To find standard deviation of given data.

Question

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Answer 1

Question

Chapter 4, Problem 1PP

a)

To determine

To find standard deviation of given data.

a)

Expert Solution

Answer to Problem 1PP

The standard deviation is 15.95

Explanation of Solution

Given:

Data is,

112, 128, 108, 129, 125, 153, 155, 132, 137

Mean = 131.0 mm Hg and the variance = 254.5

Formula:

Standard deviation=variance

Calculation:

Using formula,

Standard deviation=254.5=15.95

b)

To determine

To find sample size of given data

b)

Expert Solution

Answer to Problem 1PP

The sample size is 9.

Explanation of Solution

Given:

Data is,

112, 128, 108, 129, 125, 153, 155, 132, 137

Mean = 131.0 mm Hg and the variance = 254.5

Calculation:

The sample size is a count of selected sample data values.

Therefore, here count is 9 hence sample size is n = 9.

c)

To determine

To find the standard error of the mean

c)

Expert Solution

Answer to Problem 1PP

The standard error of the mean = 5.32

Explanation of Solution

Given:

Data is,

112, 128, 108, 129, 125, 153, 155, 132, 137

Mean = 131.0 mm Hg and the variance = 254.5

Standard deviation s = 15.95 and Sample size = n = 9

Formula:

Standard error=sn

Calculation:

Using formula,

Standard error=15.959=5.32

d)

To determine

To find the approximate 95% confidence interval for mean

d)

Expert Solution

Answer to Problem 1PP

Lower limit = 120.4 mm Hg and upper limit = 141.6 mm Hg

Explanation of Solution

Given:

Data is,

112, 128, 108, 129, 125, 153, 155, 132, 137

Mean = 131.0 mm Hg and the variance = 254.5

Standard deviation s = 15.95 and Sample size = n = 9

Standard error = SE = 5.32

Formula:

Lower limit = Mean – 2×SEUpper limit = Mean + 2×SE

Calculation:

2 SE rule of thumb for approximate 95% confidence interval for mean is,

Population mean will be within Mean – 2×SE and Mean + 2×SE

Therefore,

Lower limit = 131 – 2×5.32=120.4 mm HgUpper limit = 131 + 2×5.32 =141.6 mm Hg

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Answer 2

Question

Chapter 4, Problem 1PP

a)

To determine

To find standard deviation of given data.

a)

Expert Solution

Answer to Problem 1PP

The standard deviation is 15.95

Explanation of Solution

Given:

Data is,

112, 128, 108, 129, 125, 153, 155, 132, 137

Mean = 131.0 mm Hg and the variance = 254.5

Formula:

Standard deviation=variance

Calculation:

Using formula,

Standard deviation=254.5=15.95

b)

To determine

To find sample size of given data

b)

Expert Solution

Answer to Problem 1PP

The sample size is 9.

Explanation of Solution

Given:

Data is,

112, 128, 108, 129, 125, 153, 155, 132, 137

Mean = 131.0 mm Hg and the variance = 254.5

Calculation:

The sample size is a count of selected sample data values.

Therefore, here count is 9 hence sample size is n = 9.

c)

To determine

To find the standard error of the mean

c)

Expert Solution

Answer to Problem 1PP

The standard error of the mean = 5.32

Explanation of Solution

Given:

Data is,

112, 128, 108, 129, 125, 153, 155, 132, 137

Mean = 131.0 mm Hg and the variance = 254.5

Standard deviation s = 15.95 and Sample size = n = 9

Formula:

Standard error=sn

Calculation:

Using formula,

Standard error=15.959=5.32

d)

To determine

To find the approximate 95% confidence interval for mean

d)

Expert Solution

Answer to Problem 1PP

Lower limit = 120.4 mm Hg and upper limit = 141.6 mm Hg

Explanation of Solution

Given:

Data is,

112, 128, 108, 129, 125, 153, 155, 132, 137

Mean = 131.0 mm Hg and the variance = 254.5

Standard deviation s = 15.95 and Sample size = n = 9

Standard error = SE = 5.32

Formula:

Lower limit = Mean – 2×SEUpper limit = Mean + 2×SE

Calculation:

