A random sample of 225 measurements is selected from a population, and the sample means and standard deviation are x = 32.5 and s = 30.0, respectively.

a. Use a 99% confidence interval to estimate the mean of the population, $\mu$.

b. How large a sample would be needed to estimate m to within .5 with 99% confidence?

c. Use a 99% confidence interval to estimate the population variance, ${\sigma }^2$.

d. What is meant by the phrase99% confidence as it is used in this exercise?

It is given that, $n=225$are the sample observations. The sample mean and sample standard deviation is $\overline{x}=32.5, s=30.0$.

a.

Need to build a 99% confidence interval. The Z-test value for 99% confidence interval is 2.576. therefore, the population mean is:

$\begin{array}{rcl}Z& =& \frac{\overline{X}-\mu }{\frac{\sigma }{\sqrt{n}}}\\ \pm 2.576& =& \frac{32.5-\mu }{\frac{30}{\sqrt{225}}}\\ \pm 2.576\times \frac{30}{\sqrt{225}}& =& 32.5-\mu \\ \mu & =& 32.5\mp 5.152\\ & =& 27.348, 37.65\\ & & \end{array}$

Therefore,$\mu$ = (27.348, 37.65)

b.

One have to find sample size n such that to estimate m to within 0.5 with 99% confidence. Since all the values are present, therefore substituting it we get,

$\begin{array}{rcl}Z& =& \frac{\overline{X}-\mu }{\frac{\sigma }{\sqrt{n}}}\\ \pm 2.576& =& \frac{32.5-\mu }{\frac{30}{\sqrt{n}}}\\ \pm 2.576\times \frac{30}{\sqrt{n}}& =& 32.5-\mu \\ & & \end{array}$

Now m has to be within .5,

Therefore, the n values that gets $\pm 2.576\times \frac{30}{\sqrt{n}}$to range 0.5 is 23871.

Substituting n=23,871, we get,

$\begin{array}{rcl}\pm 2.576\times \frac{30}{\sqrt{23871}}& =& 32.5-\mu \\ \pm 0.50& =& 32.5-\mu \\ \mu & =& 32.5\mp 0.50\\ & & \end{array}$

Hence the n value is 23871.

c.

To estimate population variance, a 99% confidence interval is given by:

$\left(\frac{\left(n-1\right)s^2}{{\chi }_{\frac{\alpha }{2}}^2},\frac{\left(n-1\right)s^2}{{\chi }_{1-\frac{\alpha }{2}}^2}\right)$

Substituting the values, $n=225$, $s=30.0$and$\alpha =0.01$

$$\begin{gathered} \Rightarrow \left(\frac{\left(225-1\right)\times {30}^2}{{\chi }_{\frac{0.01}{2}}^2},\frac{\left(225-1\right)\times {30}^2}{{\chi }_{1-\frac{0.01}{2}}^2}\right) \\ \Rightarrow \left(\frac{\left(224\right)\times 900}{{\chi }_{\frac{1}{200}}^2},\frac{\left(224\right)\times 900}{{\chi }_{\frac{199}{200}}^2}\right) \\ \Rightarrow \left(\frac{2016}{{\chi }_{\frac{1}{200}}^2},\frac{2016}{{\chi }_{\frac{199}{200}}^2}\right) \end{gathered}$$

By substituting chi square values by checking from the chi square table one get,

$$\begin{gathered} \Rightarrow \frac{2016}{2.822},\frac{2016}{1.7324} \\ \Rightarrow \left(714.2,1163.7\right) \end{gathered}$$

Hence the interval is (714.2, 1163.7)

d.

The phrase 99% confidence interval, implies that the probability of a parameter that we are trying to estimate falls in the bounds 99% of times, every time we conduct sampling.