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The article "Technology, Productivity, and Industry Structure" (Technological Forecasting and Social Change [1983]: 1-13) included the accompanying data on \(x=\) research and development expenditure and \(y=\) growth rate for eight different industries. $$ \begin{array}{rrrrrrrrr} x & 2024 & 5038 & 905 & 3572 & 1157 & 327 & 378 & 191 \\ y & 1.90 & 3.96 & 2.44 & 0.88 & 0.37 & -0.90 & 0.49 & 1.01 \end{array} $$ a. Would a simple linear regression model provide useful information for predicting growth rate from research and development expenditure? Use a \(.05\) level of significance. b. Use a \(90 \%\) confidence interval to estimate the average change in growth rate associated with a 1 -unit increase in expenditure. Interpret the resulting interval.

Short Answer

Expert verified
Yes, a simple linear regression model can provide useful information for predicting growth rate from research and development expenditure, if the p-value is less than 0.05. The average increase in growth rate corresponding to a 1-unit increase in expenditure can be estimated using the 90% confidence interval for the slope.

Step by step solution

01

Calculate Pearson Correlation Coefficient

Pearson’s correlation coefficient measures the strength and direction of the relationship between the two variables. Use the formula: \[ r= \frac{n (\Sigma xy) - (\Sigma x)(\Sigma y)}{\sqrt{[n\Sigma x^2 - (\Sigma x)^2][n \Sigma y^2 - (\Sigma y)^2}}]\], where n is the number of pairs of data, \(\Sigma xy\) is the sum of the products of paired data, \(\Sigma x\) and \(\Sigma y\) are the sums of x and y respectively, and \(\Sigma x^2\) and \(\Sigma y^2\) are the sums of the squares of x and y. If r is close to 1 or -1, a linear model might be appropriate.
02

Conduct Linear Regression Analysis

Use the least squares method to get the parameters of the regression model (slope and y-intercept). The regression equation is of the form: \(y = a + bx\), where b (slope) = \[ \frac{n (\Sigma xy) - (\Sigma x)(\Sigma y)}{n\Sigma x^2 - (\Sigma x)^2}\] and a (y-intercept) = \[ \frac{\Sigma y - b (\Sigma x)}{n}\]. It can be used to predict the value of y for any given x.
03

Hypothesis Testing for Significance of Regression

Test the null hypothesis that the slope (b) equals 0 versus the alternative hypothesis that it does not equal 0, using t-test. If the p-value is less than 0.05, reject the null hypothesis. A significant p-value indicates that the model predicts the outcome variable significantly well.
04

Calculate 90% Confidence Interval for the Slope

Use the formula: \[b \pm t_{(n-2), (1-\alpha/2)} * SE(b)\], where \(SE(b) = \sqrt{\frac{\Sigma (y - \hat y)^2 / (n - 2)}{\Sigma (x - mean \ x)^2}}\]. A 90% confidence interval contains the true slope 90% of the time. The estimate of increase in growth rate corresponding to 1-unit increase in expenditure falls within this interval.
05

Interpret the Result

The resulting interval can be interpreted as follows: We can say with 90% confidence that for every 1-unit increase in research and development expenditure, the average change in the growth rate falls within the calculated interval.

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Key Concepts

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

Pearson Correlation Coefficient
Understanding the relationship between two variables is crucial in statistics, and one way to measure this relationship is by using the Pearson correlation coefficient. This coefficient, denoted as \( r \), helps determine the strength and direction of a linear relationship between paired data. The formula is:\[r = \frac{n (\Sigma xy) - (\Sigma x)(\Sigma y)}{\sqrt{[n\Sigma x^2 - (\Sigma x)^2][n \Sigma y^2 - (\Sigma y)^2}}]\]Here, \( n \) is the number of data pairs, \( \Sigma xy \) is the sum of the products of paired data points, \( \Sigma x \) and \( \Sigma y \) are the sums of x and y values, and \( \Sigma x^2 \) and \( \Sigma y^2 \) are the sums of the squares of x and y respectively. If \( r \) is close to 1, it suggests a strong positive linear relationship, while a value close to -1 suggests a strong negative linear relationship.
Values near zero indicate a weak or no linear relationship. Calculating \( r \) can help decide if a linear regression model is suitable for the dataset.
Least Squares Method
Once you've established a potential relationship using the Pearson correlation coefficient, the next step involves finding the best-fit line for your data through simple linear regression. The least squares method is a popular approach for this, minimizing the sum of squares of the vertical distances of the points from the line.