2 SE rule of thumb for approximate 95% confidence interval for mean is,

Population mean will be within Mean – 2×SE and Mean + 2×SE

Therefore,

Lower limit = 131 – 2×5.32=120.4 mm HgUpper limit = 131 + 2×5.32 =141.6 mm Hg

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Answer 3

Question

Chapter 4, Problem 1PP

a)

To determine

To find standard deviation of given data.

a)

Expert Solution

Answer to Problem 1PP

The standard deviation is 15.95

Explanation of Solution

Given:

Data is,

112, 128, 108, 129, 125, 153, 155, 132, 137

Mean = 131.0 mm Hg and the variance = 254.5

Formula:

Standard deviation=variance

Calculation:

Using formula,

Standard deviation=254.5=15.95

b)

To determine

To find sample size of given data

b)

Expert Solution

Answer to Problem 1PP

The sample size is 9.

Explanation of Solution

Given:

Data is,

112, 128, 108, 129, 125, 153, 155, 132, 137

Mean = 131.0 mm Hg and the variance = 254.5

Calculation:

The sample size is a count of selected sample data values.

Therefore, here count is 9 hence sample size is n = 9.

c)

To determine

To find the standard error of the mean

c)

Expert Solution

Answer to Problem 1PP

The standard error of the mean = 5.32

Explanation of Solution

Given:

Data is,

112, 128, 108, 129, 125, 153, 155, 132, 137

Mean = 131.0 mm Hg and the variance = 254.5

Standard deviation s = 15.95 and Sample size = n = 9

Formula:

Standard error=sn

Calculation:

Using formula,

Standard error=15.959=5.32

d)

To determine

To find the approximate 95% confidence interval for mean

d)

Expert Solution

Answer to Problem 1PP

Lower limit = 120.4 mm Hg and upper limit = 141.6 mm Hg

Explanation of Solution

Given:

Data is,

112, 128, 108, 129, 125, 153, 155, 132, 137

Mean = 131.0 mm Hg and the variance = 254.5

Standard deviation s = 15.95 and Sample size = n = 9

Standard error = SE = 5.32

Formula:

Lower limit = Mean – 2×SEUpper limit = Mean + 2×SE

Calculation:

2 SE rule of thumb for approximate 95% confidence interval for mean is,

Population mean will be within Mean – 2×SE and Mean + 2×SE

Therefore,

Lower limit = 131 – 2×5.32=120.4 mm HgUpper limit = 131 + 2×5.32 =141.6 mm Hg

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Answer 4

Question

Chapter 4, Problem 1PP

a)

To determine

To find standard deviation of given data.

a)

Expert Solution

Answer to Problem 1PP

The standard deviation is 15.95

Explanation of Solution

Given:

Data is,

112, 128, 108, 129, 125, 153, 155, 132, 137

Mean = 131.0 mm Hg and the variance = 254.5

Formula:

Standard deviation=variance

Calculation:

Using formula,

Standard deviation=254.5=15.95

b)

To determine

To find sample size of given data

b)

Expert Solution

Answer to Problem 1PP

The sample size is 9.

Explanation of Solution

Given:

Data is,

112, 128, 108, 129, 125, 153, 155, 132, 137

Mean = 131.0 mm Hg and the variance = 254.5

Calculation:

The sample size is a count of selected sample data values.

Therefore, here count is 9 hence sample size is n = 9.

c)

To determine

To find the standard error of the mean

c)

Expert Solution

Answer to Problem 1PP

The standard error of the mean = 5.32

Explanation of Solution

Given:

Data is,

112, 128, 108, 129, 125, 153, 155, 132, 137

Mean = 131.0 mm Hg and the variance = 254.5

Standard deviation s = 15.95 and Sample size = n = 9

Formula:

Standard error=sn

Calculation:

Using formula,

Standard error=15.959=5.32

d)

To determine

To find the approximate 95% confidence interval for mean

d)

Expert Solution

Answer to Problem 1PP

Lower limit = 120.4 mm Hg and upper limit = 141.6 mm Hg

Explanation of Solution

Given:

