Question

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?

Short answer

The regression equation is\(\hat y = 90.190 + 0.007X\)

The best predicted ‘interval after’ time for an eruption with a duration of 253 seconds will be approximately 92 (minutes).

There is an error of approximately 9 minutes in prediction. This is because the actual eruption with a duration of 253 seconds and an ‘interval after’ time is 83 minutes.

Step-by-step solution

Given information

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.

State the estimated regression line

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).

Compute the slope and intercept

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

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

Check the model

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.

Compute the predicted value

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.