Chapter 1.3, Problem 17E

To determine

Find the approximated value of s for the data sets in Exercise 1.2, 1.3, and 1.4.

Compare the results in each case to the actual calculated value of s.

Answer to Problem 17E

The approximated value of s for Exercise 1.2 is 7.35.

The approximated value of s for Exercise 1.3 is 3.04.

The approximated value of s for Exercise 1.4 is 2.3175.

Explanation of Solution

Calculation:

Mean:

The mean of a sample of measurements $y_1,y_2,\mathrm{...},y_n$ is defined as,

$\overline{y}=\frac{1}{n}{\displaystyle \sum _{i=1}^n{y}_i}$.

Variance:

The variance of a sample of measurements $y_1,y_2,\mathrm{...},y_n$ is defined as the sum of the square of the difference between the measurements and the mean, divided by $n-1$.

That is,

$s^2=\frac{1}{n-1}{\displaystyle \sum _{i=1}^n{\left(y_i-\overline{y}\right)}^2}$.

Standard deviation:

The standard deviation of a sample is defined as the positive square root of the variance. That is, $s=\sqrt{s^2}$.

The approximated value of the Standard deviation is obtained by one-fourth of the range.

For Exercise 1.2:

In the given problem, $n=45$.

Consider the following table for necessary calculations.

$y_i$	$y_i^2$
8.9	79.21
12.4	153.76
8.6	73.96
11.3	127.69
9.2	84.64
8.8	77.44
35.1	1232.01
6.2	38.44
7	49
7.1	50.41
11.8	139.24
10.7	114.49
7.6	57.76
9.1	82.81
9.2	84.64
8.2	67.24
9	81
8.7	75.69
9.1	82.81
10.9	118.81
10.3	106.09
9.6	92.16
7.8	60.84
11.5	132.25
9.3	86.49
7.9	62.41
8.8	77.44
8.8	77.44
12.7	161.29
8.4	70.56
7.8	60.84
5.7	32.49
10.5	110.25
10.5	110.25
9.6	92.16
8.9	79.21
10.2	104.04
10.3	106.09
7.7	59.29
10.6	112.36
8.3	68.89
8.8	77.44
9.5	90.25
8.8	77.44
9.4	88.36
${\displaystyle \sum _{i=1}^{45}{y}_i}=440.6$	${\displaystyle \sum _{i=1}^{45}{y}_i^2}=5,067.38$

Hence, ${\left({\displaystyle \sum _{i=1}^{45}{y}_i}\right)}^2=194,128.4$.

Substitute $n=45$, and ${\displaystyle \sum _{i=1}^{45}{y}_i}=440.6$ in the formula of mean.

That is,

$\begin{array}{c}\overline{y}=\frac{440.6}{45}\\ \approx 9.79\end{array}$.

Thus, the mean value is 9.79.

Substitute $n=45$, ${\left({\displaystyle \sum _{i=1}^{45}{y}_i}\right)}^2=194,128.4$, and ${\displaystyle \sum _{i=1}^{45}{y}_i^2}=5,067.38$ in the $s^2$ formula.

That is,

$\begin{array}{c}{s}^2=\frac{1}{45-1}\left[5,067.38-\frac{1}{45}\left(194,128.4\right)\right]\\ =\frac{1}{44}\left[5,067.38-4,313.964\right]\\ \approx 17.12\end{array}$

Thus, the value of s is calculated below:

$\begin{array}{c}s=\sqrt{s^2}\\ =\sqrt{17.12}\\ \approx 4.14\end{array}$

Thus, the value of s is 4.14.

The lowest and highest values of the give data are 5.7 and 35.1, respectively.

Thus, the approximated value of s is,

$\begin{array}{c}s=\frac{35.1-5.7}{4}\\ =7.35\end{array}$

Thus, it can be said that the estimated value of s is not close to the actual value for is due to the outlier.

For Exercise 1.3:

In the given problem, $n=25$.

Consider the following table for necessary calculations.

