Question

Question 2 (a The following table shows the solid reduction and oxygen demand reduction of an experiment. d Reduction, 1|Oxygen Demand Reduction, y(%) 20 16 28 35 27 10 (i) Calculate Σ, Υ ,Σ and (2 marks) (ii) Find the inear regression ne of oxygen demand solid reduction on (6 marks) iii) Estimate the oxygen demand reduction if the solid reduction is 40%. Comment on the reliability of your estimation. (3 marks) iii Find the coefficient of correlation. Hence, comment on the strength of correlation between the two variables. marks) (v) Calculate and comment the coefficient of determination (2 marks) (b)A random sample of 5 cotton fiber was taken and their absorbency values are shown below. 19.7, 21.5, 17.3, 18.9, 20.6 (i) Find the mean and standard deviation for the sample given (5 marks) (ii) Construct a 90% confidence interval for the true mean absorbency values for cotton fiber. (4 marks) (iii) Test at 2.5% significance level whether the population mean absorbency value for cotton fiber is more than 19 (6 marks)

Answer 1

Solution : ( 2 )

( i )

x	y	x * y	x * x
16	20	320	256
18	16	288	324
27	28	756	729
30	35	1050	900
33	34	1122	1089
36	37	1332	1296

$\sum{x} = 160 ~,~ \sum{y} = 170 ~,~ \sum{x * y} = 4868 ~,~ \sum{x^2} = 4594$

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Solution : ( 2 )

( ii )

x	y	x * y	x * x
16	20	320	256
18	16	288	324
27	28	756	729
30	35	1050	900
33	34	1122	1089
36	37	1332	1296

$\sum{x} = 160 ~,~ \sum{y} = 170 ~,~ \sum{x * y} = 4868 ~,~ \sum{x^2} = 4594$

ΣΥ*ΣΧ2 ㅡ Σ2. * Σ.ry 170 * 4594-160 * 4868 6 * 4594 _ (160)2 Cl

780980 -778880 27564-25600

2100 1964

$=1.069$

and

=rị * Σ.ry _ ΣΖ. * ΣΥ=6*4868-160 * 170

29208 - 27200 27564-25600

2008 1964

1.022

$\mathrm{Substitute\:}a\mathrm{\:and\:}b\mathrm{\:in\:regression\:equation\:formula}$

$\begin{aligned} y~&=~a ~+~ b * x \\y~&=~1.069 ~+~ 1.022 *x\end{aligned}$

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Solution : ( 2 )

( iii )

$y=1.069+1.022 *\left ( \frac{40}{100} \right )$

$=1.069+1.022 *\left ( \frac{2}{5} \right )$

$=1.069+\left ( \frac{1.022 *2}{5} \right )$

$=1.069+\left ( \frac{2.044}{5} \right )$

$=1.069+0.4088$

$=1.4778$

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Solution : ( 2 )

( iv )

x	y	x * y	x * x	y * y
16	20	320	256	400
18	16	288	324	256
27	28	756	729	784
30	35	1050	900	1225
33	34	1122	1089	1156
36	37	1332	1296	1369

$\mathrm{Here,}$

$\sum{X}=160 ~,$

$~ \sum{Y}=170 ~,$

$~ \sum{X * Y}=4868 ~,$

$~ \sum{X^2}=4594 ~$

and

$~ \sum{Y^2}=5190$

$\mathrm{Now,}$

$r=\frac{n*\sum{xy} - \sum{x}*\sum{y}}{\sqrt{\left[n \sum{x^2}-\left(\sum{x}\right)^2\right] * \left[n \sum{y^2}-\left(\sum{y}\right)^2\right]}}$

$=\frac{ 6*4868 - 160 * 170 }{\sqrt{\left[ 6 * 4594 - 160^2 \right] *\left[ 6 * 5190 - (170)^2 \right] }}$

$=\frac{ 29208-27200 }{\sqrt{\left(27564-25600\right)\left(31140-28900\right)}}$

$=\frac{ 29208-27200 }{\sqrt{1964*2240}}$

$=\frac{2008}{\sqrt{4399360}}$

$=\frac{2008}{2097.46514}$

$=0.95734$

$\mathrm{Hence,\quad\:the\:coefficient\:of\:correlation\: \:is\:\:}\bold{0.95734}$

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Solution : ( 2 )

( v )

$\mathrm{Coefficient\:of\:determination\: \:}r^{2}=(0.95734)^{2}=0.9164998756$

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Solution : ( 3 )

( i )

$\bold{Mean\: }(\overline{x} )\:\::$

$\mathrm{Mean\: }(\overline{x} )= \mathrm{\frac{Sum ~ of ~ terms}{Number ~ of ~ terms}=\frac{19.7 + 21.5 + 17.3 + 18.9 + 20.6 }{5}}$

$\overline{x}=\frac{ 98 }{ 5 }= 19.6$

and

$\bold{Standard\:deviation\:\:}(\sigma )\::$

data	data - mean	(data - mean)2
19.7	0.1	0.01
21.5	1.9	3.61
17.3	-2.3	5.29
18.9	-0.7	0.49
20.6	1	1

$\mathrm{Here,}$

$\sum{\left(x_i - \overline{x}\right)^2} = 10.4$

= 1.6125 3.2249 V10.4 /10.4 2 Σ(zi-T)" /10.4 5-1

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Solution : ( 3 )

( ii )

$\mathrm{90\%\:confidence\:interval\:for\:the\:mean\:is\:}$

$=\overline{x}\pm z*\left ( \frac{\sigma }{\sqrt{n}} \right )$

$=\left (\overline{x}- z*\left ( \frac{\sigma }{\sqrt{n}} \right ),\quad\overline{x}+ z*\left ( \frac{\sigma }{\sqrt{n}} \right ) \right )$

$=\left (19.6- 1.645 *\left ( \frac{1.6125 }{\sqrt{5}} \right ),\quad\:19.6+ 1.645 *\left ( \frac{1.6125 }{\sqrt{5}} \right ) \right )$

$=\left (19.6- \left ( \frac{1.645 *1.6125 }{\sqrt{5}} \right ),\quad\:19.6+ \left ( \frac{1.645 *1.6125 }{\sqrt{5}} \right ) \right )$

-(19.-2 519620702 2.6525625 2.6525625

$=\left (19.6- \frac{2.6525625}{2.23607},\quad\:19.6+ \frac{2.6525625}{2.23607} \right )$

$=\left (19.6-1.18626,\quad\:19.6+1.18626 \right )$

$=\left (18.41374,\quad\:20.78626 \right )$