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Chapter 13: Problem 61

Exercise \(13.8\) gave data on \(x=\) treadmill run time to exhaustion and \(y=20-\mathrm{km}\) ski time for a sample of 11 biathletes. Use the accompanying MINITAB output to answer the following questions. The regression equation is ski \(=-88.8-2.33\) tread \(\begin{array}{lrrrr}\text { Predictor } & \text { Coef } & \text { Stdev } & \text { t-ratio } & p \\\ \text { Constant } & 88.796 & 5.750 & 15.44 & 0.000 \\ \text { tread } & 2.3335 & 0.5911 & 3.95 & 0.003 \\ s=2.188 & \text { R-sq }=63.4 \% & \text { R-sq }(a d j)=59.3 \% & \end{array}\) Analysis of Variance \(\begin{array}{lrrrrr}\text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { p } \\ \text { Regression } & 1 & 74.630 & 74.630 & 15.58 & 0.003 \\ \text { Error } & 9 & 43.097 & 4.789 & & \\ \text { Total } & 10 & 117.727 & & & \end{array}\) a. Carry out a test at significance level \(.01\) to decide whether the simple linear regression model is useful. b. Estimate the average change in ski time associated with a 1 -minute increase in treadmill time, and do so in a way that conveys information about the precision of estimation. c. MINITAB reported that \(s_{a+b(10)}=.689\). Predict ski time for a single biathlete whose treadmill time is \(10 \mathrm{~min}\), and do so in a way that conveys information about the precision of prediction. d. MINITAB also reported that \(s_{a+b(11)}=1.029 .\) Why is this larger than \(s_{a+b(10) ?}\)

Short Answer

Expert verified
In conclusion, the linear regression model was found to be useful at the 0.01 significance level. The average increase in ski time associated with a 1-minute increase in treadmill time was estimated to be 2.3335 minutes. The predicted ski time for a biathlete with a treadmill time of 10 minutes is approximately 111.13 minutes. Lastly, \(s_{a+b(11)}\) was larger than \(s_{a+b(10)}\) because the prediction was happening further from the mean treadmill time.

Step by step solution

01

Carry out a test at significance level .01

We are asked to test whether the simple linear regression model is useful. To do this, we look at the p-value for the F-statistic in the Analysis of Variance table. In this case, the p-value is 0.003, which is less than the given significance level of 0.01. Hence, we reject the null hypothesis, which means that the linear regression model is useful.
02

Estimate the average change in ski time

We are to estimate the average change in ski time with a 1-minute increase in treadmill time. This is essentially the slope of the regression line, or the tread coefficient, which is 2.3335. Thus, for each additional minute on the treadmill, we can expect the ski time to increase on average by 2.3335 minutes.
03

Predict ski time for a single biathlete whose treadmill time is 10 min

To calculate this, we use the regression equation, substituting 10 for the treadmill time. Therefore, ski time = 88.8 + 2.33*10 = 111.13 minutes. MINITAB also reported \(s_{a+b(10)}= .689\), which represents the standard error of the forecast. So the forecasted ski time is 111.13 minutes, give or take .689 minutes.
04

Explore why \(s_{a+b(11)}\) is larger than \(s_{a+b(10)}\)

It has been mentioned that \(s_{a+b(11)}= 1.029\) and it's larger than \(s_{a+b(10)}\). The standard error of the forecast increases as we move away from the mean of the predictor variable (treadmill time). Hence, \(s_{a+b(11)}\) is larger than \(s_{a+b(10)}\) because we are predicting further from the mean treadmill time when we substitute treadmill time with 11 instead of 10.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Regression Coefficient
In simple linear regression, the regression coefficient, or slope, indicates the average change in the dependent variable (the variable we are trying to predict) for each one-unit change in the independent variable (the variable we use for prediction). In the provided exercise, the regression equation is given as ski \(-88.8 - 2.33 \times \text{tread}\). Here, the coefficient of tread, \(-2.3335\), signifies the average increase in ski time with each additional minute of treadmill running.

Thus, it tells us that for every extra minute spent on the treadmill, we can expect the ski time to increase by an average of about \(2.3335\) minutes. This coefficient is crucial for drawing predictive insights from your data. Understanding the impact of treadmill time on ski performance helps biathletes improve their training regimens.
Significance Testing
Significance testing in regression analysis helps us determine if the relationship between the predictor and response variables is statistically meaningful. In this context, we typically perform a hypothesis test using the F-statistic, which is derived from the Analysis of Variance (ANOVA).
  • The null hypothesis (\(H_0\)) states that there is no relationship between the variables, implying that the coefficient is zero.
  • The alternative hypothesis (\(H_1\)) suggests that there is a non-zero relationship.
In our step-by-step solution, the p-value for the F-statistic is \(0.003\), which is less than the significance level of \(0.01\). This leads us to reject \(H_0\), affirming that the regression model is significant. A p-value less than \(0.01\) indicates strong evidence against the null hypothesis, suggesting a useful model.
Analysis of Variance (ANOVA)
ANOVA is a statistical technique used in the context of regression to dissect the total variability found within data into components. Each component is attributed to different sources of variation: between groups and within groups. ANOVA helps validate the regression model by checking if the predictor variables contribute significantly to the model.In simple linear regression, ANOVA provides the F-statistic which allows us to compare the model to a scenario where no predictors are used. For the given exercise:
  • The ANOVA table shows a Regression SS (sum of squares) of \(74.630\), indicating the variability explained by treadmill time.
  • With \(9\) degrees of freedom, the Error MS (mean square) is \(4.789\), reflecting unexplained variability.
  • The F-statistic from these values is \(15.58\) with an associated p-value of \(0.003\), confirming significant variation explained by our model.
Standard Error of the Estimate
The standard error of the estimate gives us a measure of the accuracy of predictions made with a regression line. It quantifies the typical distance that the observed values fall from the regression line. A lower value indicates a more precise estimate.In the step-by-step solution, the standard error is \(s=2.188\). This value reflects the average prediction error of the regression model. When predicting ski time:

  • For a \(10\)-minute treadmill time, MINITAB reports a forecast standard error of \(s_{a+b(10)} = 0.689\), indicating precision in this prediction.
  • The error increases to \(1.029\) when predicting for \(11\) minutes (reported as \(s_{a+b(11)}\)). This highlights a key insight: predictions farther from the mean of the treadmill times face higher errors due to model constraints.

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Most popular questions from this chapter

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