the confidence interval for the population mean .

Question

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Answer 1

Question

Chapter 8.5, Problem 8.26E

(a)

To determine

To find: the confidence interval for the population mean.

(a)

Expert Solution

Answer to Problem 8.26E

The confidence interval is $(32.5530,35.4470)$ .

Explanation of Solution

Given:

The sample size is 38.

The level of significance ( $α$ ) is (1-0.98) i.e. 0.01.

The sample mean is 0.34.

The sample variance is 12.

Formulae used:

The $100(1−α)%$ Confidence interval for the population mean is given by:-

$(x¯− s 2 nZα2,x¯+ s 2 nZα2 )$

Calculations:

The Z Critical value at 99% level of significance is 2.58 calculated from the standard normal table.

The $100(1−α)%$ Confidence interval can be computed as:- $(x¯− s 2 n Z α 2 ,x¯+ s 2 n Z α 2 )=(34−2.58 12 38 ,34+2.58 12 38 )=(32.5530,35.4470)$

Therefore, the required confidence interval is $(32.5530,35.4470)$ .

(b)

To determine

To find: the confidence interval for the population mean.

(b)

Expert Solution

Answer to Problem 8.26E

The confidence interval is $(1047.5429,1050.4751)$ .

Explanation of Solution

Given:

The sample size is 65.

The level of significance ( $α$ ) is (1-0.90) i.e. 0.1.

The sample mean is 1049.

The sample variance is 51.

Calculations:

The Z Critical value at 90% level of significance is 1.645 calculated from the standard normal table.

The $100(1−α)%$ Confidence interval can be computed as:-

$(x¯− s 2 n Z α 2 ,x¯+ s 2 n Z α 2 )=(1049−1.645 51 65 ,1049+1.645 51 65 )=(1047.5429,1050.4751)$

Therefore, the required confidence interval is $(1047.5429,1050.4751)$ .

(c)

To determine

To find: the confidence interval for the population mean.

(c)

Expert Solution

Answer to Problem 8.26E

The confidence interval is $(65.9728,66.6272)$ .

Explanation of Solution

Given:

The sample size is 89.

The level of significance ( $α$ ) is (1-0.90) i.e. 0.1.

The sample mean is 66.3.

The sample variance is 2.48.

Calculations:

The Z Critical value at 95% level of significance is 1.96 calculated from the standard normal table.

The $100(1−α)%$ Confidence interval can be computed as:-

$(x¯− s 2 n Z α 2 ,x¯+ s 2 n Z α 2 )=(66.3−1.96 2.48 89 ,66.3−1.96 2.48 89 )=(65.9728,66.6272)$

Therefore, the required confidence interval is $(65.9728,66.6272)$ .

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Answer 2

Question

Chapter 8.5, Problem 8.26E

(a)

To determine

To find: the confidence interval for the population mean.

(a)

Expert Solution

Answer to Problem 8.26E

The confidence interval is $(32.5530,35.4470)$ .

Explanation of Solution

Given:

The sample size is 38.

The level of significance ( $α$ ) is (1-0.98) i.e. 0.01.

The sample mean is 0.34.

The sample variance is 12.

Formulae used:

The $100(1−α)%$ Confidence interval for the population mean is given by:-

$(x¯− s 2 nZα2,x¯+ s 2 nZα2 )$

Calculations:

The Z Critical value at 99% level of significance is 2.58 calculated from the standard normal table.

The $100(1−α)%$ Confidence interval can be computed as:- $(x¯− s 2 n Z α 2 ,x¯+ s 2 n Z α 2 )=(34−2.58 12 38 ,34+2.58 12 38 )=(32.5530,35.4470)$

Therefore, the required confidence interval is $(32.5530,35.4470)$ .

(b)

To determine

To find: the confidence interval for the population mean.

(b)

Expert Solution

Answer to Problem 8.26E

The confidence interval is $(1047.5429,1050.4751)$ .

