Following are measurements of tensile strength in ksi x and

Following are measurements of tensile strength in ksi (x) and Brinell hardness (y) for 10 specimens of cold drawn copper. Assume that tensile strength and Brinell hardness follow a bivariate normal distribution.

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           x                                                   y                                         

            106.2                                      35.0

            106.3                                      37.2

            105.3                                      39.8

            106.1                                      35.8

            105.4                                      41.3   

            106.3                                      40.7

            104.7                                     38.7

            105.4                                      40.2

            105.5                                      38.1

            105.1                                      41.6                           

a) Find a 95% confidence interval for p, the population correlation between tensile strength and Brinell hardness.

b) Can you conclude that p<0.3?

c) Can you conclude that p =/= 0?

Solution

We assume that the data points are a random sample from a bivariate normal distribution. In computing the correlation coefficiient, we find

Sxx = Px ^2 i ? ( Pxi) ^2/n = 111579.79 ? 6 (1056.3)^2/10 = 2.821,

Sxy = Pxiyi ?( Pxi)(Pyi)/n = 41020.79?(388.4)(1056.3)/10 = ?5.902

so R < 0 and Syy = Py ^2 i ? ( Pyi) ^2/n = 15132.6 ? (388.4) ^2/1047.144 = 47.144. Hence

R 2 = S^ 2 xy / SxxSyy = (?5.902^)2 /2.821