Can anyone explain this answer to me..specifically how to find Sum(sigma) of XiY

Question

Can anyone explain this answer to me..specifically how to find Sum(sigma) of XiYi.... Explinations on the rest of the the problem would help as well

Tire pressure (in kPa) was measured for the right and left front tires on a sample of 10 automobiles. Assume that the tire pressures follow a bivariate normal distributions. Find a 95% confidence interval for rho, the population correlation between the pressure in the right tire and the pressure in the left tire. n = 10; x = 201.3; y = 202.2; xi2 = 407381; yi2 = 410420; xiyi = 408745 r = 0.930698; W = 1/2 In 1+r/1-r = 1.663582; sigma omega = 1/(n - 3) = 0.377964 95% for is mu omega is W plusminus 1.96 sigma omega = (0.92279,2.40438). Hence 95% Cl for rho is (e2(0.92279-1/e2(0.92279)+1, e2(2.40438-1/e2(2.40438+1) = (0.7272,0.9838)

Explanation / Answer

Here used formulas of regression analysis

first we find the correlation between two tires

and lso find standerd deviation for constants

sum of (XiYi) is obtainned as follows

product of 1st X to 1st Y +product of second X to second Y +..................+product of nth X to nth Y

we also find mean of right tire and left tire

where W is correlation between two

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