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Chapter 13: Problem 11

Consider the accompanying data on \(x=\) advertising share and \(y=\) market share for a particular brand of soft drink during 10 randomly selected years. \(\begin{array}{llllllll}x & .103 & .072 & .071 & .077 & .086 & .047 & .060 & .050 & .070 & .052\end{array}\) \(\begin{array}{llllll}y & .135 & .125 & .120 & .086 & .079 & .076 & .065 & .059 & .051 & .039\end{array}\) a. Construct a scatterplot for these data. Do you think the simple linear regression model would be appropriate for describing the relationship between \(x\) and \(y\) ? b. Calculate the equation of the estimated regression line and use it to obtain the predicted market share when the advertising share is \(.09\). c. Compute \(r^{2}\). How would you interpret this value? d. Calculate a point estimate of \(\sigma .\) On how many degrees of freedom is your estimate based?

Short Answer

Expert verified
The scatter plot should show a rough linear relationship. The equation of the regression line from step 2 can be used for prediction in step 3. The value of \(r^{2}\) from step 4 indicates how much of the variation in y can be explained by x. Finally, the value of \(\sigma\) gives an idea of the standard deviation of the y-values around the regression line and is based on 8 degrees of freedom.

Step by step solution

01

Construct Scatterplot

To analyze the relationship between advertising share and market share, a scatterplot needs to be created. Each dot represents a year. The x-values are advertising shares and y-values are the corresponding market shares for each year.
02

Calculate Regression Line Equation

The equation of a linear regression line is given by \(y = a + bx\), where \(a\) is the y-intersect and \(b\) is the slope of the line. Use those formulas to calculate these coefficients: \(b = \frac{n( \Sigma xy) - (\Sigma x)(\Sigma y)}{n( \Sigma x^2) - (\Sigma x)^2}\) and \(a = \frac{\Sigma y - b (\Sigma x)}{n}\) where 'n' is the total number of years in the dataset (10 in this case), and \(\Sigma\) denotes the sum of the values.
03

Predict Market Share

Using the equation of the regression line derived from step 2, substitute \(x = 0.09\) into the equation to predict the market share when the advertising share is 0.09.
04

Compute Coefficient of Determination \(r^2\)

\(r^2\) provides a measure of how well observed outcomes are replicated by the model. It is calculated as: \(r^2 = \frac{(\Sigma xy - n \bar x \bar y)^2}{(\Sigma x^2 - n \bar x^2)(\Sigma y^2 - n \bar y^2)}\) where \(\bar x\) and \(\bar y\) are the means of x and y respectively.
05

Calculate Point Estimate of \(\sigma\)

\(\sigma\) is the standard error of the estimate, it measures the dispersion of the data points. It can be calculated as: \(\sigma = \sqrt{\frac{\Sigma(y - \hat y)^2}{n-2}}\) where \(\hat y\) are the predicted y-values from the regression equation. This estimate is based on the degrees of freedom which are equal to \(n-2\) (which is 8 in this case, since there are 10 data points).

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Key Concepts

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

Scatterplot
A scatterplot is a crucial first step in analyzing the relationship between two variables, such as advertising share and market share. Each point on the scatterplot represents an observation from the dataset, plotted with advertising share on the x-axis and market share on the y-axis.
Creating a scatterplot allows you to visually inspect the data for any noticeable patterns or trends. For instance, you can identify if there's a linear relationship, where the points tend to cluster around a straight line, which would indicate that a simple linear regression model might be appropriate.

The scatterplot will also help display any outliers or deviations from typical patterns, which could influence your analysis. By examining the graphical representation, you gain a first intuitive understanding of how strongly advertising influences market share.
Coefficient of Determination
The coefficient of determination, often represented as r虏, is a key measure in simple linear regression.
It tells us how well our regression line fits the observed data. In mathematical terms, r虏 is the proportion of the variance in the dependent variable (market share) that is predictable from the independent variable (advertising share).

Ranging between 0 and 1, an r虏 value closer to 1 indicates a strong linear relationship where most of the variability in market share can be explained by changes in advertising share. Conversely, a value closer to 0 suggests a weak relationship, implying that other factors are influencing market share.

Interpreting r虏 involves understanding that this value doesn't imply causation. A high r虏 merely suggests a strong correlation, not that advertising directly causes changes in market share.
Standard Error of the Estimate
The standard error of the estimate, denoted as 蟽, measures the accuracy of predictions made by the linear regression model. It essentially indicates how much the actual data points deviate from the estimated regression line.

Calculated using the formula \[\sigma = \sqrt{\frac{\Sigma(y - \hat y)^2}{n-2}}\]
this error gives an estimate of the typical distance between observed and predicted values. A smaller standard error of the estimate indicates that the model predictions are, on average, close to the actual data points.

Understanding this metric is crucial because it offers intuitive feedback about the precision of your regression model. If your standard error is high, it may suggest a need to revise the model or consider additional variables.
Degrees of Freedom
Degrees of freedom in the context of regression are the values that have the freedom to vary when estimating statistical parameters, such as the standard error of the estimate.

For a simple linear regression with n data points, the degrees of freedom for the error is n-2. This reduction by two accounts for estimating the slope and the intercept of the regression line. In our exercise, with 10 observations, the degrees of freedom are 8.

