Y e a r s o f e d u c a t i o n x : 17 10 9 12 14 18 H r s s p e n t w a t c h i n g t v ( y ) : 1 4 6 5 3 2 a ) N u l l H y p o t h e s i s H 0 : C o r r e l a t i o n b e t w e e n x a n d y ( ρ = ρ 0 ) A l t e r n a t i v e H y p o t h e s i s H 1 : N o c o r r e l a t i o n b e t w e e n x a n d y ( ρ ≠ ρ 0 ) b ) F o r t h e s i g n i f i c a n c e l e v e l α = 0.05 t h e c r i t i c a l v a l u e f o r t w o t a i l e d t − t e s t i s t α 2 , ν = n − 2 = t 0.05 2 , 6 − 2 = t 0.025 , 4 = 2.78 c ) T h e P e a r s o n c o r r e l a t i o n c o e f f i c i e n t r = ∑ i = 1 n ( x i y i ) − ( ∑ i = 1 n ( x i ) ) ( ∑ i = 1 n ( y i ) ) ( n ∑ i = 1 n x i 2 − ( ∑ i = 1 n x i ) 2 ) ( n ∑ i = 1 n y i 2 − ( ∑ i = 1 n y i ) 2 ) ∵ T h e n u m b e r o f s a m p l e p o i n t s a r e 6 , n = 6 ∑ i = 1 n ( x i y i ) = ∑ i = 1 6 ( x i y i ) ( 17 ) ( 1 ) + 10 ( 4 ) + 9 ( 6 ) + 12 ( 5 ) + 14 ( 3 ) + 18 ( 2 ) = 17 + 40 + 54 + 60 + 42 + 36 = 249 ⇒ ∑ i = 1 6 ( x i y i ) = 249 ∑ i = 1 6 ( x i ) = 17 + 10 + 9 + 12 + 14 + 18 = 80 ∑ i = 1 6 ( y i ) = 1 + 4 + 6 + 5 + 3 + 2 = 21 ∑ i = 1 n x i 2 = ( 17 ) 2 + ( 10 ) 2 + ( 9 ) 2 + ( 12 ) 2 + ( 14 ) 2 + ( 18 ) 2 = 1 , 134 ∑ i = 1 n y i 2 = ( 1 ) 2 + ( 4 ) 2 + ( 6 ) 2 + ( 5 ) 2 + ( 3 ) 2 + ( 2 ) 2 = 91 S u b s t i t u t i n g t h e s e v a l u e s i n c o r r e l a t i o n c o e f f i c i e n t f o r m u l a , w e g e t r = 249 − ( 80 ) ( 21 ) ( 6 ( 1134 ) − ( 80 ) 2 ) ( 6 ( 91 ) − ( 21 ) 2 ) = − 1 , 431 ( 404 ) ( 105 ) = − 0.0337 ≈ − 0.034 ∴ T h e P e a r s o n c o r r e l a t i o n c o e f f i c i e n t r = − 0.034 T h e t e s t s t a t i s t i c i s t = r ( n − 2 ) ( 1 − r 2 ) t = ( − 0.034 ) ( 6 − 2 ) ( 1 − ( − 0.034 ) 2 ) = − 0.034 ( 4 ) 0.998844 = − 0.034 ( 2.0011 ) = − 0.0680374 ⇒ t = − 0.068 S i n c e t = − 0.068 > − 2.78 , t h e r e i s n o r e a s o n t o r e j e c t t h e n u l l h y p o t h e s i s . T h e r e f o r e , t h e r e i s a c o r r e l a t i o n b e t w e e n y a n d x . d ) S i n c e , t h e c o r r e l a t i o n c o e f f i c i e n t i s n e g a t i v e , s t r e n g t h i s m o d e r a t e a n d i n t h e n e g a t i v e d i r e c t i o n . ⇒ I f y i n c r e a s e s x d e c r e a s e s . e ) T h e r e g r e s s i o n e q u a t i o n i s o b t a i n e d b y m e t h o d o f l e a s t s q u a r e s . L e t t h e l e a s t s q u a r e s r e g r e s i o n l i n e b e y = a x + b T h e n o r m a l e q u a t i o n s o f r e g r e s s i o n l i n e a r e ∑ i = 1 n ( y i ) = a ∑ i = 1 n ( x i ) + b n ∑ i = 1 n ( x i y i ) = a ∑ i = 1 n ( x i 2 ) + b ∑ i = 1 n ( x i ) S u b s t i t u