Question

1FIND THE CORRELATION COEFFICIENT OF THE FOLLOWING DATA AND COMMENT ON ITS DIRECTION AND STRENGTH.HIGH TEMP(X)= 55 58 64 68 70 75 80 85 CANS SOLD(Y)= 340 335 410 450 460 610 735 780

2. IF THE EQUATION ABOVE IS A GOOD FIT PREDICT THE NUMBER OF CANS SOLD WHEN THE HIGH TEMP(X) IS 77

3. FIND THE RESIDUALS FOR TEMP(X)= 58 and 70

4. WHAT IS THE AMOUNT OF VARIATION IN CANS SOLD(Y) THAT KNOWING HIGH TEMP(X), EXPLAINS?

Answer 1

1.

The correlation coefficient (r ) formula as below:...............(1)

Step-1 :

Calculation table as below;

Sr.no.	HIGH TEMP(X)	CANS SOLD (Y)	XY	X^2	Y^2
1	55	340	18700	3025	115600
2	58	335	19430	3364	112225
3	64	410	26240	4096	168100
4	68	450	30600	4624	202500
5	70	460	32200	4900	211600
6	75	610	45750	5625	372100
7	80	735	58800	6400	540225
8	85	780	66300	7225	608400
	555	4120	298020	39259	2330750
	Total	Total	Total	Total	Total
	69.375	515.00
	Mean	Mean
	8	8
	n	n

Hence; we have the following .............(2)

n=	8
Sum-x =	555
Sum-y =	4120
Sum-x^2 =	39259
Sum-xy =	298020
x-bar =	69.38
y-bar =	515.00
Sum-y^2 =	2330750

Substitute (2) values in (1) , we get

r =	97560			0.9704
	6047	x	1671600

So correlation coefficient (r ) = 0.9704

Comment : Its direction is Positive and strength is very high ( near to 1 )

#####

2. Regression Equation as below;

y = a + b X................(3)

where a, and b are formula as below;........(4)

Put values of (2) in (4) we get;

b=	12195	=	16.1336
	755.875

a=

515.00

-

1,119.2699

=

-604.2699

Hence, Regression Equation ( 3) becomes;

y = -604.2699 + 16.1336 x ...................(5)

To check the best fit, we run Excel Regression and get the following output;

SUMMARY OUTPUT
Regression Statistics
Multiple R	0.9704
R Square	0.9416
Adjusted R Square	0.9319
Standard Error	45.0934
Observations	8
ANOVA
	df	SS	MS	F	Significance F
Regression	1	196749.50	196749.50	96.7580	0.0001
Residual	6	12200.50	2033.42
Total	7	208950.00
	Coefficients	Standard Error	t Stat	P-value	Lower 95%	Upper 95%
Intercept	-604.2699	114.8981	-5.2592	0.0019	-885.4154	-323.1243
HIGH TEMP(X)	16.1336	1.6402	9.8366	0.0001	12.1203	20.1470

Since from the above output , we can say the fit is a Good fit, because of the following;

i) R-square = 0.9416 ( Very high )

ii) All p-values of F and t-statistic are < 0.05.

Now,

To PREDICT THE NUMBER OF CANS SOLD WHEN THE HIGH TEMP(X) IS 77,

we put x=77 in the above (5); we get;

y = -604.2699+16.1336*(77) = 638.0173 = 638 cans

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3.

FINDING THE RESIDUALS FOR TEMP(X)= 58 and 70

Put x=58 in (5) , we get

y = -604.2699+16.1336*(58) = 331

Residual = ( Y_actual - Y_predict ) = (335-331) = 4

#############

Put x=70 in (5) , we get

y = -604.2699+16.1336*(70) = 525

Residual = ( Y_actual - Y_predict ) = (460-525) = -65

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4.

THE AMOUNT OF VARIATION IN CANS SOLD(Y) THAT KNOWING HIGH TEMP(X), EXPLAINS by 94.16 %

( R-square = 94.16 from Excel output )

### End of answers

Note : since no methodology was mentioned in the question, Excel was used.

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