Chapter 8.3, Problem 69E

(a)

To determine

To calculate: The population mean for the confidence interval of 95%.

Answer to Problem 69E

Thepopulation mean is from2.8379% to 6.2747%.

Explanation of Solution

Given information:

Sample size $n=21$

Confidence interval $c=95\%$

Table for sample is:

Tire	Weight	Groove
1	45.9	35.7
2	41.9	39.2
3	37.5	31.1
4	33.4	28.1
5	31.0	24.0
6	30.5	28.7
7	30.9	25.9
8	31.9	23.3
9	30.4	23.1
10	27.3	23.7
11	20.4	20.9
12	24.5	16.1
13	20.9	19.9
14	18.9	15.2
15	13.7	11.5
16	11.4	11.2

Formula used:

Sample mean $\overline{x}=\frac{\text{sum}\text{of}\text{the}\text{samples}}{\text{total}\text{number}\text{of}\text{sample}}$

Standard deviation $s_x=\sqrt{\frac{{\displaystyle \sum {\left(x_i-\overline{x}\right)}^2}}{n-1}}$

Degree of freedom $df=n-1$

Margin of error $E=t^{*}\frac{s_x}{\sqrt{n}}$

Calculation:

Find the mean, use the formulaSample mean $\overline{x}=\frac{\text{sum}\text{of}\text{the}\text{samples}}{\text{total}\text{number}\text{of}\text{sample}}$ .

  $\begin{array}{c}\overline{x}=\frac{\text{sum}\text{of}\text{the}\text{samples}}{\text{total}\text{number}\text{of}\text{sample}}\\ =\frac{10.2+2.7+\cdots +2.2+0.2}{16}\\ =4.5563\end{array}$

Find the standard deviation, use the formulaStandard deviation $s_x=\sqrt{\frac{{\displaystyle \sum {\left(x_i-\overline{x}\right)}^2}}{n-1}}$.

  $\begin{array}{c}{s}_x=\sqrt{\frac{{\displaystyle \sum {\left(x_i-\overline{x}\right)}^2}}{n-1}}\\ =\sqrt{\frac{{\left(10.2-4.5563\right)}^2+\cdots +{\left(0.2-4.5563\right)}^2}{16-1}}\\ =\sqrt{10.4040}\\ =3.2255\end{array}$

Find the degree of freedom, use the formula $df=n-1$ .

  $\begin{array}{c}df=n-1\\ =16-1\\ =15\end{array}$

The confidence level is 95%.

Convert 95% into decimal.

  $\frac{95}{100}=0.95$

Find the value of the column.

  $\begin{array}{c}\frac{1-c}{2}=\frac{1-0.95}{2}\\ =0.025\end{array}$

From table B find critical value $t^{*}$ , use the row $df$ and column.

Thus, critical value $t^{*}\text{is}2.131$ .

Find the margin of error, use the formula $E=t^{*}\frac{s_x}{\sqrt{n}}$ .

  $\begin{array}{c}E=t^{*}\frac{s_x}{\sqrt{n}}\\ =2.131\times \frac{3.2255}{\sqrt{21}}\\ =1.7184\end{array}$

Now, find the population mean.

  $\begin{array}{c}\overline{x}-E<\mu <\overline{x}+E\\ 4.5563-1.7184<\mu <4.5563+1.7184\\ 2.8379<\mu <6.2747\end{array}$

Hence, the required population mean is from 2.8379% to 6.2747%.

(b)

To determine

Explain whether or not theconfidence interval gives convincing evidence of a difference in the two method of estimating tire wear.

Answer to Problem 69E

The confidence interval is sufficient to give convincing evidence of a difference in the two method of estimating tire wear.

Explanation of Solution

From 69(a)confidence interval contain the population mean from2.8379% to 6.2747%. Since the confidence interval lies above 0.

Hence, the confidence interval is sufficient to give convincing evidence of a difference in the two method of estimating tire wear.