Question 8.8: In an experiment to determine the effect of temperature on s......
In an experiment to determine the effect of temperature on shrinkage of a synthetic fiber, 25 specimens were subjected to various temperatures. For each specimen, the temperature in °C (x) and shrinkage in % (y) were measured, and the following summary statistics were calculated:
\sum\limits_{i=1}^{n}{(x_{i} \ −\ \overline{x})^{2}= 87.34} \sum\limits_{i=1}^{n}{(x_{i}\ −\ \overline{x})(y_{i}\ −\ \overline{y}) = 11.62} s = 0.951
Assuming that x and y follow a linear model, compute the estimated change in shrinkage due to an increase of 1°C in temperature. Should we use the linear model to predict shrinkage from temperature?
The linear model is y = \beta_{0} + \beta_{1}x + \varepsilon, and the change in shrinkage (y) due to a 1°C increase in temperature (x) is \beta_{1}. The null and alternate hypotheses are
H_{0} : \beta_{1} = 0 versus H_{1} : \beta_{1} ≠ 0
The null hypothesis says that increasing the temperature does not affect the shrinkage, while the alternate hypothesis says that is does. The quantity
\frac{\hat{\beta}_{1}\ −\ \beta_{1}}{s_{\hat{\beta}_{1}}}
has a Student’s t distribution with n − 2 = 25 − 2 = 23 degrees of freedom. Under H_{0}, \beta_{1} = 0. The test statistic is therefore
\frac{\hat{\beta}_{1}\ −\ 0}{s_{\hat{\beta}_{1}}}
We compute \hat{\beta}_{1} and s_{\hat{\beta}_{1}}:
\hat{\beta}_{1} = \frac{∑^{n}_{i=1} (x_{i}\ −\ \overline{x})(y_{i}\ −\ \overline{y})}{∑^{n}_{i=1} (x_{i}\ −\ \overline{x})^{2}} = \frac{11.62}{87.34} = 0.13304
s_{\hat{\beta}_{1}} = \frac{s}{\sqrt{ ∑^{n}_{i=1} (x_{i}\ −\ \overline{x})^{2}}} = 0.10176
The value of the test statistic is
\frac{0.13304\ −\ 0}{0.10176} = 1.307
The t table shows that the P-value is greater than 0.20. (Software yields P = 0.204.) We cannot conclude that the linear model is useful for predicting elongation from carbon content.