$y_i$	$y_i^2$
0.74	0.5476
6.47	41.8609
1.9	3.61
2.69	7.2361
0.75	0.5625
0.32	0.1024
9.99	99.8001
1.77	3.1329
2.41	5.8081
1.96	3.8416
1.66	2.7556
0.7	0.49
2.42	5.8564
0.54	0.2916
3.36	11.2896
3.59	12.8881
0.37	0.1369
1.09	1.1881
8.32	69.2224
4.06	16.4836
4.55	20.7025
0.76	0.5776
2.03	4.1209
5.7	32.49
12.48	155.7504
${\displaystyle \sum _{i=1}^{25}{y}_i}=80.63$	${\displaystyle \sum _{i=1}^{25}{y}_i^2}=500.746$

Hence, ${\left({\displaystyle \sum _{i=1}^{25}{y}_i}\right)}^2=6,501.197$.

Substitute $n=25$, and ${\displaystyle \sum _{i=1}^{25}{y}_i}=80.63$ in the formula of mean.

That is,

$\begin{array}{c}\overline{y}=\frac{80.63}{25}\\ \approx 3.23\end{array}$.

Thus, the mean value is 3.23.

Substitute $n=25$, ${\left({\displaystyle \sum _{i=1}^{25}{y}_i}\right)}^2=6,501.197$, and ${\displaystyle \sum _{i=1}^{25}{y}_i^2}=500.746$ in the $s^2$ formula.

That is,

$\begin{array}{c}{s}^2=\frac{1}{25-1}\left[500.746-\frac{1}{25}\left(6,501.197\right)\right]\\ =\frac{1}{24}\left[500.746-260.0479\right]\\ \approx 10.03\end{array}$

Thus, the value of s is calculated below:

$\begin{array}{c}s=\sqrt{s^2}\\ =\sqrt{10.13}\\ \approx 3.17\end{array}$

Thus, the value of s is 3.17.

The lowest and highest values of the give data are 0.32 and 12.48, respectively.

Thus, the approximated value of s is,

$\begin{array}{c}s=\frac{12.48-0.38}{4}\\ =3.04\end{array}$

Thus, it can be said that the estimated value of s is close to the actual value.

For Exercise 1.4:

In the given problem, $n=40$.

Consider the following table for necessary calculations.

$y_i$	$y_i^2$
11.88	141.13
7.99	63.84
7.15	51.12
7.13	50.84
6.27	39.31
6.07	36.84
5.98	35.76
5.91	34.93
5.49	30.14
5.26	27.67
5.07	25.70
4.94	24.40
4.81	23.14
4.79	22.94
4.55	20.70
4.43	19.62
4.4	19.36
4.05	16.40
3.94	15.52
3.93	15.44
3.78	14.29
3.69	13.62
3.62	13.10
3.48	12.11
3.44	11.83
3.36	11.29
3.26	10.63
3.2	10.24
3.11	9.67
3.03	9.18
2.99	8.94
2.89	8.35
2.88	8.29
2.74	7.51
2.74	7.51
2.69	7.24
2.68	7.18
2.63	6.92
2.62	6.86
2.61	6.81
${\displaystyle \sum _{i=1}^{40}{y}_i}=175.48$	${\displaystyle \sum _{i=1}^{40}{y}_i^2}=906.41$

Hence, ${\left({\displaystyle \sum _{i=1}^{40}{y}_i}\right)}^2=30,793.23$.

Substitute $n=40$, and ${\displaystyle \sum _{i=1}^{40}{y}_i}=175.48$ in the formula of mean.

That is,

$\begin{array}{c}\overline{y}=\frac{175.48}{40}\\ \approx 4.39\end{array}$.

Thus, the mean value is 4.39.

Substitute $n=40$, ${\left({\displaystyle \sum _{i=1}^{40}{y}_i}\right)}^2=30,793.23$, and ${\displaystyle \sum _{i=1}^{40}{y}_i^2}=906.41$ in the $s^2$ formula.

That is,

$\begin{array}{c}{s}^2=\frac{1}{40-1}\left[906.41-\frac{1}{40}\left(30,793.23\right)\right]\\ =\frac{1}{39}\left[906.41-769.83\right]\\ \approx 3.5021\end{array}$

Thus, the value of s is calculated below:

$\begin{array}{c}s=\sqrt{s^2}\\ =\sqrt{3.5021}\\ \approx 1.87\end{array}$

Thus, the value of s is 1.87.

The lowest and highest values of the give data are 11.88 and 2.61, respectively.

Thus, the approximated value of s is,

$\begin{array}{c}s=\frac{11.88-2.61}{4}\\ =2.3175\end{array}$

Thus, it can be said that the estimated value of s is somewhat close to the actual value.