Chapter 11.10, Problem 9E

Solution

To find:The 95% confidence interval for the population mean or population proportion and to interpret the confidence interval in context.

The 95% confidence interval for the population mean is $67.5\le \mu \le 73.5$ .

The confidence interval means that with a probability of 95% the battery life of the batteries would lie between 67.5 and 73.5 hours (approximately)

Given information:

Given, a consumer research group tested the battery life of 36 randomly chosen batteries to establish the likely battery life for the population of same type of battery as follows:

  

Calculation:

The mean of the data is given by:

  $\begin{array}{l}\overline{x}=\frac{{\displaystyle \sum x_n}}{n}\\ \overline{x}=\frac{\begin{array}{l}63.2+62.1+79.4+66.3+61.2+61.4+68.3+86.2+71.4\\ +84.6+81.8+75.2+74.5+74.5+83.6+82.2+64.3+63.6\\ +78.4+63.6+54.1+71.6+72.4+75.6+59.3+72.7+59.6\\ +85.8+64.2+73.4+60.1+81.3+74.1+47.6+71.8+68.1\end{array}}{36}\\ \overline{x}=\frac{2537.5}{36}\\ \overline{x}=70.4861\\ \overline{x}\approx 70.5\end{array}$

To determine the variance, the following table is used:

$x$	$\overline{x}$	$x-\overline{x}$	${\left(x-\overline{x}\right)}^2$
63.2	70.5	-7.3	53.3
62.1	70.5	-8.4	70.6
79.4	70.5	8.9	79.2
66.3	70.5	-4.2	17.6
61.2	70.5	-9.3	86.5
61.4	70.5	-9.1	82.8
68.3	70.5	-2.2	4.8
86.2	70.5	15.7	246.5
71.4	70.5	0.9	0.8
84.6	70.5	14.1	198.8
81.8	70.5	11.3	127.7
75.2	70.5	4.7	22.1
74.5	70.5	4	16
74.5	70.5	4	16
83.6	70.5	13.1	171.6
82.2	70.5	11.7	136.9
64.3	70.5	-6.2	38.4
63.6	70.5	-6.9	47.6
78.4	70.5	7.9	62.4
63.6	70.5	-6.9	47.6
54.1	70.5	-16.4	269
71.6	70.5	1.1	1.2
72.4	70.5	1.9	3.6
75.6	70.5	5.1	26
59.3	70.5	-11.2	125.4
72.7	70.5	2.2	4.8
59.6	70.5	-10.9	118.8
85.8	70.5	15.3	234.1
64.2	70.5	-6.3	39.7
73.4	70.5	2.9	8.4
60.1	70.5	-10.4	108.2
81.3	70.5	10.8	116.6
74.1	70.5	3.6	13
47.6	70.5	-22.9	524.4
71.8	70.5	1.3	1.7
68.1	70.5	-2.4	5.8
		Sum:	3127.9

Hence,

  $\begin{array}{l}{\sigma }^2=\frac{{\displaystyle \sum {\left(x-\overline{x}\right)}^2}}{n}\\ {\sigma }^2=\frac{3127.9}{36}\\ {\sigma }^2=86.9\end{array}$

The standard deviation is given by:

  $\begin{array}{l}\sigma =\sqrt{{\sigma }^2}\\ \sigma =\sqrt{86.9}\\ \sigma =9.3\end{array}$

The margin of error at 95% confidence level is given by:

  $ME=1.96\cdot \frac{s}{\sqrt{n}}$

Here, $s=9.3,n=36$

Plugging the values:

  $\begin{array}{l}ME=1.96\cdot \frac{9.3}{\sqrt{36}}\\ ME=3.038\end{array}$

The confidence interval for a population mean at 95% confidence level is given by:

  $\overline{x}-ME\le \mu \le \overline{x}+ME$

Here, $\overline{x}=70.5,ME=3.038$

Plugging the values:

  $\begin{array}{l}70.5-3.038\le \mu \le 70.5+3.038\\ 67.462\le \mu \le 73.538\\ 67.5\le \mu \le 73.5\left(\text{approx}\text{.}\right)\end{array}$

The confidence interval means that the battery life of 95% of the batteries would lie between 67.5 and 73.5 hours (approximately)

Conclusion:

The 95% confidence interval for the population meanis $67.5\le \mu \le 73.5$ .

The confidence interval means that with a probability of 95% the battery lifeof the batteries would lie between 67.5 and 73.5 hours (approximately)