Question 8.10: In a study of the relationship between the permeability (y) ......

We calculate the following quantities:

\overline{x} = 21.7548  \overline{y} = 7.849  \sum\limits_{i=1}^{n} (x_{i}\ −\ \overline{x})^{2} = 1886.48  \sum\limits_{i=1}^{n} (y_{i}\ −\ \overline{y} )^{2} =325.993

\sum\limits_{i=1}^{n}(x_{i}\ −\ \overline{x})(y_{i}\ −\ \overline{ y}) = −566.121  \hat{\beta}_{0} = 14.3775  \hat{\beta}_{1} = −0.300094  s = 1.80337

The estimate of the mean permeability for skin specimens with a resistance of 25 kΩ is

ŷ = 14.3775 − 0.300094(25) = 6.875

     The standard deviation of ŷ is estimated to be

s_{\hat{y}} = s\sqrt{\frac{1}{n} + \frac{(x\ −\ \overline{x})^{2}}{ ∑^{n}_{i=1} (x_{i}\ −\ \overline{x})^{2}}}

                    = 1.80337\sqrt{\frac{1}{50} + \frac{(25\ −\ 21.7548)^{2} }{1886.48}}

                                   = 0.28844

There are n − 2 = 50 − 2 = 48 degrees of freedom. The t value is therefore t_{48,0.25} = 2.011. (This value is not found in Table A.3 but can be obtained on many calculators or with computer software. Alternatively, since there are more than 30 degrees of freedom, one could use z = 1.96.) The 95% confidence interval is

6.785 ± (2.011)(0.28844) = (6.295, 7.455)