.85

6. Business Statistics (BUS 505) Assignment 9
= 10.8452<m <21.8948
So, 99% confidence interval range is from 10.8452 to 21.8948
(b) narrower range.
Answer the question: 13
(a) x =1/ n åXi=11/25(1508) =60.32
1 x2 nx2
n i
Sx2 = å -
{ }
( -
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)a/ 2 <m < x +
n
t Sx (n-1)a / 2 [ n -1= 24] [ n-1 t
n
a / 2 =1.711]
Þ 60.32-
1.711´13.94
25
<m <60.32+
1.711´13.94
25
Þ55.55<m <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
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, Sx = 4780 n = 25
a 90% confidence interval is: 100(1 - a ) = .90 Þa = .10 Þ =
.05
2
So, we know that, x -
t Sx (n-1)a / 2 <m < x +
n
t Sx (n-1)a/ 2 [ n -1=24] [ n-1 t
n
a / 2 =1.711]
= 42740 -
(1.711)(4780)
25
(1.711)(4780)
<m < 42740 + 25
= 42740-(1635.716) <m <42740+ (1635.716)
= 41104.284 <m < 44375.716
So, 90% confidence interval range is from 41104.284 to 44375.716
Answer the question: 19
Page 6 of 16

7. Business Statistics (BUS 505) Assignment 9
Here, n=189 X=132 Px=
x =
n
132 =.698
189
(a) 90% confidence interval for population proportion is;
Px(1-Px) < P < Ṗ+Z a / 2 n
Ṗ-Za / 2 n
Px(1-Px) [ Z a / 2 =1.645]
Þ.698-1.645
.698(1-.698) < P < .698+1.645
189
.698(1-.698)
189
Þ.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 =.4799 W= [.5-.458] = .
323
042
We know,
W = 2Za / 2 Px(1- Px)
n
Þ
.042 = 2Za / 2 .4799(1-.4799)
323
Þ Za / 2 = .755 =.7764
Þ1-a / 2 =.7764
Þa =.4472Þ1-a =.5528
So, The Confidence Interval is 55.28 %.
Answer the question: 21
Here, n=134 X=82 Px=
x =
n
82 =.612
134
95% confidence interval for population proportion is;
Px(1-Px) < P < Ṗ+Z a / 2 n
Ṗ-Za / 2 n
Px(1-Px) [ Z a / 2 =1.955]
Þ.612-1.955
.612(1-.612) < P < .612+1.955
134
.612(1-.612)
134
Þ.53 <P< .694
Answer the question: 22
Here, n=95 X=29 Px=
x =
n
29 =.3053
95
(a) 99% confidence interval for population proportion is;
Page 7 of 16

8. Business Statistics (BUS 505) Assignment 9
Px(1-Px) < P < Ṗ+Z a / 2 n
Ṗ-Za / 2 n
Px(1-Px) [ Z a / 2 =2.575]
Þ.3053-2.575
.3053(1-.3053) < P < .3053+2.575
95
.3053(1-.3053)
95
Þ.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 =.333
96
80% confidence interval for population proportion is;
Px(1-Px) < P < Ṗ+Z a / 2 n
Ṗ-Za / 2 n
Px(1-Px) [ Z a / 2 =1.285]
Þ.333-1.285
.333(1-.3053) < P < .333+1.285
95
.333(1-.333)
96
Þ.2712<P< .3948
Answer the question: 24
Here, n=198 X= 98 Px=
x =
n
98 =.495 W= [.545-.445] = .10
198
We know,
W = 2Za / 2 Px(1- Px)
n
Þ
.10 = 2Za / 2 .495(1-.495)
198
Þ Za / 2 =1.41=.9207
Þ1-a / 2 =.9207
Þa =.1586Þ1-a =.8414
So, The Confidence Interval is 84.14 %.
Answer the question: 25
Given that, n =15 =.880 x s so, s2 x = .7744
a =
100(1 - a ) = .95 Þa = .05 Þ .025
2
We know,
x n s x
n s x
- < < -
( 1)
,1 / 2
( 1)
, / 2
2
2
2
2
2
v
c a
s
v
c a -
- < < -
(15 1).7744
(15 1).7744
= 14,.025
2
14,1 .025
2
2 c
-
s
c
x
(15 -1).7744 -
=
<s 2 x < = .4151 <s2x <1.9257
(15 1).7744
5.63
26.12
Page 8 of 16