t i n g , n = 6 , ∑ i = 1 6 ( y i ) = 21 , ∑ i = 1 6 ( x i ) = 80 , ∑ i = 1 6 ( x i ) 2 = 1134 ∑ i = 1 6 ( x i y i ) = 249 , w e g e t 21 = 80 a + 6 b 249 = 1134 a + 80 b S o l v i n g t h e s e t w o e q u a t i o n s w e g e t a = − 93 202 = − 0.4604 & b = 1947 202 = 9.6386 T h e r e g r e s s i o n l i n e i s y = − 0.4604 x + 9.6386 g ) S u b s t i t u t i n g x = 11 i n t h e r e g r e s s i o n l i n e w e g e t y = − 0.4604 ( 11 ) + 9.6386 = 4.5742 T h e r e f o r e , i f t h e n u m b e r o f s t u d y y e a r s i s 11 , t h e n u m b e r o f h o u r s o f w a t c h i n g T V i s y = 4.5742 h r s . S t r e n g t h o f t h e c o r r e l a t i o n i s m o d e r a t e . Years\ of \ education \ x: \quad\quad \ 17 \quad 10\quad 9\quad 12\quad 14\quad 18\\
Hrs\ spent\ watching \ tv (y):\quad 1 \quad\ 4 \quad 6 \ \quad 5 \ \quad 3\ \quad 2\\
a)\ Null \ Hypothesis \ H_{0}: Correlation \ between \ x \ and \ y\ (\rho=\rho_{0})\\
\quad Alternative \ Hypothesis\ H_{1}: No \ correlation \ between \ x \ and \ y\\ (\rho\ne\rho_{0})\\
b) For \ the \ significance \ level \ \alpha=0.05\ the \ critical \ value \ for \ two \ tailed \\ t-test\ is \
t_{\frac{\alpha}{2},\nu=n-2}=t_{\frac{0.05}{2},6-2}=t_{0.025,4}=2.78\\
c) The \ Pearson \ correlation \ coefficient \ r= \frac{\sum_{i=1}^{n}(x_iy_i)-(\sum_{i=1}^{n}(x_i))(\sum_{i=1}^{n}(y_i))}{\sqrt{(n\sum_{i=1}^{n}x_{i}^2-(\sum_{i=1}^{n}x_{i})^2)(n\sum_{i=1}^{n}y_{i}^2-(\sum_{i=1}^{n}y_{i})^2)}}\\
\because The \ number \ of \ sample \ points \ are \ 6, \ n=6\\
\sum_{i=1}^{n}(x_iy_i)=\sum_{i=1}^{6}(x_iy_i)(17)(1)+10(4)+9(6)+12(5)+14(3)+18(2)\\
\quad \quad \quad \quad=17+40+54+60+42+36\\ \quad\ \quad \ \quad=249\\
\Rightarrow \sum_{i=1}^{6}(x_iy_i)=249\\
\sum_{i=1}^{6}(x_i)=17+10+9+12+14+18=80\\
\sum_{i=1}^{6}(y_i)=1+4+6+5+3+2=21\\
\sum_{i=1}^{n}x_{i}^2=(17)^2+(10)^2+(9)^2+(12)^2+(14)^2+(18)^2=1,134\\
\sum_{i=1}^{n}y_{i}^2=(1)^2+(4)^2+(6)^2+(5)^2 +(3)^2+(2)^2=91\\
Substituting \ these\ values \ in\ correlation \ coefficient \ formula ,\ we \ get \\
r= \frac{249-(80)(21)}{\sqrt{(6(1134)-(80)^2)(6(91)-(21)^2)}}\\
\quad = \frac{-1,431}{(404)(105)}\\
\quad=-0.0337\approx-0.034\\
\therefore The \ Pearson \ correlation\ coefficient \ r= -0.034\\
The\ test \ statistic \ is \ t=r\sqrt{\frac{(n-2)}{(1-r^2)}}\\
t=(-0.034)\sqrt{\frac{(6-2)}{(1-(-0.034)^2)}}\\
=-0.034\sqrt{\frac{(4)}{0.998844}}\\
=-0.034(2.0011)\\
=-0.0680374\\
\Rightarrow t=-0.068\\
Since \ t=-0.068>-2.78, there \ is \ no \ reason \ to \ reject \ the \ null \ hypothesis.\\
Therefore, \ there \ is \ a \ correlation \ between \ y \ and x.
