# recorded temperatures (in F) a

recorded temperatures (in F) and precipitation in (inches) at aplace are as follows: Temperature(x) 86, 81,83,89,90,74,64Precipitation(y) 3.4,1.8,3.5,3.6,3.7,1.5,0.2 A Find the samplecorrelation coeficcient, r B. What is the meaning of the value of rthat you have calculated in part A? C. Find the significance of thepopulation correlation coefficient, p at a=0.05 D. Find the valueof the coefficient of variation r2 E. What is the meaning of r2value you have calculated in part D? F. Find y, when x=70 G. Findthe standard error of the estimate Has to be done in excell

a)

Computational Table:

 X Y X2 Y2 XY 86 3.4 7396 11.56 292.4 81 1.8 6561 3.24 145.8 83 3.5 6889 12.25 290.5 89 3.6 7921 12.96 320.4 90 3.7 8100 13.69 333 74 1.5 5476 2.25 111 64 0.2 4096 0.04 12.8 Total 567 17.7 46439 55.99 1505.9

A) Find Correlation Coefficient (r):

r = 0.95

B)

There is High degree of Positive correlation coefficientbetween Temperature(X) & Precipitation (Y)

C)

Hypothesis:

(No linear correlation)

(Positive linear correlation)

Test statistic:

Degrees of freedom = n-2 = 7-2 = 5

P-value: 0.0009 ………….From t table

Conclusion:

p-value < , i.e 0.0009 <0.05, That is Reject Ho at 5% level ofsignificance.

Therefore, population correlation coefficient statisticallysignificant at the 5% level of significance.

D)

Coefficient of Determination (R2) =r2 = 0.952 = 0.9063

E)

This result means that 95% of the variation in thedependent variable is accounted for by the variation in theindependent variable. The rest of the variation, 0.05, or 5% isunexplained. This value is called the coefficient ofnon-determination and is found by subtracting the coefficient ofdetermination from 1.

F)

For slope:

b = 0.14

For Intercept:

a = -8.89

Therefore, the least square regression line would be,

Find Y when X = 70

Therefore, above equation becomes……..

G)

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