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The article "Readability of Liquid Crystal Displays: A Response Surface" (Human Factors [1983]: \(185-190\) ) used the estimated regression equation to describe the relationship between \(y=\) error percentage for subjects reading a four-digit liquid crystal display and the independent variables \(x_{1}=\) level of backlight, \(x_{2}=\) character subtense, \(x_{3}=\) viewing angle, and \(x_{4}=\) level of ambient light. From a table given in the article, SSRegr \(=19.2\), SSResid = \(20.0\), and \(n=30\). a. Does the estimated regression equation specify a useful relationship between \(y\) and the independent variables? Use the model utility test with a \(.05\) significance level. b. Calculate \(R^{2}\) and \(s_{e}\) for this model. Interpret these values. c. Do you think that the estimated regression equation would provide reasonably accurate predictions of error rate? Explain.

Short Answer

Expert verified
The short answer to the questions would depend on the actual results of the calculations in Steps 1-5. For instance, in terms of a useful relationship (part a), if the calculated F statistic is greater than the critical F value, then yes, there is a useful relationship. Likewise, the \(R^{2}\) and \(s_{e}\) values (part b) would be the actual results of the calculations in Step 3. Whether the equation would provide reasonably accurate predictions (part c) would depend on the combined assessment of the model utility test, \(R^{2}\) value, and \(s_{e}\) value.

Step by step solution

01

Model Utility Test

Firstly, it's necessary to perform the model utility test to determine if there's a useful relationship between the dependent variable \(y\) and the independent variables. The F statistic is calculated as \((SSRegr / p) / (SSResid / (n - p - 1))\), where \(SSRegr\) is the regression sum of squares, \(SSResid\) is the residual sum of squares, \(p\) is the number of predictor variables (which is 4 in this case), and \(n\) is the number of observations (which is 30). Here, \(SSRegr = 19.2\), \(SSResid = 20.0\), \(p = 4\), and \(n = 30\). So, the F statistic calculation would be \((19.2 / 4) / (20.0 / (30 - 4 - 1))\).
02

Test of Significance

After calculating the F statistic, it'd be necessary to compare it with the critical F value for a .05 significance level and degrees of freedom \(p\) and \(n - p - 1\). If the calculated F statistic is greater than the critical F value, then the null hypothesis that all regression coefficients are zero is rejected. This would indicate that the estimated regression equation specifies a useful relationship between the error percentage and the independent variables.
03

Calculation of \(R^{2}\) and \(s_{e}\)

Next, the coefficient of determination (\(R^{2}\)) is calculated as \(R^{2} = SSRegr / (SSRegr + SSResid)\). Here, \(SSRegr = 19.2\) and \(SSResid = 20.0\). Thus, \(R^{2} = 19.2 / (19.2 + 20.0)\). The standard error of estimate (\(s_{e}\)) is calculated as \(s_{e} = \sqrt{SSResid / (n - p - 1)}\). So, \(s_{e} = \sqrt{20.0 / (30 - 4 - 1)}\).
04

Interpretation of \(R^{2}\) and \(s_{e}\)

The \(R^{2}\) value represents the proportion of the variance in the dependent variable that is predictable from the independent variables. The \(s_{e}\) value is a measure of the standard deviation of the observed \(y\) values about the predicted \(y\) values.
05

Predictive Assessment

Finally, the usefulness of the estimated regression equation for predicting the error rate would depend on the results of the model utility test, the magnitude of the \(R^{2}\) value, and the size of the \(s_{e}\) value. An equation that passes the utility test, has a high \(R^{2}\) value, and a small \(s_{e}\) value would typically be deemed a good predictor.

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Key Concepts

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

Model Utility Test
In regression analysis, the model utility test is a statistical method used to assess whether a set of independent variables has a statistically significant relationship with the dependent variable. This test is crucial in determining whether it's worth using the model for prediction or if the observed results could have occurred by chance.

In our example, the F statistic is calculated to ascertain if the independent variables relating to the readability of liquid crystal displays significantly predict the error percentage. If the calculated F value exceeds the critical F value at a given significance level (in this case, .05), we reject the null hypothesis, which assumes the model has no utility, and we can conclude that there is a useful relationship between the variables.

For students, understanding the model utility test is essential as it helps determine if further analysis is warranted and ensures that predictions made by the regression model are based on statistically relevant relationships.
Coefficient of Determination
The coefficient of determination, denoted as \( R^2 \), is an essential metric in regression analysis as it provides an insight into the amount of variation in the dependent variable that can be explained by the independent variables. It ranges from 0 to 1, with higher values indicating a better fit of the model to the data.

In practical terms, if the \( R^2 \) value is close to 1, it suggests that a large proportion of the variance in the error percentage is accounted for by the variables like backlight level, character subtense, viewing angle, and ambient light level. It is a core tool for students to evaluate how well their regression model captures the underlying data patterns and makes reliable predictions.
Standard Error of Estimate
The standard error of estimate, represented by \( s_e \), quantifies the typical distance between the observed data points and the estimated regression line. In essence, it measures the accuracy with which the regression line predicts the dependent variable.

Calculating \( s_e \) gives students a numerical value to assess the precision of the model's estimates. A small value for \( s_e \) indicates that the model has a high predictive accuracy because the observed values tend to be close to the predicted values. Conversely, a large \( s_e \) may suggest the model's predictions are often far from the actual data points, signaling that the model might not be the best fit for the data.
Predictive Assessment in Statistics
Predictive assessment is the evaluation of a statistical model's capability to accurately forecast the value of a dependent variable. For a regression model, predictive power is determined by combining several diagnostic measures, including the F statistic from the model utility test, the coefficient of determination (\( R^2 \)), and the standard error of estimate (\( s_e \)).

