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United International University, MBA Faculty: Rashed Mohammad Saadullah Assistant Professor School of Business and Economics
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Interval Estimation (Page No. 289)
Answer the question: 1
Given that, ; ;
(a) 95% confidence interval is: 100(1 - ) = .95 = .05
So, we know that,
= [ ]
= 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:
[w = 2.99 – 2.81 = .18]
=
= 1 = .8413
So, 1 - = .8413 = .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,
(a) 99% confidence interval is: 100(1 - ) = .999 = .01
So, we know that,
= [ ]
= 3.99< <4.15So, 99% confidence interval range is from 3.97 to 4.17
(b) Narrower(c) Narrower(d) Wider
Answer the question: 3
Given that,
90% confidence interval is: 100(1 - ) = .90 = .10
So, we know that,
= [ ]
= 14.67< <18.85So, 90% confidence interval range is from 14.67 to18.85
b) Wider Answer the question: 4
Here, n=9
(a) 80% confidence interval is: 100(1 - ) = 80 = .20
= [ ]
= 174.02< <201.78So, 80% confidence interval range is from 174.02 to 201.78.
(b) The probability content associated the interval from 185.8 to 210 is:
[w = 210 – 185.8 = 44.2]
44.2= =4.09 =2.046 Fz( )=Fz(2.05)
1- =.9798 =.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 =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
[ ]
= 3.92- <μ< 3.92+
= 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 =3.81 Sx =1.34
The 90% Confidence Interval for the population mean100 %( 1-α) =90% α = .10 /2=.05 Z /2=1.65
(a) So, we know that, [ ]
= 3.81- <μ<3.81+
=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, 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: 8Here, n=352 Sx=11.28 =60.41
= [ ]
=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 =6.06 W=.2
(a) We know,
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%
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 =157.82 V= (n-1)=(9-1)=8
= [ ]
=
=127.93 <μ< 187.71
Answer the question: 11
(a) 1/7(523) =74.7143
= 1/6{39321-(7) (74.7143) }
=40.90
=6.3953
(b) 95% confidence interval is: 100(1 -
) = .95 = .05
So, we know that, - < < + [ = 6] [ =2.447]
= 74.7143- < < 74.7143+
= 74.7143-5.9149 < < 74.7143+5.9149 =68.7994 < < 80.6292
So, 95% confidence interval range is from 68.7994 to 80.6292
Answer the question: 12
(a) 1/10(163.7) =16.37
= 1/9{2939.85 - (10) (16.37) }
= 28.8979
= 5.3757
i
1 79 6241
2 73 5329
3 68 4624
4 77 5929
5 86 7396
6 71 5041
7 69 4761
523 39321
i
1 18.2 331.24 2 25.9 670.813 6.3 39.694 11.8 139.245 15.4 237.166 20.3 412.097 16.8 282.248 19.5 380.259 12.3 151.29
10 17.2 295.84
163.7
2939.85
Here, 5.3757 = 16.37 n=10
99% confidence interval is: 100(1 - ) = .99 = .01 /2=.005
So, we know that, - < < + [ = 9] [
=3.250]
=16.37- < < 16.37+
=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) 1/25(1508) =60.32
= 1/24{95628 - (25) (60.32) }
= 194.39 = 13.94
(b) So, we know that, - < < + [ = 24] [
=1.711]
60.32- < <60.32+
55.55< <65.09Answer 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,
90% confidence interval is: 100(1 - ) = .90 = .10
So, we know that, - < < + [ =24] [
=1.711]
= 42740 - < < 42740 +
= 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= = =.698
(a) 90% confidence interval for population proportion is;
Ṗ-Z < P < Ṗ+Z [ Z =1.645]
.698-1.645 < P < .698+1.645
.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= = =.4799 W= [.5-.458] = .042
We know,
=.7764 =.7764 =.4472 1- =.5528
So, The Confidence Interval is 55.28 %.Answer the question: 21
Here, n=134 X=82 Px= = =.612
95% confidence interval for population proportion is;
Ṗ-Z < P < Ṗ+Z [ Z =1.955]
.612-1.955 < P < .612+1.955
.53 <P< .694Answer the question: 22
Here, n=95 X=29 Px= = =.3053
(a) 99% confidence interval for population proportion is;
Ṗ-Z < P < Ṗ+Z [ Z =2.575]
.3053-2.575 < P < .3053+2.575
.1836 <P< .4270(b) If confidence is decreases than rang will be narrower.
Answer the question: 23
Here, n=96 X=32 Px= = =.333
80% confidence interval for population proportion is;
Ṗ-Z < P < Ṗ+Z [ Z =1.285]
.333-1.285 < P < .333+1.285
.2712<P< .3948Answer the question: 24
Here, n=198 X= 98 Px= = =.495 W= [.545-.445] = .10
We know,
=.9207 =.9207 =.1586 1- =.8414
So, The Confidence Interval is 84.14 %.
Answer the question: 25
Given that, so,
100(1 - ) = .95 = .05
We know,
=
=
= .4151 < <1.9257
Answer the question: 26
1/7(523) =74.7143
i
1 79 62412 73 53293 68 46244 77 59295 86 73966 71 50417 69 4761
523 39321
= 1/6{39321-(7) (74.7143) }
=40.90
100(1 - ) = .80 = .20
We know,
=
=
= 23.064 < <111.545
Answer the question: 27
Here, 1/10(163.7) =16.37
= 1/9{2939.85 - (10) (16.37) }
= 28.89
100(1 - ) = .90 = .10
We know,
=
=
= 15.370 < <78.097
i
1 18.2 331.24 2 25.9 670.813 6.3 39.694 11.8 139.245 15.4 237.166 20.3 412.097 16.8 282.248 19.5 380.259 12.3 151.29
10 17.2 295.84
163.7
2939.85
So, the confidence interval for population standard deviation is
= 3.92 < < 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,
And = = 194.393
95% confidence interval for population standard deviation is: 100(1 - ) = .95 = .05
We know,
=
=
= 118.53 < <376.24
So, the confidence interval for population standard deviation is = 10.89 < < 19.40
Answer the question: 31
Here, Sx=10.4
(a) 90% confidence interval for population standard deviation is: 100(1 - ) = .95
= .05
We know,
=
=
= 66.64 < <212.08
Answer the question: 32
Given that, so,
a) So, 95% confidence interval for variance is: 100(1 - ) = .95 = .05
=
=
= 2.99 < <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
=
=
= 2.49 < <19.16
Confidence interval for variance range is from 2.49 to 19.16 So, it is wider than a.
Answer the question: 33
Given that,
[xi = 19.8, 21.2, 18.6, 20.4, 21.6, 19.8, 19.9, 20.3, and 20.8]
And = = .788
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
We know,
=
=
= .406 < <2.892
Hence, the confidence interval for population variance range is from .406 to 2.892