Explanation of Solution

Given:

The sample size is 65.

The level of significance ( $α$ ) is (1-0.90) i.e. 0.1.

The sample mean is 1049.

The sample variance is 51.

Calculations:

The Z Critical value at 90% level of significance is 1.645 calculated from the standard normal table.

The $100(1−α)%$ Confidence interval can be computed as:-

$(x¯− s 2 n Z α 2 ,x¯+ s 2 n Z α 2 )=(1049−1.645 51 65 ,1049+1.645 51 65 )=(1047.5429,1050.4751)$

Therefore, the required confidence interval is $(1047.5429,1050.4751)$ .

(c)

To determine

To find: the confidence interval for the population mean.

(c)

Expert Solution

Answer to Problem 8.26E

The confidence interval is $(65.9728,66.6272)$ .

Explanation of Solution

Given:

The sample size is 89.

The level of significance ( $α$ ) is (1-0.90) i.e. 0.1.

The sample mean is 66.3.

The sample variance is 2.48.

Calculations:

The Z Critical value at 95% level of significance is 1.96 calculated from the standard normal table.

The $100(1−α)%$ Confidence interval can be computed as:-

$(x¯− s 2 n Z α 2 ,x¯+ s 2 n Z α 2 )=(66.3−1.96 2.48 89 ,66.3−1.96 2.48 89 )=(65.9728,66.6272)$

Therefore, the required confidence interval is $(65.9728,66.6272)$ .

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Answer 3

Question

Chapter 8.5, Problem 8.26E

(a)

To determine

To find: the confidence interval for the population mean.

(a)

Expert Solution

Answer to Problem 8.26E

The confidence interval is $(32.5530,35.4470)$ .

Explanation of Solution

Given:

The sample size is 38.

The level of significance ( $α$ ) is (1-0.98) i.e. 0.01.

The sample mean is 0.34.

The sample variance is 12.

Formulae used:

The $100(1−α)%$ Confidence interval for the population mean is given by:-

$(x¯− s 2 nZα2,x¯+ s 2 nZα2 )$

Calculations:

The Z Critical value at 99% level of significance is 2.58 calculated from the standard normal table.

The $100(1−α)%$ Confidence interval can be computed as:- $(x¯− s 2 n Z α 2 ,x¯+ s 2 n Z α 2 )=(34−2.58 12 38 ,34+2.58 12 38 )=(32.5530,35.4470)$

Therefore, the required confidence interval is $(32.5530,35.4470)$ .

(b)

To determine

To find: the confidence interval for the population mean.

(b)

Expert Solution

Answer to Problem 8.26E

The confidence interval is $(1047.5429,1050.4751)$ .

Explanation of Solution

Given:

The sample size is 65.

The level of significance ( $α$ ) is (1-0.90) i.e. 0.1.

The sample mean is 1049.

The sample variance is 51.

Calculations:

The Z Critical value at 90% level of significance is 1.645 calculated from the standard normal table.

The $100(1−α)%$ Confidence interval can be computed as:-

$(x¯− s 2 n Z α 2 ,x¯+ s 2 n Z α 2 )=(1049−1.645 51 65 ,1049+1.645 51 65 )=(1047.5429,1050.4751)$

Therefore, the required confidence interval is $(1047.5429,1050.4751)$ .

(c)

To determine

To find: the confidence interval for the population mean.

(c)

Expert Solution

Answer to Problem 8.26E

The confidence interval is $(65.9728,66.6272)$ .

Explanation of Solution

Given:

The sample size is 89.

The level of significance ( $α$ ) is (1-0.90) i.e. 0.1.

The sample mean is 66.3.

The sample variance is 2.48.

Calculations:

The Z Critical value at 95% level of significance is 1.96 calculated from the standard normal table.