This concept is critical because it influences the calculation of the standard error and subsequent statistical tests. Fewer degrees of freedom can increase the uncertainty of your estimates and make it harder to draw strong conclusions from your data.

Understanding degrees of freedom helps you gauge the reliability of your regression analysis and ensures you're using the proper statistics for evaluation.

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Most popular questions from this chapter

A simple linear regression model was used to describe the relationship between sales revenue \(y\) (in thousands of dollars) and advertising expenditure \(x\) (also in thousands of dollars) for fast-food outlets during a 3 -month period. A sample of 15 outlets yielded the accompanying summary quantities. $$ \begin{aligned} &\sum x=14.10 \quad \sum y=1438.50 \quad \sum x^{2}=13.92 \\ &\sum y^{2}=140,354 \quad \sum x y=1387.20 \\ &\sum(y-\vec{y})^{2}=2401.85 \quad \sum(y-\hat{y})^{2}=561.46 \end{aligned} $$ a. What proportion of observed variation in sales revenue can be attributed to the linear relationship between revenue and advertising expenditure? b. Calculate \(s_{e}\) and \(s_{b}\) c. Obtain a \(90 \%\) confidence interval for \(\beta\), the average change in revenue associated with a \(\$ 1000\) (that is, 1 -unit) increase in advertising expenditure.

A sample of \(n=10,000(x, y)\) pairs resulted in \(r=.022\). Test \(H_{0}: \rho=0\) versus \(H_{a^{\circ}} \rho \neq 0\) at significance level .05. Is the result statistically significant? Comment on the practical significance of your analysis.

A regression of \(x=\) tannin concentration \((\mathrm{mg} / \mathrm{L})\) and \(y=\) perceived astringency score was considered in Examples \(5.2\) and \(5.6\). The perceived astringency was computed from expert tasters rating a wine on a scale from 0 to 10 and then standardizing the rating by computing a \(z\) -score. Data for 32 red wines (given in Example 5.2) was used to compute the following summary statistics and estimated regression line: $$ \begin{aligned} &n=32 \quad \bar{x}=.6069 \quad \sum(x-\bar{x})^{2}=1.479 \\ &\text { SSResid }=1.936 \quad \hat{y}=-1.59+2.59 x \end{aligned} $$ a. Calculate a \(95 \%\) confidence interval for the mean astringency rating for red wines with a tannin concentration of \(.5 \mathrm{mg} / \mathrm{L}\). b. When two \(95 \%\) confidence intervals are computed, it can be shown that the simultaneous confidence level is at least \([100-2(5)] \%=90 \%\). That is, if both intervals are computed for a first sample, for a second sample, for a third sample, and so on, in the long run at least \(90 \%\) of the samples will result in intervals which both capture the values of the corresponding population characteristics. Calculate confidence intervals for the mean astringency rating when the tannin concentration is \(.5 \mathrm{mg} / \mathrm{L}\) and when the tannin concentration is \(.7 \mathrm{mg} / \mathrm{L}\) in such a way that the simultaneous confidence level is at least \(90 \%\). c. If two \(99 \%\) confidence intervals were computed, what do you think could be said about the simultaneous confidence level? d. If a \(95 \%\) confidence interval were computed for the mean astringency rating when \(x=.5\), another confidence interval was computed for \(x=.6\), and yet another one for \(x=.7\), what do you think would be the simultaneous confidence level for the three resulting intervals?

Exercise \(13.21\) gave data on \(x=\) nerve firing frequency and \(y=\) pleasantness rating when nerves were stimulated by a light brushing stoke on the forearm. The \(x\) values and the corresponding residuals from a simple linear regression are as follows: a. Construct a standardized residual plot. Does the plot exhibit any unusual features? b. A normal probability plot of the standardized residuals follows. Based on this plot, do you think it is reasonable to assume that the error distribution is approximately normal? Explain.

The accompanying data were read from a plot (and are a subset of the complete data set) given in the article . The data represent the mean response times for a group of individuals with closed-head injury (CHI) and a matched control group without head injury on 10 different tasks. Each observation was based on a different study, and used different subjects, so it is reasonable to assume that the observations are independent. \begin{tabular}{ccc} & \multicolumn{2}{l} { Mean Response Time } \\ \cline { 2 - 3 } Study & Control & CHI \\ \hline 1 & 250 & 303 \\ 2 & 360 & 491 \\ 3 & 475 & 659 \\ 4 & 525 & 683 \\ 5 & 610 & 922 \\ 6 & 740 & 1044 \\ 7 & 880 & 1421 \\ 8 & 920 & 1329 \\ 9 & 1010 & 1481 \\ 10 & 1200 & 1815 \\ \hline \end{tabular} a. Fit a linear regression model that would allow you to predict the mean response time for those suffering a closed-head injury from the mean response time on the same task for individuals with no head injury. b. Do the sample data support the hypothesis that there is a useful linear relationship between the mean response time for individuals with no head injury and the mean response time for individuals with CHI? Test the appropriate hypotheses using \(\alpha=.05\). c. It is also possible to test hypotheses about the \(y\) intercept in a linear regression model. For these data, the null hypothesis \(H_{0}: \alpha=0\) cannot be rejected at the \(.05\) significance level, suggesting that a model with a \(y\) intercept of 0 might be an appropriate model. Fitting such a model results in an estimated regression equation of \(\mathrm{CHI}=1.48(\) Control \()\) Interpret the estimated slope of \(1.48\).
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