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Exercise \(13.10\) presented information from a study in which \(y\) was the hardness of molded plastic and \(x\) was the time elapsed since termination of the molding process. Summary quantities included \(n=15 \quad b=2.50 \quad\) SSResid \(=1235.470\) $$ \sum(x-\bar{x})^{2}=4024.20 $$ a. Calculate the estimated standard deviation of the statistic \(b\). b. Obtain a \(95 \%\) confidence interval for \(\beta\), the slope of the true regression line. c. Does the interval in Part (b) suggest that \(\beta\) has been precisely estimated? Explain.

Short Answer

Expert verified
The estimated standard deviation of the statistic \(b\), the 95% confidence interval for the slope \(\beta\), and the interpretation of the precision of the estimated slope \(\beta\) depend on calculations and may vary. To determine them, use the formulas from steps 1 and 2 with the given values of SSResid, n, and \(b\). The interpretation in step 3 depends on the confidence interval resulting from step 2.

Step by step solution

01

Calculate the Standard Error of the Estimate

The standard error of the estimation \(b\) is derived as follows: \[ SE(b) = \sqrt{ \frac{SSResid}{(n-2) \sum (x-\bar{x})^2}} \] The values of \( SSResid = 1235.470 \), \( n = 15 \) and \( \sum (x-\bar{x})^2 = 4024.20 \) are given and can be substituted into the above formula to calculate the value of the standard error of the estimate of \( b \).
02

Calculate the Confidence Interval for \(\beta\)

The 95% confidence interval for the slope \( \beta \) of the regression line is given by the following expression: \[ b \pm t(0.025, n-2)SE(b) \] where \( b = 2.50 \) is the estimated value of the slope, \( t(0.025, n-2) \) is the t-value corresponding to a 0.025 level of significance and \( 13 \) degrees of freedom (15 samples - 2 parameters estimated = 13), and \( SE(b) \) is the standard error of the estimate of \( b \) calculated in the step 1. The t-value \( t(0.025, n-2) \) can be either looked up in a statistical table or calculated by a statistical software. Then the lower and upper limits of the confidence interval can be obtained by solving the above equation.
03

Interpretation of the Confidence Interval for \(\beta\)

The precision of the estimate of the slope \( \beta \) is interpreted based on the confidence interval derived in step 2. If the confidence interval includes zero, we can conclude that \( \beta \) is not significantly different from zero, meaning that the predictor variable \(x\) does not significantly affect the response variable \(y\). Otherwise, if the confidence interval does not contain zero, then \( \beta \) is significantly different from zero and \(x\) does have a significant effect on \(y\). Additionally, a narrower confidence interval indicates a more precise estimation.

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Key Concepts

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

Standard Error of Estimate
The standard error of estimate is a statistical term that assesses the accuracy with which the coefficients in a regression equation represent the data. In simpler terms, it measures how much the observed data points deviate from the regression line's predictions. This error helps in determining how well the regression model fits the data. A small standard error indicates a tight fit, while a larger error suggests a looser fit.

When calculating the standard error of the estimated coefficient, like the slope (\b), we use the formula: \[ SE(b) = \sqrt{ \frac{SSResid}{(n-2) \sum (x-\bar{x})^2}} \]
In this formula, SSResid represents the sum of squares of the residuals, which are differences between the observed and estimated values. The term \((x-\bar{x})^2\) collectively denotes the squared differences between each value of the independent variable and its mean. Here, SSResid is given as 1235.470, with 15 observations, and the sum of squared differences is 4024.20.
With these values, we can say the precision of our slope estimate is directly tied to this standard error—the smaller it is, the more confident we are about our estimate of \b.
Confidence Interval
A confidence interval gives a range of values within which we can say with a certain level of confidence—typically 95%—that the true parameter lies. It is not just a single estimate, but rather an interval that, based on the sample data, is likely to cover the true population parameter. For regression slope estimation, this means we can say with 95% confidence where the true slope of the regression line (represented by \(\beta\)) is likely to be.

To compute the confidence interval for \(\beta\), we start with our estimate of the slope, denote it as \b, and then consider the variability of this estimate, reflected by the standard error. Using a t-distribution, since our sample size is relatively small and we do not necessarily know the standard deviation of the population, we adjust the margin of error to get the interval: \[ b \tpm t(0.025, n-2)SE(b) \]
Here, the t-value depends on our desired confidence level and the degrees of freedom (\(n-2\)), reflecting how many data points contribute to the estimation after accounting for the number of estimated parameters. The closer the interval is to our estimated value without being too wide, the more reassurance we have that our estimation is meaningful and robust.
Regression Slope Estimation
The slope in a regression analysis represents the change in the dependent variable (\(y\)) for a one-unit increase in the independent variable (\(x\)). Essentially, it tells us the rate at which the predictor affects the response variable. When we calculate the slope (\b), we are estimating this effect based on sample data. This estimation, while useful, is subject to the variability of the data and the sample size.