9. Business Statistics (BUS 505) Assignment 9
Answer the question: 26
n = 7
x =1/ nå i= X 1/7(523) =74.7143
1 x2 nx2
n i
Sx2 = å -
{ }
( -
1)
= 1/6{39321-(7) (74.7143) 2 }
=40.90
a =
100(1 - a ) = .80 Þa = .20 Þ .1
2
x n s x
n s x
- < < -
( 1)
( 1)
We know, , / 2
2
,1 / 2
2
2
2
2
v
c a
s
v
c a -
- < < -
(7 1)40.90
(7 1)40.90
= 6,.1
2
6,1 .1
2
2 c
-
s
c
x
(7 -1)40.90 <s 2 x < -
(7 1)40.90
= 10.64
2.20
= 23.064 <s2x <111.545
Answer the question: 27
Page 9 of 16
i i c
2
i c
1 79 6241
2 73 5329
3 68 4624
4 77 5929
5 86 7396
6 71 5041
7 69 4761
å i= X 523 å =
Xi 2 39321

10. Business Statistics (BUS 505) Assignment 9
Here, n =10
x =1/ nå i= X 1/10(163.7) =16.37
1 x2 nx2
n i
Sx2 = å -
{ }
( -
1)
= 1/9{2939.85 - (10) (16.37) 2 }
= 28.89
a =
100(1 - a ) = .90 Þa = .10 Þ .05
2
We know,
x n s x
n s x
- < < -
( 1)
,1 / 2
( 1)
, / 2
2
2
2
2
2
v
c a
s
v
c a -
- < < -
(10 1)28.89
(10 1)28.89
= 9,.05
2
9,1 .05
2
2 c
-
s
c
x
=
(10 -1)28.89 <s 2 x < -
(10 1)28.89
3.33
16.92
= 15.370 <s2x <78.097
So, the confidence interval for population standard deviation is 15.370 <sx < 78.097
= 3.92 < sx < 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.
Page 10 of 16
i i c
2
i c
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
å i= X 163.
7
å =
Xi2 2939.85

11. Business Statistics (BUS 505) Assignment 9
Answer the question: 28
Here, n = 25
60.32
= = 1508 = å
25
n
xi
x And Sx2 =
2 2
-
å -
xi nx =
1
n
95628 25(60.32)2
- = 194.393
-
25 1
95% confidence interval for population standard deviation is: 100(1 - a ) = .95 Þa = .
Þ a 05 =
.025
2
We know,
x n s x
n s x
- < < -
( 1)
,1 / 2
( 1)
, / 2
2
2
2
2
2
v
c a
s
v
c a -
- < < -
(25 1)194.393
(25 1)194.393
= 24,.025
2
24,1 .025
2
2 c
-
s
c
x
=
(25 -1)194.393 <s 2 x < -
(25 1)194.393
12.40
39.36
= 118.53 <s2x <376.24
So, the confidence interval for population standard deviation is
118.53 <sx < 376.24 = 10.89 < sx < 19.40
Answer the question: 31
Here, n =18 Sx=10.4
(a) 90% confidence interval for population standard deviation is: 100(1 - a ) = .95
a =
Þa = .05 Þ .05
2
We know,
x n s x
n s x
- < < -
( 1)
,1 / 2
( 1)
, / 2
2
2
2
2
2
v
c a
s
v
c a -
=
(18 1)(10.4)
17,1 .05
(18 1)(10.4)
17,.05
2
2
2
2
2
-
- < < -
c
s
c
x
=
(17)108.16 <s 2x <
= 66.64 <s2x <212.08
(17)108.16
8.67
27.59
Answer the question: 32
Given that, n =15 = 2.36% x s so, s2 x = 5.57%
a) So, 95% confidence interval for variance is: 100(1 - a ) = .95 Þa = .05 Þ
.025
a =
2
- < < -
(15 1)5.57
(15 1)5.57
= 14,.025
2
14,1 .025
2
2 c
-
s
c
x
Page 11 of 16

12. Business Statistics (BUS 505) Assignment 9
(15 -1)5.57 -
=
<s 2 x < = 2.99 <s2x <13.85
(15 1)5.57
5.63
26.12
Hence, confidence interval for variance range is from 2.99 to 13.85
b) So, 99% confidence interval for variance is: 100(1 - a ) = .99 Þa = .01 Þ
.005
a =
2
- < < -
(15 1)5.57
(15 1)5.57
= 14,.005
2
14,1 .005
2
2 c
-
s
c
x
(15 -1)5.57 -
=
<s 2 x < = 2.49 <s2x <19.16
(15 1)5.57
4.07
31.34
Confidence interval for variance range is from 2.49 to 19.16
So, it is wider than a.
Answer the question: 33
= = 182.4 = å
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 Sx2 =
( ) 2
-
å -
xi x =
1
n
6.3001
-
9 1
= .788
So, å( - ) 2 xi x
= .2209 + .8649 + 2.7889 + .0169 + 1.7689 + .2209 + .1369 + .0009 + .2809 = 6.3001
So, 90% confidence interval for population variance is: 100(1 - a ) = .90 Þa = .10 Þ
.05
a =
2
We know,
x n s x
n s x
- < < -
( 1)
,1 / 2
( 1)
, / 2
2
2
2
2
2
v
c a
s
v
c a -
- < < -
(9 1).788
(9 1).788
= 8,.05
2
8,1 .05
2
2 c
-
s
c
x
(9 -1).788 -
=
<s 2 x < = .406 <s2x <2.892
(9 1).788
2.18
15.51
Hence, the confidence interval for population variance range is from .406 to 2.892
Estimating the sample size
Answer the question: 55
Given that, sx =.2 so,s 2 x =.04
Page 12 of 16