The linear regression equation is: \[ y = a + bx \]where \( b \) is the slope, calculated using:\[ b = \frac{n (\Sigma xy) - (\Sigma x)(\Sigma y)}{n\Sigma x^2 - (\Sigma x)^2} \]and \( a \) is the y-intercept:\[ a = \frac{\Sigma y - b (\Sigma x)}{n} \]With this equation, you can predict the value of \( y \) (growth rate) for any value of \( x \) (R&D expenditure). This method ensures the line has the best possible fit through the data by reducing prediction errors.
Hypothesis Testing
In evaluating the significance of your regression model, hypothesis testing is an essential step. For simple linear regression, the focus is on testing the significance of the slope \( b \). Here you test:
  • Null hypothesis \( (H_0) \): The slope \( b \) is equal to zero. There is no relationship between the variables.
  • Alternative hypothesis \( (H_a) \): The slope \( b \) is not equal to zero. A relationship exists.
The t-test is used to determine if \( b \) significantly differs from zero. If the p-value is less than the given level of significance (like 0.05), you reject the null hypothesis. This implies that the relationship described by the regression model is statistically significant. Ensuring the significance of \( b \) means your predictions and interpretations have real-world relevance.
Confidence Interval
Confidence intervals provide a range within which you can expect the true parameter value to lie, with a certain level of confidence. For regression, this often involves estimating parameters like the slope \( b \). A 90% confidence interval can be calculated as:\[ b \pm t_{(n-2), (1-\alpha/2)} * SE(b) \]where \( SE(b) \) is the standard error of the slope, given by:\[ SE(b) = \sqrt{\frac{\Sigma (y - \hat y)^2 / (n - 2)}{\Sigma (x - \text{mean} \ x)^2}} \]Here, \( \hat y \) are predicted values. This interval suggests that the true slope will fall within this range 90% of the time.
It allows you to estimate how much the growth rate might change for every 1-unit increase in expenditure. Confidence intervals provide a range of plausible slopes, contributing to a deeper understanding of the regression relationship.

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Most popular questions from this chapter

The accompanying figure is from the article "Root and Shoot Competition Intensity Along a Soil Depth Gradient" (Ecology [1995]: \(673-682\) ). It shows the relationship between above-ground biomass and soil depth within the experimental plots. The relationship is described by the linear equation: biomass \(=-9.85+\) \(25.29(\) soil depth \()\) and \(r^{2}=.65 ; P \leq 0.001 ; n=55 .\) Do you think the simple linear regression model is appropriate here? Explain. What would you expect to see in a plot of the standardized residuals versus \(x\) ?

The article "Improving Fermentation Productivity with Reverse Osmosis" (Food Technology [1984]: \(92-96\) ) gave the following data (read from a scatterplot) on \(y=\) glucose concentration \((\mathrm{g} / \mathrm{L})\) and \(x=\) fermentation time (days) for a blend of malt liquor. $$ \begin{array}{lrrrrrrrr} x & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ y & 74 & 54 & 52 & 51 & 52 & 53 & 58 & 71 \end{array} $$ a. Use the data to calculate the estimated regression line. b. Do the data indicate a linear relationship between \(y\) and \(x\) ? Test using a \(.10\) significance level. c. Using the estimated regression line of Part (a), compute the residuals and construct a plot of the residuals versus \(x\) (that is, of the \((x\), residual \()\) pairs. d. Based on the plot in Part (c), do you think that a linear model is appropriate for describing the relationship between \(y\) and \(x\) ? Explain.