Students gauge prediction quality by looking at the collective interpretative strength of these measures. A model passing the utility test, with a high \( R^2 \) and low \( s_e \), is often considered to have good predictive capacity. Hence, students must learn to critically analyze these diagnostic statistics to evaluate whether a regression model can be used for reliable predictions, as these skills are fundamental for statisticians and data analysts.

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Most popular questions from this chapter

This exercise requires the use of a computer package. The authors of the article "Absolute Versus per Unit Body Length Speed of Prey as an Estimator of Vulnerability to Predation" (Animal Behaviour [1999]: \(347-\) 352) found that the speed of a prey (twips/s) and the length of a prey (twips \(\times 100\) ) are good predictors of the time (s) required to catch the prey. (A twip is a measure of distance used by programmers.) Data were collected in an experiment where subjects were asked to "catch" an animal of prey moving across his or her computer screen by clicking on it with the mouse. The investigators varied the length of the prey and the speed with which the prey moved across the screen. The following data are consistent with summary values and a graph given in the article. Each value represents the average catch time over all subjects. The order of the various speed-length combinations was randomized for each subject. $$ \begin{array}{ccc} \begin{array}{c} \text { Prey } \\ \text { Length } \end{array} & \begin{array}{l} \text { Prey } \\ \text { Speed } \end{array} & \begin{array}{l} \text { Catch } \\ \text { Time } \end{array} \\ \hline 7 & 20 & 1.10 \\ 6 & 20 & 1.20 \\ 5 & 20 & 1.23 \\ 4 & 20 & 1.40 \\ 3 & 20 & 1.50 \\ 3 & 40 & 1.40 \\ 4 & 40 & 1.36 \\ 6 & 40 & 1.30 \\ 7 & 40 & 1.28 \\ 7 & 80 & 1.40 \\ 6 & 60 & 1.38 \\ 5 & 80 & 1.40 \\ 7 & 100 & 1.43 \\ 6 & 100 & 1.43 \\ 7 & 120 & 1.70 \\ 5 & 80 & 1.50 \\ 3 & 80 & 1.40 \\ 6 & 100 & 1.50 \\ 3 & 120 & 1.90 \\ & & \\ \hline \end{array} $$ a. Fit a multiple regression model for predicting catch time using prey length and speed as predictors. b. Predict the catch time for an animal of prey whose length is 6 and whose speed is 50 . c. Is the multiple regression model useful for predicting catch time? Test the relevant hypotheses using \(\alpha=.05\). d. The authors of the article suggest that a simple linear regression model with the single predictor \(x=\frac{\text { length }}{\text { speed }}\) might be a better model for predicting catch time. Calculate the \(x\) values and use them to fit this linear regression model. e. Which of the two models considered (the multiple regression model from Part (a) or the simple linear regression model from Part (d)) would you recommend for predicting catch time? Justify your choice.

The article "Impacts of On-Campus and Off-Campus Work on First-Year Cognitive Outcomes" (Journal of College Student Development \([1994]: 364-370\) ) reported on a study in which \(y=\) spring math comprehension score was regressed against \(x_{1}=\) previous fall test score, \(x_{2}=\) previous fall academic motivation, \(x_{3}=\) age, \(x_{4}=\) number of credit hours, \(x_{5}=\) residence \((1\) if on campus, 0 otherwise), \(x_{6}=\) hours worked on campus, and \(x_{7}=\) hours worked off campus. The sample size was \(n=210\), and \(R^{2}=.543\). Test to see whether there is a useful linear relationship between \(y\) and at least one of the predictors.

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If we knew the width and height of cylindrical tin cans of food, could we predict the volume of these cans with precision and accuracy? a. Give the equation that would allow us to make such predictions. b. Is the relationship between volume and its predictors, height and width, a linear one? c. Should we use an additive multiple regression model to predict a volume of a can from its height and width? Explain. d. If you were to take logarithms of each side of the equation in Part (a), would the relationship be linear?

A number of investigations have focused on the problem of assessing loads that can be manually handled in a safe manner. The article "Anthropometric, Muscle Strength, and Spinal Mobility Characteristics as Predictors in the Rating of Acceptable Loads in Parcel Sorting" (Ergonomics [1992]: \(1033-1044\) ) proposed using a regression model to relate the dependent variable \(y=\) individual's rating of acceptable load \((\mathrm{kg})\) to \(k=3\) independent (predictor) variables: \(x_{1}=\) extent of left lateral bending \((\mathrm{cm})\) $$ \begin{aligned} &x_{2}=\text { dynamic hand grip endurance (sec) } \\ &x_{3}=\text { trunk extension ratio }(\mathrm{N} / \mathrm{kg}) \end{aligned} $$ Suppose that the model equation is $$ y=30+.90 x_{1}+.08 x_{2}-4.50 x_{3}+e $$ and that \(\sigma=5\). a. What is the population regression function? b. What are the values of the population regression \(\underline{\mathrm{co}}\) efficients? c. Interpret the value of \(\beta_{1}\). d. Interpret the value of \(\beta_{3}\). e. What is the mean value of rating of acceptable load when extent of left lateral bending is \(25 \mathrm{~cm}\), dynamic hand grip endurance is \(200 \mathrm{sec}\), and trunk extension ratio is \(10 \mathrm{~N} / \mathrm{kg}\) ? f. If repeated observations on rating are made on different individuals, all of whom have the values of \(x_{1}, x_{2}\), and \(x_{3}\) specified in Part (e), in the long run approximately what percentage of ratings will be between \(13.5 \mathrm{~kg}\) and \(33.5 \mathrm{~kg} ?\)

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