d) Since, \ the \ correlation \ coefficient \ is \ negative, \ strength \ is \ moderate\ and \\
in \ the \ negative \ direction.\Rightarrow If \ y \ increases \ x\ decreases.\\
e) The \ regression \ equation \ is \ obtained \ by \ method \ of \ least \ squares.\\
Let \ the \ least \ squares \ regresion \ line \ be \ y=ax+b\\
The \ normal\ equations \ of \ regression \ line \ are \\
\sum_{i=1}^{n}(y_i)=a\sum_{i=1}^{n}(x_i)+bn\\
\sum_{i=1}^{n}(x_iy_i)=a\sum_{i=1}^{n}(x_i^2)+b\sum_{i=1}^{n}(x_i)\\
Substituting, \ n=6, \ \sum_{i=1}^{6}(y_i)=21,\ \sum_{i=1}^{6}(x_i)=80,\ \sum_{i=1}^{6}(x_i)^2=1134\\ \sum_{i=1}^{6}(x_iy_i)=249, we \ get \\
21=80a+6b \\ 249=1134a+80b\\
Solving \ these \ two \ equations \ we \ get \ a=-\frac{93}{202} =-0.4604\ \& \ b=\frac{1947}{202}=9.6386\\
The\ regression \ line \ is \ y= -0.4604x+9.6386\\
g ) Substituting \ x=11 \ in \ the \ regression \ line\ we \ get \\
y= -0.4604(11)+9.6386= 4.5742\\
Therefore, if \ the \ number \ of \ study\ years \ is\ 11, \ the \ number \ of \ hours\ of watching \ TV \ is \ y=4.5742 hrs.\\
Strength \ of\ the\ correlation \ is\ moderate. Y e a rs o f e d u c a t i o n x : 17 10 9 12 14 18 Hrs s p e n t w a t c hin g t v ( y ) : 1 4 6 5 3 2 a ) N u ll Hy p o t h es i s H 0 : C orre l a t i o n b e tw ee n x an d y ( ρ = ρ 0 ) A lt er na t i v e Hy p o t h es i s H 1 : N o corre l a t i o n b e tw ee n x an d y ( ρ = ρ 0 ) b ) F or t h e s i g ni f i c an ce l e v e l α = 0.05 t h e cr i t i c a l v a l u e f or tw o t ai l e d t − t es t i s t 2 α , ν = n − 2 = t 2 0.05 , 6 − 2 = t 0.025 , 4 = 2.78 c ) T h e P e a rso n corre l a t i o n coe ff i c i e n t r = ( n ∑ i = 1 n x i 2 − ( ∑ i = 1 n x i ) 2 ) ( n ∑ i = 1 n y i 2 − ( ∑ i = 1 n y i ) 2 ) ∑ i = 1 n ( x i y i ) − ( ∑ i = 1 n ( x i )) ( ∑ i = 1 n ( y i )) ∵ T h e n u mb er o f s am pl e p o in t s a re 6 , n = 6 ∑ i = 1 n ( x i y i ) = ∑ i = 1 6 ( x i y i ) ( 17 ) ( 1 ) + 10 ( 4 ) + 9 ( 6 ) + 12 ( 5 ) + 14 ( 3 ) + 18 ( 2 ) = 17 + 40 + 54 + 60 + 42 + 36 = 249 ⇒ ∑ i = 1 6 ( x i y i ) = 249 ∑ i = 1 6 ( x i ) = 17 + 10 + 9 + 12 + 14 + 18 = 80 ∑ i = 1 6 ( y i ) = 1 + 4 + 6 + 5 + 3 + 2 = 