The $100(1−α)%$ Confidence interval can be computed as:-

$(x¯− s 2 n Z α 2 ,x¯+ s 2 n Z α 2 )=(66.3−1.96 2.48 89 ,66.3−1.96 2.48 89 )=(65.9728,66.6272)$

Therefore, the required confidence interval is $(65.9728,66.6272)$ .

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Answer 4

Question

Chapter 8.5, Problem 8.26E

(a)

To determine

To find: the confidence interval for the population mean.

(a)

Expert Solution

Answer to Problem 8.26E

The confidence interval is $(32.5530,35.4470)$ .

Explanation of Solution

Given:

The sample size is 38.

The level of significance ( $α$ ) is (1-0.98) i.e. 0.01.

The sample mean is 0.34.

The sample variance is 12.

Formulae used:

The $100(1−α)%$ Confidence interval for the population mean is given by:-

$(x¯− s 2 nZα2,x¯+ s 2 nZα2 )$

Calculations:

The Z Critical value at 99% level of significance is 2.58 calculated from the standard normal table.

The $100(1−α)%$ Confidence interval can be computed as:- $(x¯− s 2 n Z α 2 ,x¯+ s 2 n Z α 2 )=(34−2.58 12 38 ,34+2.58 12 38 )=(32.5530,35.4470)$

Therefore, the required confidence interval is $(32.5530,35.4470)$ .

(b)

To determine

To find: the confidence interval for the population mean.

(b)

Expert Solution

Answer to Problem 8.26E

The confidence interval is $(1047.5429,1050.4751)$ .

Explanation of Solution

Given:

The sample size is 65.

The level of significance ( $α$ ) is (1-0.90) i.e. 0.1.

The sample mean is 1049.

The sample variance is 51.

Calculations:

The Z Critical value at 90% level of significance is 1.645 calculated from the standard normal table.

The $100(1−α)%$ Confidence interval can be computed as:-

$(x¯− s 2 n Z α 2 ,x¯+ s 2 n Z α 2 )=(1049−1.645 51 65 ,1049+1.645 51 65 )=(1047.5429,1050.4751)$

Therefore, the required confidence interval is $(1047.5429,1050.4751)$ .

(c)

To determine

To find: the confidence interval for the population mean.

(c)

Expert Solution

Answer to Problem 8.26E

The confidence interval is $(65.9728,66.6272)$ .

Explanation of Solution

Given:

The sample size is 89.

The level of significance ( $α$ ) is (1-0.90) i.e. 0.1.

The sample mean is 66.3.

The sample variance is 2.48.

Calculations:

The Z Critical value at 95% level of significance is 1.96 calculated from the standard normal table.

The $100(1−α)%$ Confidence interval can be computed as:-

$(x¯− s 2 n Z α 2 ,x¯+ s 2 n Z α 2 )=(66.3−1.96 2.48 89 ,66.3−1.96 2.48 89 )=(65.9728,66.6272)$

Therefore, the required confidence interval is $(65.9728,66.6272)$ .

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Answer 5

Chapter 8.5, Problem 8.26E

(a)

To determine

To find: the confidence interval for the population mean.

(a)

Expert Solution

Answer to Problem 8.26E

The confidence interval is $(32.5530,35.4470)$ .

Explanation of Solution

Given:

The sample size is 38.

The level of significance ( $α$ ) is (1-0.98) i.e. 0.01.

The sample mean is 0.34.

The sample variance is 12.

Formulae used:

The $100(1−α)%$ Confidence interval for the population mean is given by:-

$(x¯− s 2 nZα2,x¯+ s 2 nZα2 )$

Calculations:

The Z Critical value at 99% level of significance is 2.58 calculated from the standard normal table.

The $100(1−α)%$ Confidence interval can be computed as:- $(x¯− s 2 n Z α 2 ,x¯+ s 2 n Z α 2 )=(34−2.58 12 38 ,34+2.58 12 38 )=(32.5530,35.4470)$

Therefore, the required confidence interval is $(32.5530,35.4470)$ .

