91影视

Data are given on two variables, 鈥淥verhead Width鈥� and 鈥淲eight鈥�.

Let x denote the variable 鈥淥verhead Width.鈥�

Let y denote the variable 鈥淲eight鈥�.

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:

\(\begin{aligned}{c}{b_0} &= \frac{{\left( {\sum y } \right)\left( {\sum {{x^2}} } \right) - \left( {\sum x } \right)\left( {\sum {xy} } \right)}}{{n\left( {\sum {{x^2}} } \right) - {{\left( {\sum x } \right)}^2}}}\\ &= \frac{{\left( {1108} \right)\left( {439} \right) - \left( {81} \right)\left( {9639} \right)}}{{8\left( {439} \right) - {{\left( {81} \right)}^2}}}\\ &= - 156.878788\\ \approx - 156.88\end{aligned}\)

The value of the slope coefficient is computed below:

\(\begin{aligned}{c}{b_1} &= \frac{{n\left( {\sum {xy} } \right) - \left( {\sum x } \right)\left( {\sum y } \right)}}{{n\left( {\sum {{x^2}} } \right) - {{\left( {\sum x } \right)}^2}}}\\ &= \frac{{\left( 6 \right)\left( {9639} \right) - \left( {51} \right)\left( {1108} \right)}}{{6\left( {439} \right) - {{\left( {51} \right)}^2}}}\\ &= 40.181818\\ &\approx 40.18\end{aligned}\)

Thus, the regression equation becomes:

\(\begin{aligned}{c}\hat y &= - 156.878788 + 40.181818x\\ \approx - 156.88 + 40.18x\end{aligned}\)

The mean value of observed y is computed below:

\(\begin{aligned}{c}\bar y &= \frac{{\sum y }}{n}\\ &= \frac{{1108}}{6}\\ &= 184.6667\end{aligned}\)

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:

\(\sum {{{\left( {\hat y - \bar y} \right)}^2}} = 8880.1818\)

Thus, the explained variation is equal to 8880.182.

The value of the unexplained variation is shown below:

\(\sum {{{\left( {y - \hat y} \right)}^2}} = 991.1515\)

Thus, the unexplained variation is equal to 991.1515.

Substituting the value of\({x_0} = 9\)in the regression equation, the predicted value is obtained as follows:

\(\begin{aligned}{c}\hat y &= - 156.878788 + 40.181818x\\ &= - 156.878788 + 40.181818\left( 9 \right)\\ &= 204.7576\end{aligned}\)

The following formula is used to compute the level of significance

\(\begin{aligned}{c}{\rm{Confidence}}\;{\rm{Level}} &= 99\% \\100\left( {1 - \alpha } \right) &= 99\\1 - \alpha &= 0.99\\ &= 0.01\end{aligned}\)

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

\(\begin{aligned}{c}df &= n - 2\\ &= 6 - 2\\ &= 4\end{aligned}\)

The value of the t-multiplier for a level of significance equal to 0.01and degrees of freedom equal to 4 is equal to 4.6041.

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

\(\begin{array}{c}{s_e} = \sqrt {\frac{{\sum {{{\left( {y - \hat y} \right)}^2}} }}{{n - 2}}} \\ = \sqrt {\frac{{991.1515}}{{6 - 2}}} \\ = 15.74128\end{array}\)

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

\(\begin{array}{c}\bar x = \frac{{\sum x }}{n}\\ = \frac{{51}}{6}\\ = 8.5\end{array}\)

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

\(\begin{aligned}{c}E &= {t_{\frac{\alpha }{2}}}{s_e}\sqrt {1 + \frac{1}{n} + \frac{{n{{\left( {{x_0} - \bar x} \right)}^2}}}{{n\left( {\sum {{x^2}} } \right) - {{\left( {\sum x } \right)}^2}}}} \\ &= \left( {4.6041} \right)\left( {15.74128} \right)\sqrt {1 + \frac{1}{6} + \frac{{6{{\left( {9 - 8.5} \right)}^2}}}{{6\left( {439} \right) - {{\left( {51} \right)}^2}}}} \\ &= 79.791718\end{aligned}\)

Thus, the prediction interval becomes:

\(\begin{aligned}{c}PI &= \left( {\hat y - E,\hat y + E} \right)\\ &= \left( {204.7576 - 79.791718,204.7576 + 79.791718} \right)\\ &= \left( {124.966,284.549} \right)\\ &\approx \left( {124.97,284.55} \right)\end{aligned}\)

Therefore, the 99% prediction interval for the overhead width for the given value of weight equal to 9.0 cm is (124.97 cm, 284.55 cm).