21 ∑ i = 1 n x i 2 = ( 17 ) 2 + ( 10 ) 2 + ( 9 ) 2 + ( 12 ) 2 + ( 14 ) 2 + ( 18 ) 2 = 1 , 134 ∑ i = 1 n y i 2 = ( 1 ) 2 + ( 4 ) 2 + ( 6 ) 2 + ( 5 ) 2 + ( 3 ) 2 + ( 2 ) 2 = 91 S u b s t i t u t in g t h ese v a l u es in corre l a t i o n coe ff i c i e n t f or m u l a , w e g e t r = ( 6 ( 1134 ) − ( 80 ) 2 ) ( 6 ( 91 ) − ( 21 ) 2 ) 249 − ( 80 ) ( 21 ) = ( 404 ) ( 105 ) − 1 , 431 = − 0.0337 ≈ − 0.034 ∴ T h e P e a rso n corre l a t i o n coe ff i c i e n t r = − 0.034 T h e t es t s t a t i s t i c i s t = r ( 1 − r 2 ) ( n − 2 ) t = ( − 0.034 ) ( 1 − ( − 0.034 ) 2 ) ( 6 − 2 ) = − 0.034 0.998844 ( 4 ) = − 0.034 ( 2.0011 ) = − 0.0680374 ⇒ t = − 0.068 S in ce t = − 0.068 > − 2.78 , t h ere i s n o re a so n t o re j ec t t h e n u ll h y p o t h es i s . T h ere f ore , t h ere i s a corre l a t i o n b e tw ee n y an d x . d ) S in ce , t h e corre l a t i o n coe ff i c i e n t i s n e g a t i v e , s t re n g t h i s m o d er a t e an d in t h e n e g a t i v e d i rec t i o n . ⇒ I f y in cre a ses x d ecre a ses . e ) T h e re g ress i o n e q u a t i o n i s o b t ain e d b y m e t h o d o f l e a s t s q u a res . L e t t h e l e a s t s q u a res re g res i o n l in e b e y = a x + b T h e n or ma l e q u a t i o n s o f re g ress i o n l in e a re ∑ i = 1 n ( y i ) = a ∑ i = 1 n ( x i ) + bn ∑ i = 1 n ( x i y i ) = a ∑ i = 1 n ( x i 2 ) + b ∑ i = 1 n ( x i ) S u b s t i t u t in g , n = 6 , ∑ i = 1 6 ( y i ) = 21 , ∑ i = 1 6 ( x i ) = 80 , ∑ i = 1 6 ( x i ) 2 = 1134 ∑ i = 1 6 ( x i y i ) = 249 , w e g e t 21 = 80 a + 6 b 249 = 1134 a + 80 b S o l v in g t h ese tw o e q u a t i o n s w e g e t a = − 202 93 = − 0.4604 & b = 202 1947 = 9.6386 T h e re g ress i o n l in e i s y = − 0.4604 x + 9.6386 g ) S u b s t i t u t in g x = 11 in t h e re g ress i o n l in e w e g e t y = − 0.4604 ( 11 ) + 9.6386 = 4.5742 T h ere f ore , i f t h e n u mb er o f s t u d y ye a rs i s 11 , t h e n u mb er o f h o u rs o f w a t c hin g T V i s y = 4.5742 h rs . St re n g t h o f t h e corre l a t i o n i s m o d er a t e .
#340153 on Dec 2023
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