Data is,

112, 128, 108, 129, 125, 153, 155, 132, 137

Mean = 131.0 mm Hg and the variance = 254.5

Standard deviation s = 15.95 and Sample size = n = 9

Standard error = SE = 5.32

Formula:

Lower limit = Mean – 2×SEUpper limit = Mean + 2×SE

Calculation:

2 SE rule of thumb for approximate 95% confidence interval for mean is,

Population mean will be within Mean – 2×SE and Mean + 2×SE

Therefore,

Lower limit = 131 – 2×5.32=120.4 mm HgUpper limit = 131 + 2×5.32 =141.6 mm Hg

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Answer 5

Chapter 4, Problem 1PP

a)

To determine

To find standard deviation of given data.

a)

Expert Solution

Answer to Problem 1PP

The standard deviation is 15.95

Explanation of Solution

Given:

Data is,

112, 128, 108, 129, 125, 153, 155, 132, 137

Mean = 131.0 mm Hg and the variance = 254.5

Formula:

Standard deviation=variance

Calculation:

Using formula,

Standard deviation=254.5=15.95

b)

To determine

To find sample size of given data

b)

Expert Solution

Answer to Problem 1PP

The sample size is 9.

Explanation of Solution

Given:

Data is,

112, 128, 108, 129, 125, 153, 155, 132, 137

Mean = 131.0 mm Hg and the variance = 254.5

Calculation:

The sample size is a count of selected sample data values.

Therefore, here count is 9 hence sample size is n = 9.

c)

To determine

To find the standard error of the mean

c)

Expert Solution

Answer to Problem 1PP

The standard error of the mean = 5.32

Explanation of Solution

Given:

Data is,

112, 128, 108, 129, 125, 153, 155, 132, 137

Mean = 131.0 mm Hg and the variance = 254.5

Standard deviation s = 15.95 and Sample size = n = 9

Formula:

Standard error=sn

Calculation:

Using formula,

Standard error=15.959=5.32

d)

To determine

To find the approximate 95% confidence interval for mean

d)

Expert Solution

Answer to Problem 1PP

Lower limit = 120.4 mm Hg and upper limit = 141.6 mm Hg

Explanation of Solution

Given:

Data is,

112, 128, 108, 129, 125, 153, 155, 132, 137

Mean = 131.0 mm Hg and the variance = 254.5

Standard deviation s = 15.95 and Sample size = n = 9

Standard error = SE = 5.32

Formula:

Lower limit = Mean – 2×SEUpper limit = Mean + 2×SE

Calculation:

2 SE rule of thumb for approximate 95% confidence interval for mean is,

Population mean will be within Mean – 2×SE and Mean + 2×SE

Therefore,

Lower limit = 131 – 2×5.32=120.4 mm HgUpper limit = 131 + 2×5.32 =141.6 mm Hg

Answer 6

Chapter 4, Problem 1PP

a)

To determine

To find standard deviation of given data.

a)

Expert Solution

Answer to Problem 1PP

The standard deviation is 15.95

Explanation of Solution

Given:

Data is,

112, 128, 108, 129, 125, 153, 155, 132, 137

Mean = 131.0 mm Hg and the variance = 254.5

Formula:

Standard deviation=variance

Calculation:

Using formula,

Standard deviation=254.5=15.95

Answer 7

Chapter 4, Problem 1PP

a)

To determine

To find standard deviation of given data.

Answer 8

a)

Expert Solution

Answer to Problem 1PP

The standard deviation is 15.95

Answer 9

a)

Expert Solution

Answer 10

a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given:

Data is,

112, 128, 108, 129, 125, 153, 155, 132, 137

Mean = 131.0 mm Hg and the variance = 254.5

Formula:

Standard deviation=variance

Calculation:

Using formula,

Standard deviation=254.5=15.95

Answer 13

b)

To determine

To find sample size of given data

b)

Expert Solution

Answer to Problem 1PP

The sample size is 9.

Explanation of Solution

Given:

Data is,

112, 128, 108, 129, 125, 153, 155, 132, 137

Mean = 131.0 mm Hg and the variance = 254.5

Calculation:

The sample size is a count of selected sample data values.

Therefore, here count is 9 hence sample size is n = 9.

