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.

Crickets and Temperature The table below lists numbers of cricket chirps in 1 minute and the temperature in °F. For the prediction interval, use 1000 chirps in 1 minute and use a 90% confidence level.

Chirps in 1 min	882	1188	1104	864	1200	1032	960	900
Temperature (°F)	69.7	93.3	84.3	76.3	88.6	82.6	71.6	79.6

Data are given on two variables, “Chirps in 1 min” and “Temperature”.

Let x denote the variable “Chirps in 1 min”.

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

\({b_1}\) is the slope coefficient

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

The value of the y-intercept is computed below:

\(\begin{aligned}{c}{b_0} &= \frac{{\left( {\sum y } \right)\left( {\sum {{x^2}} } \right) - \left( {\sum x } \right)\left( {\sum {xy} } \right)}}{{n\left( {\sum {{x^2}} } \right) - {{\left( {\sum x } \right)}^2}}}\\ &= \frac{{\left( {646} \right)\left( {8391204} \right) - \left( {8130} \right)\left( {663245.4} \right)}}{{8\left( {8391204} \right) - {{\left( {8130} \right)}^2}}}\\ &= 27.62835082\\ &\approx 27.63\end{aligned}\)

The value of the slope coefficient is computed below:

\(\begin{aligned}{c}{b_1} &= \frac{{n\left( {\sum {xy} } \right) - \left( {\sum x } \right)\left( {\sum y } \right)}}{{n\left( {\sum {{x^2}} } \right) - {{\left( {\sum x } \right)}^2}}}\\ &= \frac{{\left( 8 \right)\left( {663245.4} \right) - \left( {8130} \right)\left( {646} \right)}}{{8\left( {8391204} \right) - {{\left( {8130} \right)}^2}}}\\ &= 0.05227222\\ &\approx 0.05\end{aligned}\)

Thus, the regression equation becomes:

\(\begin{aligned}{c}\hat y &= 27.62835082 + 0.05227222x\\ &\approx 27.63 + 0.05x\end{aligned}\)

The mean value of observed y is computed below:

\(\begin{aligned}{c}\bar y &= \frac{{\sum y }}{n}\\ &= \frac{{646}}{8}\\ &= 80.75\end{aligned}\)

The following table shows the predicted values (upon substituting the values of x in the regression equation) and other important calculations are done below:

The value of the explained variation is shown below:

\(\sum {{{\left( {\hat y - \bar y} \right)}^2}} = 352.7277\)

Thus, the explained variation is equal to 352.7278.

The value of the unexplained variation is shown below:

\(\sum {{{\left( {y - \hat y} \right)}^2}} = 109.3722\)

Thus, the unexplained variation is equal to 109.3722.

Substituting the value of\({x_0} = 1000\)in the regression equation, the predicted value is obtained as follows:

\(\begin{aligned}{c}\hat y &= 27.62835082 + 0.05227222x\\ &= 27.62835082 + 0.05227222\left( {1000} \right)\\ &= 79.90057\end{aligned}\)

The following formula is used to compute the level of significance

\(\begin{aligned}{c}{\rm{Confidence}}\;{\rm{Level}} &= 90\% \\100\left( {1 - \alpha } \right) &= 90\\1 - \alpha &= 0.90\\ &= 0.10\end{aligned}\)

The degrees of freedom for computing the value of the t-multiplier are shown below:

\(\begin{aligned}{c}df &= n - 2\\ &= 8 - 2\\ &= 6\end{aligned}\)

The value of the t-multiplier for level of significance equal to 0.10 and degrees of freedom equal to 6 is equal to 1.9432.

The value of the standard error of the estimate is computed below:

\(\begin{array}{c}{s_e} = \sqrt {\frac{{\sum {{{\left( {y - \hat y} \right)}^2}} }}{{n - 2}}} \\ = \sqrt {\frac{{109.3722}}{{8 - 2}}} \\ = 4.269508\end{array}\)

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

\(\begin{aligned}{c}\bar x &= \frac{{\sum x }}{n}\\ &= \frac{{8130}}{8}\\ &= 1016.25\end{aligned}\)

Substitute the values obtained above to calculate the value of margin of error (E) as shown:

\(\begin{aligned}{c}E &= {t_{\frac{\alpha }{2}}}{s_e}\sqrt {1 + \frac{1}{n} + \frac{{n{{\left( {{x_0} - \bar x} \right)}^2}}}{{n\left( {\sum {{x^2}} } \right) - {{\left( {\sum x } \right)}^2}}}} \\ &= \left( {1.9432} \right)\left( {4.269508} \right)\sqrt {1 + \frac{1}{8} + \frac{{8{{\left( {1000 - 1016.25} \right)}^2}}}{{8\left( {8391204} \right) - {{\left( {8130} \right)}^2}}}} \\ &= 8.807772\end{aligned}\)

Thus, the prediction interval becomes:

\(\begin{aligned}{c}PI &= \left( {\hat y - E,\hat y + E} \right)\\ &= \left( {79.90057 - 8.807772,79.90057 + 8.807772} \right)\\ &= \left( {71.0928,88.7083} \right)\\ \approx \left( {71.09,88.71} \right)\end{aligned}\)

Therefore, the 90% prediction interval for the temperature for the given value of number of chirps in 1 min equal to 1000 is (71.09, 88.71).