Chapter 8.3, Problem 14E

To determine

To find:

A 99% confidence interval for the mean

Answer to Problem 14E

Solution:

A 99% confidence interval for the mean face amount of an individual life insurance policy is in the interval, $\left(165852.02,177665.22\right)$.

Explanation of Solution

Approach:

Given a sample with standard deviation $s$, while estimating a population parameter, it is better to consider a range of values using samples of the same size (say $n$), also known as an interval estimate.

The true population parameters probability lies in this interval range which is known as the confidence level $\left(c=1-\alpha \right)$.

At a certain level of confidence, the interval in which the maximum error can be observed is known as the confidence interval.

Estimating parameters of a large population or a population following normal distribution is done using the Student’s $t$-distribution to calculate the margin of error for mean of the population when its standard deviation is not known.

The $t$-distribution is described by only one measure that is the degrees of freedom, with the area under the tails decreasing with higher degrees of freedom, eventually matching the standard normal distribution when degrees of freedom is infinitely many.

The degrees of freedom is the attribute used to obtain the appropriate curve for the $t$-distribution.

The value of $t$ such that there is an area of $\raisebox{1ex}{$\alpha $}\!\left/ \!\raisebox{-1ex}{$2$}\right.$ to the left of $-t$ and an area of $\raisebox{1ex}{$\alpha $}\!\left/ \!\raisebox{-1ex}{$2$}\right.$ to the right of $t$ is denoted by $t_{\raisebox{1ex}{$\alpha $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$ as depicted above.

Hence this is a case of two-tailed t-distribution.

The maximum distance from point estimate that the confidence interval covers is margin of error and is given by:

$E=\frac{t_{\raisebox{1ex}{$\alpha $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\times s}{\sqrt{n}}$,

Given the sample mean $\left(\overline{x}\right)$, the population mean confidence interval is calculated as:

$\left(\overline{x}-E,\overline{x}+E\right)$.

Calculation:

The sample means is given by:

$\overline{x}=\frac{{\displaystyle \sum _{i=1}^n{X}_i}}{n_1}$

Substituting the values of life insurance policies in the formula above

$\begin{array}{rll}\overline{x}& =& \frac{\left(\begin{array}{l}150000+150000+1500000+1500000+1500000+150000+151000+152000+152000+153000+\\ 153000+154000+155000+158000+159000+159000+160000+160000+160000+162000+\\ 163000+163000+163000+163000+165000+165000+168000+168000+168000+171000+\\ 171000+172000+172000+172000+173000+174000+175000+175000+175000+176000+182000+\\ 182000+183000+185000+185000+190000+190000+195000+195000+195000+196000+\\ 200000+200000+200000+200000+202000+202000\end{array}\right)}{58}\\ =\frac{9962000}{58}\\ \overline{x}=171758.62\end{array}$