13. Business Statistics (BUS 505) Assignment 9
a) Here, n = 25 and, sx =.2
We know that,
w Z x s a / 2 2 = [w = 12.12 – 11.98 = .14]
n
Or,
.14 a / 2 Z =
2 .2
25
Or, a/ 2 Z .4 = .7
Or, a/ 2 Z =1.75
So, 1 - a / 2 = .9599 or, a / 2 = .0401 or, a = .0802
So, the confidence level is: 1 - a
= 1 - .0802
= .9198 or 91.98%
b) Here, L =.07
a =
So, 99% confidence interval is: 100(1 - a ) = .99 Þa = .01 Þ .005
2
We know that,
2 2
n Z = ( )
s x a
/ 2 2
L
2
(Z ) .04
.005
(.07)
= 2
= 54.128
So, 54.128 sample observations are needed to achieve this observation.
Answer the question: 56
Given that, n = 595 x =3.38 Sx =1.80
a =
a) 90% confidence interval is: 100(1 - a ) = .90 Þa = .10 Þ .05
2
So, we know that,
Z sx
a a
n
x 2 x
2
Z sx
- < m < +
n
=
Z Z 1.80
- m [ Z = 1.645 .05 ]
3.38 .05 .05 ´
595
3.38
1.80
595
< < +
´
= 3.259<m <3.501
So, 90% confidence interval range is from 3.259 to 3.501
Page 13 of 16

14. Business Statistics (BUS 505) Assignment 9
b) Wider
Answer the question: 58
Given that, n =16 x =32.0 Sx =6.4 so, s2 x = 40.96
a =
a) 90% confidence interval is: 100(1 - a ) = .90 Þa = .10 Þ .05
2
So, we know that, x -
t Sx (n-1)a / 2 <m < x +
n
t Sx (n-1)a / 2 [ = v,a/ 2 t 15,.05 t =1.753]
n
= 32.0 -
(1.753)(6.4)
16
(1.753)(6.4)
<m < 32.0 + 16
= 29.2 <m < 34.8
So, 90% confidence interval range is from 29.2 to 34.8
b) 90% confidence for population standard deviation is: 100(1 - a ) = .90 Þa = .10 Þ
.05
a =
2
We know that,
x n s x
n s x
- < < -
( 1)
,1 / 2
( 1)
, / 2
2
2
2
2
v
c a
s
v
c a -
=
- < < -
(16 1)40.96
2 2 -
15,1 .05
(16 1)40.96
15,.05
c
s
c
x
=
- < < -
(16 1)40.96
2 2 -
15,1 .05
(16 1)40.96
15,.05
c
s
c
x
= 24.576 <sx < 84.628
= 4.957 < s x < 9.199
Hence, the confidence for population standard deviation range is from 4.957 to 9.199
Answer the question: 59
Given that, n = 6
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15. Business Statistics (BUS 505) Assignment 9
= = 97.6 = å
a) 16.27
9
n
xi
x [xi = 12.2, 18.4, 23.1, 11.7, 8.2, and 24.0]
And Sx2 =
( ) 2
-
å -
xi x =
1
n
213.5
-
6 1
= 42.7
So, å( - ) 2 xi x = 16.56 + 4.54 + 46.65 + 20.88 + 65.12 + 59.75 = 213.5
b) Sx2 =42.7 so, Sx = 6.53
99% confidence interval for population mean is: 100(1 - a ) = .99 Þa = .01 Þ
.005
a =
2
So, we know that, x -
t Sx (n-1)a / 2 <m < x +
n
t Sx (n-1)a / 2 [ = v,a/ 2 t 5,.005 t = 4.032]
n
= 16.27 -
(4.032)(6.53)
6
(4.032)(6.53)
<m < 32.0 + 6
= 5.52 <m < 27.02
So, 90% confidence interval range is from 5.52 to 27.02
c) 99% confidence interval for population variance is: 100(1 - a ) = .99 Þa = .01 Þ
.005
a =
2
We know,
x n s x
n s x
- < < -
( 1)
,1 / 2
( 1)
, / 2
2
2
2
2
2
v
c a
s
v
c a -
- < < -
(6 1)42.7
(6 1)42.7
= 5,.005
2
5,1 .005
2
2 c
-
s
c
x
=
(6 -1)42.7 <s 2 x < -
(6 1)42.7
.412
16.75
= 12.75 <s2x <518.20
Hence, confidence interval for variance range is from 12.75 to 518.20
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16. Business Statistics (BUS 505) Assignment 9
d) Narrower.
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