A crucial part of evaluating the reliability of our slope estimation is examining its confidence interval. If the confidence interval does not include zero, it indicates a statistically significant relationship between the predictor and response variable. For instance, if our confidence interval for \(\beta\) ranges from 1.5 to 3.5, we'd say that for every one unit increase in \(x\), \(y\) increases between 1.5 to 3.5 units, on average, and we are 95% confident in this interval.

The width of the confidence interval also provides insight into the estimation's precision. A wide interval may signal high variability or a small sample size, indicating less precision, while a narrow interval suggests higher precision. For the given problem, after calculating the standard error and using it to find the confidence interval, we then examine whether this interval is narrow enough to suggest that the slope was precisely estimated, and infer the strength of the relationship between \(x\) and \(y\).

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Most popular questions from this chapter

Exercise \(5.48\) described a regression situation in which \(y=\) hardness of molded plastic and \(x=\) amount of time elapsed since termination of the molding process. Summary quantities included \(n=15\), SSResid = \(1235.470\), and \(\mathrm{SSTo}=25,321.368\) a. Calculate a point estimate of \(\sigma .\) On how many degrees of freedom is the estimate based? b. What percentage of observed variation in hardness can be explained by the simple linear regression model relationship between hardness and elapsed time?

A study was carried out to relate sales revenue \(y\) (in thousands of dollars) to 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. $$ \begin{aligned} &\sum x=14.10 \quad \sum y=1438.50 \quad \sum x^{2}=13.92 \\ &\sum y^{2}=140,354 \quad \sum x y=1387.20 \\ &\sum(y-\bar{y})^{2}=2401.85 \quad \sum(y-\hat{y})^{2}=561,46 \end{aligned} $$ a. What proportion of observed variation in sales revenue can be attributed to the linear relationship between revenue and advertising expenditure? b. Calculate \(s\), 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.

In Exercise \(13.17\), we considered a regression of \(y=\) oxygen consumption on \(x=\) time spent exercising. Summary quantities given there yield $$ \begin{aligned} &n=20 \quad \bar{x}=2.50 \quad S_{x x}=25 \\ &b=97.26 \quad a=592.10 \quad s_{e}=16.486 \end{aligned} $$ a. Calculate \(s_{a+b(2.0)}\) the estimated standard deviation of the statistic \(a+b(2.0)\). b. Without any further calculation, what is \(s_{a+b(3.0)}\) and what reasoning did you use to obtain it? c. Calculate the estimated standard deviation of the statistic \(a+b(2.8)\). d. For what value \(x^{*}\) is the estimated standard deviation of \(a+b x^{*}\) smallest, and why?

The accompanying data on \(x=\) advertising share and \(y=\) market share for a particular brand of cigarettes during 10 randomly selected years are from the article "Testing Alternative Econometric Models on the Existence of Advertising Threshold Effect" (Journal of Marketing Research [1984]: 298-308). \(\begin{array}{lllllllllll}x & .103 & .072 & .071 & .077 & .086 & .047 & .060 & .050 & .070 & .052\end{array}\) \(\begin{array}{llllllllll}y & .135 & .125 & .120 & .086 & .079 & .076 & .065 & .059 & .051 & .039\end{array}\) 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?

The accompanying data on \(x=\) treadmill run time to exhaustion (min) and \(y=20-\mathrm{km}\) ski time (min) were taken from the article "Physiological Characteristics and Performance of Top U.S. Biathletes" (Medicine and Science in Sports and Exercise [1995]: \(1302-1310)\) : \(\begin{array}{rrrrrrr}x & 7.7 & 8.4 & 8.7 & 9.0 & 9.6 & 9.6 \\ y & 71.0 & 71.4 & 65.0 & 68.7 & 64.4 & 69.4 \\ x & 10.0 & 10.2 & 10.4 & 11.0 & 11.7 & \\\ y & 63.0 & 64.6 & 66.9 & 62.6 & 61.7 & \end{array}\) $$ \begin{aligned} &\sum x=106.3 \quad \sum x^{2}=1040.95 \\ &\sum y=728.70 \quad \sum x y=7009.91 \quad \sum y^{2}=48390.79 \end{aligned} $$ a. Does a scatterplot suggest that the simple linear regression model is appropriate? b. Determine the equation of the estimated regression line, and draw the line on your scatterplot. c. What is your estimate of the average change in ski time associated with a 1 -min increase in treadmill time? d. What would you predict ski time to be for an individual whose treadmill time is \(10 \mathrm{~min} ?\) e. Should the model be used as a basis for predicting ski time when treadmill time is 15 min? Explain. f. Calculate and interpret the value of \(r^{2}\). g. Calculate and interpret the value of \(s_{e}\).

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