0. Suppose want to investigate the linear relationship between screen size and T

Question

0. Suppose want to investigate the linear relationship between screen size and The following table shows the data from a random sample of 8 devices. Battery Life 3.2 2.9 3.1 4.2 3.8 4.5 4.4 3.9 Screen Size 14.7 16.9 14.4 12.5 13.7 13.9 11.6 12.6 a) Calculate the slope and y-intercept for the regression equation. b) Calculate the SST, SSR and SSE c) Calculate the correlation coefficient and the coefficient of determination. d) using =0.05, perform a hypothesis test to determine if the correlation coefficient is greater than zero. e) using =0.05, test the significance of the population coefficient of determination.

Explanation / Answer

a.

calculation procedure for regression

mean of X = X / n = 13.7875

mean of Y = Y / n = 3.75

(Xi - Mean)^2 = 18.7691

(Yi - Mean)^2 = 2.66

(Xi-Mean)*(Yi-Mean) = -5.645

b1 = (Xi-Mean)*(Yi-Mean) / (Xi - Mean)^2

= -5.645 / 18.7691

= -0.3008

bo = Y / n - b1 * X / n

bo = 3.75 - -0.3008*13.7875 = 7.8967

value of regression equation is, Y = bo + b1 X

Y'=7.8967-0.3008* X

slope = -0.3008, y intercept =7.8967

b.

calculation procedure for correlation

sum of (x) = x = 110.3

sum of (y) = y = 30

sum of (x^2)= x^2 = 1539.53

sum of (y^2)= y^2 = 115.16

sum of (x*y)= x*y = 407.98

to caluclate value of r( x,y) = covariance ( x,y ) / sd (x) * sd (y)

covariance ( x,y ) = [ x*y - N *(x/N) * (y/N) ]/n-1

= 407.98 - [ 8 * (110.3/8) * (30/8) ]/8- 1

= -0.7056

and now to calculate r( x,y) = -0.7056/ (SQRT(1/8*407.98-(1/8*110.3)^2) ) * ( SQRT(1/8*407.98-(1/8*30)^2)

=-0.7056 / (1.5317*0.5766)

=-0.7989

value of correlation is =-0.7989

coeffcient of determination = r^2 = 0.6383

properties of correlation

1. If r = 1 Corrlation is called Perfect Positive Corrlelation

2. If r = -1 Correlation is called Perfect Negative Correlation

3. If r = 0 Correlation is called Zero Correlation

& with above we conclude that correlation ( r ) is = -0.7987< 0, perfect nagative correlation

c.

correlation ( r ) is = -0.7987< 0

coeffcient of determination = r^2 = 0.6383

d.

Given that,
value of r =-0.7987
number (n)=8
null, Ho: =0
alternate, H1: >0
level of significance, = 0.05
from standard normal table,right tailed t /2 =1.943
since our test is right-tailed
reject Ho, if to > 1.943
we use test statistic (t) = r / sqrt(1-r^2/(n-2))
to=-0.7987/(sqrt( ( 1--0.7987^2 )/(8-2) )
to =-3.251
|to | =3.251
critical value
the value of |t | at los 0.05% is 1.943
we got |to| =3.251 & | t | =1.943
make decision
hence value of | to | > | t | and here we reject Ho
ANSWERS
---------------
null, Ho: =0
alternate, H1: >=0
test statistic: -3.251
critical value: 1.943
decision: reject Ho
we have enough evidence to support the claim

Line of Regression Y on X i.e Y = bo + b1 X X Y (Xi - Mean)^2 (Yi - Mean)^2 (Xi-Mean)*(Yi-Mean) 14.7 3.2 0.8327 0.3025 -0.5019 16.9 2.9 9.6877 0.7225 -2.6456 14.4 3.1 0.3752 0.4225 -0.3981 12.5 4.2 1.6577 0.2025 -0.5794 13.7 3.8 0.0077 0.0025 -0.0044 13.9 4.5 0.0127 0.5625 0.0844 11.6 4.4 4.7852 0.4225 -1.4219 12.6 3.9 1.4102 0.0225 -0.1781

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