(b)

To determine

To find: the confidence interval for the population mean.

(b)

Expert Solution

Answer to Problem 8.26E

The confidence interval is $(1047.5429,1050.4751)$ .

Explanation of Solution

Given:

The sample size is 65.

The level of significance ( $α$ ) is (1-0.90) i.e. 0.1.

The sample mean is 1049.

The sample variance is 51.

Calculations:

The Z Critical value at 90% level of significance is 1.645 calculated from the standard normal table.

The $100(1−α)%$ Confidence interval can be computed as:-

$(x¯− s 2 n Z α 2 ,x¯+ s 2 n Z α 2 )=(1049−1.645 51 65 ,1049+1.645 51 65 )=(1047.5429,1050.4751)$

Therefore, the required confidence interval is $(1047.5429,1050.4751)$ .

(c)

To determine

To find: the confidence interval for the population mean.

(c)

Expert Solution

Answer to Problem 8.26E

The confidence interval is $(65.9728,66.6272)$ .

Explanation of Solution

Given:

The sample size is 89.

The level of significance ( $α$ ) is (1-0.90) i.e. 0.1.

The sample mean is 66.3.

The sample variance is 2.48.

Calculations:

The Z Critical value at 95% level of significance is 1.96 calculated from the standard normal table.

The $100(1−α)%$ Confidence interval can be computed as:-

$(x¯− s 2 n Z α 2 ,x¯+ s 2 n Z α 2 )=(66.3−1.96 2.48 89 ,66.3−1.96 2.48 89 )=(65.9728,66.6272)$

Therefore, the required confidence interval is $(65.9728,66.6272)$ .

Answer 6

Chapter 8.5, Problem 8.26E

(a)

To determine

To find: the confidence interval for the population mean.

(a)

Expert Solution

Answer to Problem 8.26E

The confidence interval is $(32.5530,35.4470)$ .

Explanation of Solution

Given:

The sample size is 38.

The level of significance ( $α$ ) is (1-0.98) i.e. 0.01.

The sample mean is 0.34.

The sample variance is 12.

Formulae used:

The $100(1−α)%$ Confidence interval for the population mean is given by:-

$(x¯− s 2 nZα2,x¯+ s 2 nZα2 )$

Calculations:

The Z Critical value at 99% level of significance is 2.58 calculated from the standard normal table.

The $100(1−α)%$ Confidence interval can be computed as:- $(x¯− s 2 n Z α 2 ,x¯+ s 2 n Z α 2 )=(34−2.58 12 38 ,34+2.58 12 38 )=(32.5530,35.4470)$

Therefore, the required confidence interval is $(32.5530,35.4470)$ .

Answer 7

Chapter 8.5, Problem 8.26E

(a)

To determine

To find: the confidence interval for the population mean.

Answer 8

(a)

Expert Solution

Answer to Problem 8.26E

The confidence interval is $(32.5530,35.4470)$ .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given:

The sample size is 38.

The level of significance ( $α$ ) is (1-0.98) i.e. 0.01.

The sample mean is 0.34.

The sample variance is 12.

Formulae used:

The $100(1−α)%$ Confidence interval for the population mean is given by:-

$(x¯− s 2 nZα2,x¯+ s 2 nZα2 )$

Calculations:

The Z Critical value at 99% level of significance is 2.58 calculated from the standard normal table.

The $100(1−α)%$ Confidence interval can be computed as:- $(x¯− s 2 n Z α 2 ,x¯+ s 2 n Z α 2 )=(34−2.58 12 38 ,34+2.58 12 38 )=(32.5530,35.4470)$

Therefore, the required confidence interval is $(32.5530,35.4470)$ .

Answer 13

(b)

To determine

To find: the confidence interval for the population mean.

(b)

Expert Solution

Answer to Problem 8.26E

The confidence interval is $(1047.5429,1050.4751)$ .

Explanation of Solution

Given:

The sample size is 65.

