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