Exercises 13–28 use the same data sets as Exercises 13–28

in Section 10-1. In each case, find the regression equation, letting the first variable be the predictor (x) variable. Find the indicated predicted value by following the prediction procedure summarized in Figure 10-5 on page 493.

Find the best predicted temperature at a time when a cricket

chirps 3000 times in 1 minute. What is wrong with this predicted temperature?

Values of two variables are given namely, chirps in 1 min and temperature.

Let x represent the chirps in 1 min.

Let y represent the temperature

Themean value of xis given as,

\(\begin{array}{c}\bar x = \frac{{\sum\limits_{i = 1}^n {{x_i}} }}{n}\\ = \frac{{882 + 1188 + .... + 900}}{8}\\ = 1016.25\end{array}\)

Therefore, the mean value of x is 1016.25.

Themean value of yis given as,

\(\begin{array}{c}\bar y = \frac{{\sum\limits_{i = 1}^n {{y_i}} }}{n}\\ = \frac{{69.7 + 93.3 + .... + 79.6}}{{12}}\\ = 80.75\end{array}\)

Therefore, the mean value of y is 80.75.

The standard deviation of x is given as,

\(\begin{array}{c}{s_x} = \sqrt {\frac{{\sum\limits_{i = 1}^n {{{({x_i} - \bar x)}^2}} }}{{n - 1}}} \\ = \sqrt {\frac{{{{\left( {882 - 1016.25} \right)}^2} + {{\left( {1188 - 1016.25} \right)}^2} + ..... + {{\left( {900 - 1016.25} \right)}^2}}}{{8 - 1}}} \\ = 135.8\end{array}\)

Therefore, the standard deviation of x is 135.8.

The standard deviation of y is given as,

\(\begin{array}{c}{s_y} = \sqrt {\frac{{\sum\limits_{i = 1}^n {{{({y_i} - \bar y)}^2}} }}{{n - 1}}} \\ = \sqrt {\frac{{{{\left( {69.7 - 80.75} \right)}^2} + {{\left( {93.3 - 80.75} \right)}^2} + ..... + {{\left( {79.6 - 80.75} \right)}^2}}}{{8 - 1}}} \\ = 8.125\end{array}\)

Therefore, the standard deviation of yis 8.125.

The correlation coefficient is given as,

\(r = \frac{{n\left( {\sum {xy} } \right) - \left( {\sum x } \right)\left( {\sum y } \right)}}{{\sqrt {\left( {\left( {n\sum {{x^2}} } \right) - {{\left( {\sum x } \right)}^2}} \right)\left( {\left( {n\sum {{y^2}} } \right) - {{\left( {\sum y } \right)}^2}} \right)} }}\)

The calculations required to compute the correlation coefficient are as follows:

The correlation coefficient is given as,

\(\begin{array}{l}r = \frac{{n\left( {\sum {xy} } \right) - \left( {\sum x } \right)\left( {\sum y } \right)}}{{\sqrt {\left( {\left( {n\sum {{x^2}} } \right) - {{\left( {\sum x } \right)}^2}} \right)\left( {\left( {n\sum {{y^2}} } \right) - {{\left( {\sum y } \right)}^2}} \right)} }}\\ = \frac{{8\left( {663245.4} \right) - \left( {8130} \right)\left( {646} \right)}}{{\sqrt {\left( {\left( {8 \times 8391204} \right) - {{\left( {8130} \right)}^2}} \right)\left( {\left( {8 \times 52626.6} \right) - {{\left( {646} \right)}^2}} \right)} }}\\ = 0.8737\end{array}\)

Therefore, the correlation coefficient is 0.8737.

The slopeof the regression line is given as,

\(\begin{array}{c}{b_1} = r\frac{{{s_Y}}}{{{s_X}}}\\ = 0.8737 \times \frac{{8.125}}{{135.8}}\\ = 0.0523\end{array}\)

Therefore, the value of slope is 0.052.

The interceptis computed as,

\(\begin{array}{c}{b_0} = \bar y - {b_1}\bar x\\ = 80.75 - \left( {0.0523 \times 1016.25} \right)\\ = 27.6\end{array}\)

Therefore, the value of intercept is 27.6.

Theregression equationis given as,

\(\begin{array}{c}\hat y = {b_0} + {b_1}x\\ = 27.6 + 0.052x\end{array}\)

Thus, the regression equation is \(\hat y = 27.6 + 0.052x\).

Referring to exercise 22 of section 10-1,

1)The scatter plot shows an approximate linear relationship between the variables.

2)The P-value is 0.005.

As the P-value is less than the level of significance (0.05), this implies the null hypothesis is rejected.

Therefore, the correlation is significant.

Referring to figure 10-5, the criteria for a good regression model are satisfied.

The best predicted temperature at a time when a cricket chirps 3000 times in 1 minute is computed as,

\(\begin{array}{c}\hat y = 27.6 + 0.052\left( {3000} \right)\\ = 183.6\end{array}\)

Therefore, the best predicted temperature at a time when a cricket chirps 3000 times in 1 minute is 183.6 F.