Chapter 4, Problem 1PP

a)

To determine

To find standard deviation of given data.

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:

  $Standarddeviation=\sqrt{variance}$

Calculation:

Using formula,

  $Standarddeviation=\sqrt{254.5}=15.95$

b)

To determine

To find sample size of given data

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

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:

  $Standarderror=\frac{s}{\sqrt{n}}$

Calculation:

Using formula,

  $Standarderror=\frac{15.95}{\sqrt{9}}=5.32$

d)

To determine

To find the approximate 95% confidence interval for mean

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:

  $\begin{array}{l}Lowerlimit=Mean–2\times SE\\ Upperlimit=Mean+2\times SE\end{array}$

Calculation:

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

Population mean will be within $Mean–2\times SEandMean+2\times SE$

Therefore,

  $\begin{array}{l}Lowerlimit=\text{ 131 }–2\times 5.32=120.4mmHg\\ Upperlimit=\text{131 + }2\times 5.32=141.6mmHg\end{array}$