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.

Using the listed duration and interval after times, find the best predicted “interval after” time for an eruption with a duration of 253 seconds. How does it compare to an actual eruption with a duration of 253 seconds and an interval after time of 83 minutes?

The given data render the ‘interval after’ (in minutes) and eruption with the duration as follows.

The actual eruption with a duration of 253 seconds has an ‘interval after’ value of 83 minutes.

The formula for the estimated regression line is

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

where

\({b_0}\)is the Y-intercept,

\({b_1}\)is the slope,

\(x\)is the explanatory variable, and

\(\hat y\)is the response variable (predicted value).

Let X denote the duration (in seconds), and Y denote the ‘interval after’ (in minutes).

The calculations required to compute the slope and intercept are as follows.

The sample size \(\left( n \right) = 7\).

The slope is computed as

\(\begin{array}{c}{b_1} = \frac{{n\left( {\sum {xy} } \right) - \left( {\sum x } \right)\left( {\sum y } \right)}}{{n\left( {\sum {{x^2}} } \right) - {{\left( {\sum x } \right)}^2}}}\\ = \frac{{7 \times 145348 - 1584 \times 642}}{{7 \times 369212 - {{1584}^2}}}\\ = 0.006735\\ \approx 0.00673\end{array}\).

The intercept is computed as

\(\begin{array}{c}{b_0} = \frac{{\left( {\sum y } \right)\left( {\sum {{x^2}} } \right) - \left( {\sum x } \right)\left( {\sum {xy} } \right)}}{{n\left( {\sum {{x^2}} } \right) - {{\left( {\sum x } \right)}^2}}}\\ = \frac{{642 \times 369212 - 1584 \times 145348}}{{7 \times 369212 - {{1584}^2}}}\\ = 90.19027\\ \approx 90.190\end{array}\).

So, the estimated regression equation is

\(\begin{array}{c}\hat y = {b_0} + {b_1}x\\ = 90.2 + 0.00673x\end{array}\)

Refer to exercise 21 of section 10-1 for the following result.

1) The scatter plot does not show an approximate linear relationship between the variables.

2)The P-value is 0.921.

As theP-value is greater than the level of significance (0.05), the null hypothesis fails to be rejected.

Therefore, the correlation is not significant.

Referring to figure 10-5, the criteria for a good regression model are not satisfied.

As the model is bad, the best-predicted value of a variable is its sample mean.

The best-predicted interval after times for an eruption with a duration of 253 seconds is required to be obtained.

As this is a bad model, the sample mean of the response variable will be used to predict the value.

The sample meanfor the response variable is

\(\begin{array}{c}\bar y = \frac{{\sum {{y_i}} }}{n}\\ = \frac{{91 + 81 + ... + 91}}{7}\\ = 91.7143\end{array}\).

Therefore, the best-predicted interval after time for an eruption with a duration of 253 seconds will be approximately 91.7 minutes.

Thus, the actual value of 83 minutes differs significantly from 91.7 minutes.