## Related questions with answers

The amount of catalyst $(x)$ and the yield $(y)$ of a chemical experiment are analyzed using a simple linear regression model. There are 30 observations $(x_i , y_i )$, and it is found that the fitted model is

$y = 51.98 + 3.44x$

Suppose that the sum of squared residuals is $\sum_{i=1}^{30} e_i^2 = 329.77$, and that $\sum_{i=1}^{30} x_i = 603.36, \sum_{i=1}^{30} x_i^2 = 12578.22$. If an additional experiment is planned with a catalyst level 22, give a range of values with $95\%$ confidence for the yield that you think will be obtained for that experiment.

Solution

VerifiedGiven informations:

$y=51.98+3.44x$

$\sum_{i=1}^{30} e_i^2=329.77,\sum_{i=1}^{30} x_i=603.36,\sum_{i=1}^{ 30}x_i^2=12578.22$

A $99\%$ prediction interval for a catalyst level $x^*=22$ is required. We will be needed $\overline{x},\hat{ \sigma}, S_{xx}$

$\begin{gather*} \overline{x}=\frac{\sum_{i=1}^{30}x_i}{30}=21.112\\ \hat{\sigma}=\frac{\sum_{i=1}^{30} e_i^2}{28}=11.778\\ S_{xx}=\sum_{i=1}^{30} x_i^2-\frac{(\sum_{i=1}^{30} x_i)^2}{30}=443.44 \end{gather*}$

Now, we can find prediction interval.

$\begin{gather*} t_{\frac{\alpha}{2},n-2}=t_{0.025,28}=2.048\\ s.e.(\hat{\beta_0}+\hat{\beta_1}x)=\sigma\sqrt{\frac{n+1}{n}+\frac{(x^*-\overline{x})^2}{S_{xx}}}=3.49\\ \beta_0+\beta_1x^* \in (\hat{\beta_0}+\hat{\beta_1}x^*-t_{\frac{\alpha}{2},n-2}s.e.(\hat{\beta_0}+\hat{\beta_1}x),\hat{\beta_0}+\hat{\beta_1}x^*+t_{\frac{\alpha}{2},n-2}s.e.(\hat{\beta_0}+\hat{\beta_1}x))\\ \beta_0+\beta_1x^* \in (120.49,134.83) \end{gather*}$

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