Carbon aerosols have been identified as a contributing factor in a number of air quality problems. In a chemical analysis of diesel engine exhaust, \(x=\) mass \(\left(\mu \mathrm{g} / \mathrm{cm}^{2}\right)\) and \(y=\) elemental carbon \(\left(\mu \mathrm{g} / \mathrm{cm}^{2}\right)\) were recorded ("Comparison of Solvent Extraction and Thermal Optical Carbon Analysis Methods: Application to Diesel Vehicle Exhaust Aerosol" Environmental Science Technology \([1984]: 231-234\) ). The estimated regression line for this data set is \(\hat{y}=31+.737 x\). The accompanying table gives the observed \(x\) and \(y\) values and the corresponding standardized residuals. \(\begin{array}{lrrrrr}x & 164.2 & 156.9 & 109.8 & 111.4 & 87.0 \\ y & 181 & 156 & 115 & 132 & 96 \\ \text { St. resid. } & 2.52 & 0.82 & 0.27 & 1.64 & 0.08\end{array}\) \(\begin{array}{lrrrrr}x & 161.8 & 230.9 & 106.5 & 97.6 & 79.7 \\ y & 170 & 193 & 110 & 94 & 77 \\ \text { St. resid. } & 1.72 & -0.73 & 0.05 & -0.77 & -1.11 \\ x & 118.7 & 248.8 & 102.4 & 64.2 & 89.4 \\ y & 106 & 204 & 98 & 76 & 89 \\ \text { St. resid. } & -1.07 & -0.95 & -0.73 & -0.20 & -0.68 \\ x & 108.1 & 89.4 & 76.4 & 131.7 & 100.8 \\ y & 102 & 91 & 97 & 128 & 88 \\ \text { St. resid. } & -0.75 & -0.51 & 0.85 & 0.00 & -1.49 \\ x & 78.9 & 387.8 & 135.0 & 82.9 & 117.9 \\ y & 86 & 310 & 141 & 90 & 130 \\ \text { St. resid. } & -0.27 & -0.89 & 0.91 & -0.18 & 1.05\end{array}\) a. Construct a standardized residual plot. Are there any unusually large residuals? Do you think that there are any influential observations? b. Is there any pattern in the standardized residual plot that would indicate that the simple linear regression model is not appropriate? c. Based on your plot in Part (a), do you think that it is reasonable to assume that the variance of \(y\) is the same at each \(x\) value? Explain.

Exercise \(5.46\) presented data on \(x=\) squawfish length and \(y=\) maximum size of salmonid consumed, both in \(\mathrm{mm}\). Use the accompanying MINITAB output along with the values \(\bar{x}=343.27\) and \(S_{x x}=69,112.18\) to answer the following questions. The regression equation is Size \(=-89.1=0.729\) length \(\begin{array}{lrrrr}\text { Predictor } & \text { Coef } & \text { Stdev } & \text { t-ratio } & p \\ \text { Constant } & 89.09 & 16.83 & 5.29 & 0.000 \\\ \text { length } & 0.72907 & 0.04778 & 15.26 & 0.000 \\ s=12.56 & R-s q=96.3 \% & R-s q(a d j)=95.9 \% \text { Analysis of } V a\end{array}\) Variance \(\begin{array}{lrrrrr}\text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { p } \\ \text { Regression } & 1 & 36736 & 36736 & 232.87 & 0.000 \\ \text { Error } & 9 & 1420 & 158 & & \\\ \text { Total } & 10 & 38156 & & & \end{array}\) a. Does there appear to be a useful linear relationship between length and size? b. Does it appear that the average change in maximum size associated with a 1 -mm increase in length is less than \(.8 \mathrm{~mm}\) ? State and test the appropriate hypotheses. c. Estimate average maximum size when length is 325 \(\mathrm{mm}\) in a way that conveys information about the precision of estimation. d. How would the estimate when length is \(250 \mathrm{~mm}\) compare to the estimate of Part (c)? Answer without actually calculating the new estimate.

Explain the difference between a confidence interval and a prediction interval. How can a prediction level of \(95 \%\) be interpreted?

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