The level of significance ( $α$ ) is (1-0.90) i.e. 0.1.

The sample mean is 1049.

The sample variance is 51.

Calculations:

The Z Critical value at 90% level of significance is 1.645 calculated from the standard normal table.

The $100(1−α)%$ Confidence interval can be computed as:-

$(x¯− s 2 n Z α 2 ,x¯+ s 2 n Z α 2 )=(1049−1.645 51 65 ,1049+1.645 51 65 )=(1047.5429,1050.4751)$

Therefore, the required confidence interval is $(1047.5429,1050.4751)$ .

Answer 14

(b)

To determine

To find: the confidence interval for the population mean.

Answer 15

(b)

Expert Solution

Answer to Problem 8.26E

The confidence interval is $(1047.5429,1050.4751)$ .

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Given:

The sample size is 65.

The level of significance ( $α$ ) is (1-0.90) i.e. 0.1.

The sample mean is 1049.

The sample variance is 51.

Calculations:

The Z Critical value at 90% level of significance is 1.645 calculated from the standard normal table.

The $100(1−α)%$ Confidence interval can be computed as:-

$(x¯− s 2 n Z α 2 ,x¯+ s 2 n Z α 2 )=(1049−1.645 51 65 ,1049+1.645 51 65 )=(1047.5429,1050.4751)$

Therefore, the required confidence interval is $(1047.5429,1050.4751)$ .

Answer 20

(c)

To determine

To find: the confidence interval for the population mean.

(c)

Expert Solution

Answer to Problem 8.26E

The confidence interval is $(65.9728,66.6272)$ .

Explanation of Solution

Given:

The sample size is 89.

The level of significance ( $α$ ) is (1-0.90) i.e. 0.1.

The sample mean is 66.3.

The sample variance is 2.48.

Calculations:

The Z Critical value at 95% level of significance is 1.96 calculated from the standard normal table.

The $100(1−α)%$ Confidence interval can be computed as:-

$(x¯− s 2 n Z α 2 ,x¯+ s 2 n Z α 2 )=(66.3−1.96 2.48 89 ,66.3−1.96 2.48 89 )=(65.9728,66.6272)$

Therefore, the required confidence interval is $(65.9728,66.6272)$ .

Answer 21

(c)

To determine

To find: the confidence interval for the population mean.

Answer 22

(c)

Expert Solution

Answer to Problem 8.26E

The confidence interval is $(65.9728,66.6272)$ .

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Given:

The sample size is 89.

The level of significance ( $α$ ) is (1-0.90) i.e. 0.1.

The sample mean is 66.3.

The sample variance is 2.48.

Calculations:

The Z Critical value at 95% level of significance is 1.96 calculated from the standard normal table.

The $100(1−α)%$ Confidence interval can be computed as:-

$(x¯− s 2 n Z α 2 ,x¯+ s 2 n Z α 2 )=(66.3−1.96 2.48 89 ,66.3−1.96 2.48 89 )=(65.9728,66.6272)$

Therefore, the required confidence interval is $(65.9728,66.6272)$ .

Answer 27

Ch. 8.4 - Prob. 8.3E Ch. 8.4 - Prob. 8.4E Ch. 8.4 - Prob. 8.5E Ch. 8.4 - Prob. 8.6E Ch. 8.4 - Prob. 8.7E Ch. 8.4 - Prob. 8.8E Ch. 8.4 - Prob. 8.9E Ch. 8.4 - Prob. 8.10E Ch. 8.4 - Prob. 8.11E Ch. 8.4 - Prob. 8.12E

the confidence interval for the population mean .

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To find: the confidence interval for the population mean.

Answer to Problem 8.26E

Explanation of Solution

To find: the confidence interval for the population mean.

Answer to Problem 8.26E

Explanation of Solution

To find: the confidence interval for the population mean.

Answer to Problem 8.26E

Explanation of Solution

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