Chapter A1, Problem A1.12QAP

Interpretation Introduction

Interpretation:

The table for sets of the replicate measurements is given below-

A	B	C	D	E	F
3.5	70.24	0.812	2.7	70.65	0.514
3.1	70.22	0.792	3.0	70.63	0.503
3.1	70.10	0.794	2.6	70.64	0.486
3.3		0.900	2.8	70.21	0.497
2.5			3.2		0.472

The value of mean and standard deviation is to be calculated for the above data sets. The 95% confidence interval for each data is to be determined and the value of interval mean is also to be determined.

Concept introduction:

The formula for the mean is given as-

$mean(\overline{x})=\frac{A_1+A+\mathrm{....................}+A_n}{N}=\frac{\sum x_i}{N}$ .................. (1)

The formula for the standard deviation is given as-

$standarddeviation=\frac{\sum {(x-\overline{x})}^2}{N-1}$ ..................….. (2)

Answer to Problem A1.12QAP

For set A −

Mean= 3.1

Standard deviation = 0.37

95% degree interval = $\pm 0.46$

For set B −

Mean= 70.19

Standard deviation = 0.08

95% degree interval = $\pm 0.20$

For set C −

Mean= 0.82

Standard deviation = 0.05

95% degree interval = $\pm 0.80$

For set D −

Mean= 2.86

Standard deviation = 0.25

95% degree interval = $\pm 0.31$

For set E −

Mean= 70.53

Standard deviation = 0.22

95% degree interval = $\pm 0.35$

For set F −

Mean= 0.494

Standard deviation = 0.016

95% degree interval = $\pm 0.020$

Explanation of Solution

The formula that will be used-

$mean(\overline{x})=\frac{A_1+A+\mathrm{....................}+A_n}{N}=\frac{\sum x_i}{N}$

$standarddeviation=\frac{\sum {(x-\overline{x})}^2}{N-1}$

$CI,\mu =\overline{x}\pm \frac{ts}{\sqrt{N}}$

Therefore, mean value of set A −

$mean(\overline{x})=\frac{3.5+3.1+3.1+3.3+2.5}{5}=3.1$

The standard deviation for the set A is calculated as-

Set A	$x_i$	$x_i{}^2$
1	3.5	12.25
2	3.1	9.61
3	3.1	9.61
4	3.3	10.89
5	2.5	6.25
	$\sum x_i$ = 15.5	${(\sum x_i)}^2$ = 48.61

From above table-

$\frac{{(48.61)}^2}{5}=\frac{240.25}{5}=48.05$

The standard deviation for the set A is calculated as-

$s=\sqrt{\frac{48.61-48.05}{5-1}}$

$s=\sqrt{\frac{0.56}{4}}=0.37$

Similarly for set B-

Therefore mean value of set B −

$mean(\overline{x})=\frac{70.24+70.22+70.10}{3}=70.19$

The standard deviation for the set A is calculated as-

Set A	$x_i$	$x_i{}^2$
1	70.24	4933.6576
2	70.22	4930.8484
3	70.10	4914.01
	$\sum x_i$ = 210.56	${(\sum x_i)}^2$ = 14778.516

From above table-

$\frac{{(210.56)}^2}{3}=\frac{44335.5136}{3}=14778.504$

The standard deviation for the set A is calculated as-

$s=\sqrt{\frac{14778.516-14778.504}{3-1}}$

$s=\sqrt{\frac{0.012}{2}}=0.077\approx 0.08$

Similarly for set C-

Therefore mean value of set C −

$mean(\overline{x})=\frac{0.812+0.792+0.794+0.900}{4}=0.82$

The standard deviation for the set C is calculated as-

Set C	$x_i$	$x_i{}^2$
1	0.812	0.659344
2	0.792	0.627264
3	0.794	0.630436
4	0.900	0.81
	$\sum x_i$ = 3.298	${(\sum x_i)}^2$ = 2.727044

From above table-

$\frac{{(3.298)}^2}{4}=\frac{10.876804}{4}=2.719201$

The standard deviation for the set C is calculated as-

$s=\sqrt{\frac{2.727044-2.719201}{4-1}}$

$s=\sqrt{\frac{0.007843}{3}}=0.05$

Similarly for set D-

Therefore mean value of set D −

$mean(\overline{x})=\frac{2.7+3.0+2.6+2.8+3.2}{5}=2.86$

The standard deviation for the set D is calculated as-

Set D	$x_i$	$x_i{}^2$
1	2.7	7.29
2	3.0	9.00
3	2.6	6.79
4	2.8	7.64
5	3.2	10.24
	$\sum x_i$ = 14.3	${(\sum x_i)}^2$ = 41.16