Answer 14

b)

To determine

To find sample size of given data

Answer 15

b)

Expert Solution

Answer to Problem 1PP

The sample size is 9.

Answer 16

b)

Expert Solution

Answer 17

b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Given:

Data is,

112, 128, 108, 129, 125, 153, 155, 132, 137

Mean = 131.0 mm Hg and the variance = 254.5

Calculation:

The sample size is a count of selected sample data values.

Therefore, here count is 9 hence sample size is n = 9.

Answer 20

c)

To determine

To find the standard error of the mean

c)

Expert Solution

Answer to Problem 1PP

The standard error of the mean = 5.32

Explanation of Solution

Given:

Data is,

112, 128, 108, 129, 125, 153, 155, 132, 137

Mean = 131.0 mm Hg and the variance = 254.5

Standard deviation s = 15.95 and Sample size = n = 9

Formula:

Standard error=sn

Calculation:

Using formula,

Standard error=15.959=5.32

Answer 21

c)

To determine

To find the standard error of the mean

Answer 22

c)

Expert Solution

Answer to Problem 1PP

The standard error of the mean = 5.32

Answer 23

c)

Expert Solution

Answer 24

c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Given:

Data is,

112, 128, 108, 129, 125, 153, 155, 132, 137

Mean = 131.0 mm Hg and the variance = 254.5

Standard deviation s = 15.95 and Sample size = n = 9

Formula:

Standard error=sn

Calculation:

Using formula,

Standard error=15.959=5.32

Answer 27

d)

To determine

To find the approximate 95% confidence interval for mean

d)

Expert Solution

Answer to Problem 1PP

Lower limit = 120.4 mm Hg and upper limit = 141.6 mm Hg

Explanation of Solution

Given:

Data is,

112, 128, 108, 129, 125, 153, 155, 132, 137

Mean = 131.0 mm Hg and the variance = 254.5

Standard deviation s = 15.95 and Sample size = n = 9

Standard error = SE = 5.32

Formula:

Lower limit = Mean – 2×SEUpper limit = Mean + 2×SE

Calculation:

2 SE rule of thumb for approximate 95% confidence interval for mean is,

Population mean will be within Mean – 2×SE and Mean + 2×SE

Therefore,

Lower limit = 131 – 2×5.32=120.4 mm HgUpper limit = 131 + 2×5.32 =141.6 mm Hg

Answer 28

d)

To determine

To find the approximate 95% confidence interval for mean

Answer 29

d)

Expert Solution

Answer to Problem 1PP

Lower limit = 120.4 mm Hg and upper limit = 141.6 mm Hg

Answer 30

d)

Expert Solution

Answer 31

d)

Expert Solution

Answer 32

Expert Solution

Answer 33

Explanation of Solution

Given:

Data is,

112, 128, 108, 129, 125, 153, 155, 132, 137

Mean = 131.0 mm Hg and the variance = 254.5

Standard deviation s = 15.95 and Sample size = n = 9

Standard error = SE = 5.32

Formula:

Lower limit = Mean – 2×SEUpper limit = Mean + 2×SE

Calculation:

2 SE rule of thumb for approximate 95% confidence interval for mean is,

Population mean will be within Mean – 2×SE and Mean + 2×SE

Therefore,

Lower limit = 131 – 2×5.32=120.4 mm HgUpper limit = 131 + 2×5.32 =141.6 mm Hg

Answer 34

Ch. 4 - Prob. 1PP Ch. 4 - Prob. 2PP Ch. 4 - Prob. 3PP Ch. 4 - Prob. 4PP Ch. 4 - Prob. 5PP Ch. 4 - Prob. 6PP Ch. 4 - Prob. 7PP Ch. 4 - Prob. 8PP Ch. 4 - Prob. 9PP Ch. 4 - Prob. 10PP

To find standard deviation of given data.

Concept explainers

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To find standard deviation of given data.

Answer to Problem 1PP

Explanation of Solution

To find sample size of given data

Answer to Problem 1PP

Explanation of Solution

To find the standard error of the mean

Answer to Problem 1PP

Explanation of Solution

To find the approximate 95% confidence interval for mean

Answer to Problem 1PP

Explanation of Solution

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Chapter 4 Solutions