${\mathit{x}}_{\mathit{i}}$	$\stackrel{-}{\mathit{x}}$	${\mathit{x}}_{\mathit{i}}-\mathit{x}$	${({\mathit{x}}_{\mathit{i}}-\mathit{x})}^2$
150, 000	171758.62	-21, 758.62	473437544.3
150, 000	171758.62	-21, 758.62	473437544.3
150, 000	171758.62	-21, 758.62	473437544.3
150, 000	171758.62	-21, 758.62	473437544.3
150, 000	171758.62	-21, 758.62	473437544.3
150, 000	171758.62	-21, 758.62	473437544.3
151, 000	171758.62	-20, 758.62	430920304.3
152, 000	171758.62	-19, 758.62	390403064.3
152, 000	171758.62	-19, 758.62	390403064.3
153, 000	171758.62	-18, 758.62	351885824.3
153, 000	171758.62	-18, 758.62	351885824.3
154, 000	171758.62	-17, 758.62	315368584.3
155, 000	171758.62	-16, 758.62	280851344.3
158, 000	171758.62	-13, 758.62	189299624.3
158, 000	171758.62	-12, 758.62	162782384.3
158, 000	171758.62	-12, 758.62	162782384.3
160, 000	171758.62	-11, 758.62	138265144.3
160, 000	171758.62	-11, 758.62	138265144.3
160, 000	171758.62	-11, 758.62	138265144.3
162, 000	171758.62	-9, 758.62	95230664.3
163, 000	171758.62	-8, 758.62	76713424.3
163, 000	171758.62	-8, 758.62	76713424.3
163, 000	171758.62	-8, 758.62	76713424.3
165, 000	171758.62	-6, 758.62	45678944.3
165, 000	171758.62	-6, 758.62	45678944.3
168, 000	171758.62	-3, 758.62	14127224.3
168, 000	171758.62	-3, 758.62	14127224.3
168, 000	171758.62	-3, 758.62	14127224.3
171, 000	171758.62	-758.62	575504.3044
171, 000	171758.62	-758.62	575504.3044
172, 000	171758.62	241.38	58264.3044
172, 000	171758.62	241.38	58264.3044
172, 000	171758.62	241.38	58264.3044
173, 000	171758.62	1, 241.38	1541024.304
174, 000	171758.62	2, 241.38	5023784.304
175, 000	171758.62	3, 241.38	10506544.3
175, 000	171758.62	3, 241.38	10506544.3
175, 000	171758.62	3, 241.38	10506544.3
176, 000	171758.62	4, 241.38	17989304.3
182, 000	171758.62	10, 241.38	104885864.3
182, 000	171758.62	10, 241.38	104885864.3
183, 000	171758.62	11, 241.38	126368624.3
185, 000	171758.62	13, 241.38	175334144.3
185, 000	171758.62	13, 241.38	175334144.3
190, 000	171758.62	18, 241.38	332747944.3
190, 000	171758.62	18, 241.38	332747944.3
195, 000	171758.62	23, 241.38	540161744.3
195, 000	171758.62	23, 241.38	540161744.3
195, 000	171758.62	23, 241.38	540161744.3
196, 000	171758.62	24, 241.38	587644504.3
200, 000	171758.62	28, 241.38	797575544.3
200, 000	171758.62	28, 241.38	797575544.3
200, 000	171758.62	28, 241.38	797575544.3
200, 000	171758.62	28, 241.38	797575544.3
200, 000	171758.62	28, 241.38	797575544.3
202, 000	171758.62	30, 241.38	914541064.3
202, 000	171758.62	30, 241.38	914541064.3
163, 000	171758.62	-8, 758.62	76713424.3

Form the above table we get,

${\displaystyle \sum _{i=1}^n{(x_i-\overline{x})}^2}=16252620690$

Now the standard division is,

$\begin{array}{l}S=\sqrt{\frac{{\displaystyle \sum _{i=1}^n{(x_i-\overline{x})}^2}}{n-1}}\\ S=\sqrt{\frac{16252620690}{57}}\\ S=16885.9\end{array}$

Level of confidence is 99%

$c=0.99$

Where $\frac{\alpha }{2}$ is calculated as:

$\begin{array}{l}\frac{\alpha }{2}=\frac{\left(1-c\right)}{2}\\ =\frac{\left(1-0.99\right)}{2}\\ =0.005\end{array}$

Critical t value at $\frac{\alpha }{2}=0.005$ using t table and degree of freedom as $\left(58-1=57\right)$ is 2.664.

Now substitute these values in margin of error.

Then, the margin of error is calculated as:

$\begin{array}{rll}E& =& \frac{t_{\frac{\alpha }{2}}\times s}{\sqrt{n}}\\ & =& \frac{2.664\times 16885.9}{\sqrt{58}}\\ & =& 5906.69\end{array}$

Then, interval is:

$\begin{array}{rll}\left(\overline{x}-E,\overline{x}+E\right)& =& \left(171758.62-5906.69,171758.62+5906.69\right)\\ \approx \left(165852.02,177665.22\right)\end{array}$

Conclusion:

A 95% confidence interval for the mean fastball pitching speed of all high school is in the interval $\left(165852.02,177665.22\right)$.