From above table-

$\frac{{(14.3)}^2}{5}=\frac{204.49}{5}=40.898$

The standard deviation for the set A is calculated as-

$s=\sqrt{\frac{41.16-40.898}{5-1}}$

$s=\sqrt{\frac{0.262}{4}}=0.25$

Similarly for set E-

Therefore mean value of set E −

$mean(\overline{x})=\frac{70.65+70.63+70.64+70.21}{4}=70.53$

The standard deviation for the set E is calculated as-

Set E	$x_i$	$x_i{}^2$
1	70.65	4991.4225
2	70.63	4988.5969
3	70.64	4990.0096
4	70.21	4929.4441
	$\sum x_i$ = 282.13	${(\sum x_i)}^2$ = 19899.4731

From above table-

$\frac{{(282.13)}^2}{4}=\frac{79597.3369}{4}=19899.3342$

The standard deviation for the set A is calculated as-

$s=\sqrt{\frac{19899.4731-19899.3342}{4-1}}$

$s=\sqrt{\frac{0.1389}{3}}=0.215\approx 0.22$

Similarly for set F-

Therefore mean value of set F −

$mean(\overline{x})=\frac{0.514+0.503+0.486+0.497+0.472}{5}=0.494$

The standard deviation for the set F is calculated as-

Set A	$x_i$	$x_i{}^2$
1	0.514	0.264196
2	0.503	0.253009
3	0.486	0.236196
4	0.497	0.247009
5	0.472	0.222784
	$\sum x_i$ = 210.56	${(\sum x_i)}^2$ = 1.223194

From above table-

$\frac{{(210.56)}^2}{3}=\frac{44335.5136}{3}=14778.504$

The standard deviation for the set F is calculated as-

$s=\sqrt{\frac{1.223194-1.222156}{5-1}}$

$s=\sqrt{\frac{0.001038}{4}}=0.016$

The 95% confidence interval for each data is calculated as-

$CI,\mu =\overline{x}\pm \frac{ts}{\sqrt{N}}$ ................... Equ (3)

Where,

N = degree of freedom

$\mu$ = confidence interval

$\overline{x}$ = mean

s = standard deviation

The value of t-

Degree of freedom	95%
1	12.7
2	4.30
3	3.18
4	2.78
5	2.57

For set A-

Given that-

N = 5

$\overline{x}$ = 3.1

s = 0.37

t = 2.78

Put the above values in Equ (3)

$CI,\mu =3.1\pm \frac{2.78\times 0.37}{\sqrt{5}}$

$CI,\mu =3.1\pm 0.46$

$confidenceinterval(A)=\pm 0.46$

For set B-

Given that-

N = 3

$\overline{x}$ = 70.19

s = 0.08

t = 4.30

Put the above values in Equ (3)

$CI,\mu =70.19\pm \frac{4.30\times 0.08}{\sqrt{3}}$

$CI,\mu =70.19\pm 0.19\approx 70.19\pm 0.20$

$confidenceinterval(B)=\pm 0.20$

For set C-

Given that-

N = 4

$\overline{x}$ = 0.82

s = 0.05

t = 3.18

Put the above values in Equ (3)

$CI,\mu =0.82\pm \frac{3.18\times 0.05}{\sqrt{4}}$

$CI,\mu =0.82\pm 0.795\approx 0.82\pm 0.80$

$confidenceinterval(C)=\pm 0.80$

For set D-

Given that-

N = 5

$\overline{x}$ = 2.86

s = 0.25

t = 2.78

Put the above values in Equ (3)

$CI,\mu =2.86\pm \frac{2.78\times 0.25}{\sqrt{5}}$

$CI,\mu =2.86\pm 0.31$

$confidenceinterval(D)=\pm 0.31$

For set E-

Given that-

N = 4

$\overline{x}$ = 70.53

s = 0.22

t = 3.18

Put the above values in Equ (3)

$CI,\mu =70.53\pm \frac{3.18\times 0.22}{\sqrt{4}}$

$CI,\mu =70.53\pm 0.35$

$confidenceinterval(E)=\pm 0.35$

For set F-

Given that-

N = 5

$\overline{x}$ = 0.494

s = 0.016

t = 2.78

Put the above values in Equ (3)

$CI,\mu =0.494\pm \frac{2.78\times 0.016}{\sqrt{5}}$

$CI,\mu =0.494\pm 0.020$

$confidenceinterval(F)=\pm 0.020$

Conclusion

Thus, this can be concluded that the results for each set are obtained by using the mean standard deviation and 95% confidence interval formula. Therefore,

For set A −

Mean= 3.1

Standard deviation = 0.37

95% degree interval = $\pm 0.46$

For set B −

Mean= 70.19

Standard deviation = 0.08

95% degree interval = $\pm 0.20$

For set C −

Mean= 0.82

Standard deviation = 0.05

95% degree interval = $\pm 0.80$

For set D −

Mean= 2.86

Standard deviation = 0.25

95% degree interval = $\pm 0.31$

For set E −

Mean= 70.53

Standard deviation = 0.22

95% degree interval = $\pm 0.35$

For set F −

Mean= 0.494

Standard deviation = 0.016

95% degree interval = $\pm 0.020$