Variation and Prediction Intervals. In Exercises 17–20, find the (a) explained variation, (b) unexplained variation, and (c) indicated prediction interval. In each case, there is sufficient evidence to support a claim of a linear correlation, so it is reasonable to use the regression equation when making predictions.

Altitude and Temperature Listed below are altitudes (thousands of feet) and outside air temperatures (°F) recorded by the author during Delta Flight 1053 from New Orleans to Atlanta. For the prediction interval, use a 95% confidence level with the altitude of 6327 ft (or 6.327 thousand feet).

Altitude (thousands of feet)	3	10	14	22	28	31	33
Temperature (°F)	57	37	24	-5	-30	-41	-54

Data are given on two variables “Altitude (thousands of feet)” and “Temperature (in degrees Fahrenheit).”

Let x denote the variable “Altitude.”

Let y denote the variable “Temperature.”

The regression equation of y on x has the following notation:

\(\hat y = {b_0} + {b_1}x\),

where

\({b_0}\)is the intercept term and

\({b_1}\)is the slope coefficient.

The following calculations are done to compute the intercept and the slope coefficient:

The y-intercept is computed below:

\(\begin{array}{c}{b_0} = \frac{{\left( {\sum y } \right)\left( {\sum {{x^2}} } \right) - \left( {\sum x } \right)\left( {\sum {xy} } \right)}}{{n\left( {\sum {{x^2}} } \right) - {{\left( {\sum x } \right)}^2}}}\\ = \frac{{\left( { - 12} \right)\left( {3623} \right) - \left( {141} \right)\left( { - 3126} \right)}}{{7\left( {3623} \right) - {{\left( {141} \right)}^2}}}\\ = 72.49817518\end{array}\)

The slope coefficient is computed below:

\(\begin{array}{c}{b_1} = \frac{{n\left( {\sum {xy} } \right) - \left( {\sum x } \right)\left( {\sum y } \right)}}{{n\left( {\sum {{x^2}} } \right) - {{\left( {\sum x } \right)}^2}}}\\ = \frac{{\left( 7 \right)\left( { - 3126} \right) - \left( {141} \right)\left( { - 12} \right)}}{{7\left( {3623} \right) - {{\left( {141} \right)}^2}}}\\ = - 3.68430656\end{array}\)

Thus, the regression equation becomes

\(\hat y = 72.49817518 - 3.68430656x\).

The following table shows the predicted values (upon substituting the values of x in the regression equation):

The mean value of observed y is computed below:

\(\begin{array}{c}\bar y = \frac{{\sum y }}{n}\\ = \frac{{ - 12}}{7}\\ = - 1.71429\end{array}\)

Some important calculations are done below:

The explained variation is \(\sum {{{\left( {\hat y - \bar y} \right)}^2}} = 10626.59\).

Thus, the explained variation is equal to 10626.59.

The unexplained variation is \(\sum {{{\left( {y - \hat y} \right)}^2}} = 68.83577\).

Thus, the unexplained variation is equal to 68.83577.

Substitute\({x_0} = 6.327\)in the regression equation to get the predicted value.

\(\begin{array}{c}\hat y = 72.49817518 - 3.68430656x\\ = 72.49817518 - 3.68430656\left( {6.327} \right)\\ = 49.18757\\ \approx 49\end{array}\)

The prediction interval is obtained using the formula

\(\begin{array}{c}PI = \hat y \pm E\\ = \hat y \pm {t_{\frac{\alpha }{2}}}{s_e}\sqrt {1 + \frac{1}{n} + \frac{{n{{\left( {{x_0} - \bar x} \right)}^2}}}{{n\left( {\sum {{x^2}} } \right) - {{\left( {\sum x } \right)}^2}}}} \end{array}\).

The following formula is used to compute the level of significance.

\(\begin{array}{c}{\rm{Confidence}}\;{\rm{Leve}}l = 95\% \\100\left( {1 - \alpha } \right) = 95\\1 - \alpha = 0.95\\\alpha = 0.05\end{array}\)

The degrees of freedom for computing the t-multiplier are shownbelow:

\(\begin{array}{c}df = n - 2\\ = 7 - 2\\ = 5\end{array}\)

The two-tailed value of the t-multiplier for the level of significance (0.05) and degrees of freedom (5) is 2.5706.

The standard error of the estimate is computed below:

\(\begin{array}{c}{s_e} = \sqrt {\frac{{\sum {{{\left( {y - \hat y} \right)}^2}} }}{{n - 2}}} \\ = \sqrt {\frac{{68.83577}}{{7 - 2}}} \\ = 3.710411\end{array}\)

The value of \(\bar x\)is computed as follows:

\(\begin{array}{c}\bar x = \frac{{\sum x }}{n}\\ = \frac{{141}}{7}\\ = 20.14286\end{array}\)

Substitute the values obtained above to calculate the margin of error (E).

\(\begin{array}{c}E = {t_{\frac{\alpha }{2}}}{s_e}\sqrt {1 + \frac{1}{n} + \frac{{n{{\left( {{x_0} - \bar x} \right)}^2}}}{{n\left( {\sum {{x^2}} } \right) - {{\left( {\sum x } \right)}^2}}}} \\ = \left( {2.5706} \right)\left( {3.710411} \right)\sqrt {1 + \frac{1}{7} + \frac{{7{{\left( {6.327 - 20.14286} \right)}^2}}}{{7\left( {3623} \right) - \left( {19881} \right)}}} \\ = 11.23167664\end{array}\)

Thus, the prediction interval (PI) becomes as shown:

\(\begin{array}{c}PI = \left( {\hat y - E,\hat y + E} \right)\\ = \left( {49.18757 - 11.23167664,49.18757 + 11.23167664} \right)\\ = (37.9559,60.4192)\\ \approx \left( {38.0,60.4} \right)\end{array}\)

Therefore, the 95% prediction interval for the temperature (in degrees Fahrenheit) for the given altitude of 6.327 is (38.0, 60.4).