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Is the plant density of a species related to the altitude at which data are collected? Let \(Y\) denote the species density and \(X\) denote the altitude. A fit of a simple linear regression model using 14 observations yielded \(\hat{y}=21.6-7.79 x\) and \(r^{2}=.61\) a. What is the value of the correlation coefficient \(r ?\) b. What proportion of the variation in densities is explained by the linear model using altitude as the independent variable? c. Is there sufficient evidence at the \(\alpha=.05\) to indicate that plant densities decrease with an increase in altitude?

Short Answer

Expert verified
a: \( r = -0.78 \). b: 61% variation explained. c: Likely evidence of decrease, more data needed for p-value.

Step by step solution

01

Interpreting R-squared

The R-squared value given is 0.61, which represents the proportion of variation in the dependent variable (plant density \( Y \)) that is explained by the independent variable (altitude \( X \)) in the regression model. Thus, 61% of the variation in species density is explained by the model.
02

Calculating the Correlation Coefficient

The correlation coefficient \( r \) can be calculated from the R-squared value using the formula \( r = \sqrt{r^2} \). Since the slope of the regression line is negative, \( r \) will also be negative. Therefore, \( r = -\sqrt{0.61} \approx -0.78 \). This value reflects a strong negative correlation between plant density and altitude.
03

Hypothesize and Set Alpha Level

The hypothesis to test whether plant densities decrease with an increase in altitude is: \( H_0: \beta = 0 \) (no relationship) and \( H_a: \beta < 0 \) (negative relationship). The significance level \( \alpha \) is 0.05.
04

Conducting the t-test for the Slope

The slope of the regression line is \(-7.79\). A t-test evaluates if this value is significantly different from zero. Using the standard formula \( t = \frac{b - 0}{SE_b} \) where \( b = -7.79 \) and \( SE_b \) is the standard error of the slope. Since \( SE_b \) isn’t provided, the t-statistic and p-value are not calculable without additional data. However, statistical software should compute them when available.
05

Decision Making and Conclusion

If the p-value of the t-test is less than 0.05, we reject the null hypothesis. This would indicate that there is sufficient evidence to say plant densities significantly decrease as altitude increases. Without the exact p-value, we infer based on the negative slope and correlation; plant density likely decreases with altitude.

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Key Concepts

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

Correlation Coefficient
The correlation coefficient, denoted by \( r \), measures the strength and direction of the linear relationship between two variables. In the context of our exercise, \( r \) quantifies how species density (\( Y \)) changes with altitude (\( X \)). If \( r \) is positive, as one variable increases, so does the other. If \( r \) is negative, as one variable increases, the other decreases.
To find \( r \) when you have an \( R^2 \) value, use the formula \( r = \pm\sqrt{R^2} \). Given an \( R^2 \) of 0.61 in this exercise, \( r = \pm\sqrt{0.61} \).
Since the regression slope here is negative (-7.79), it's crucial to consider that in calculating \( r \), making \( r = -0.78 \). This indicates a strong negative correlation, implying a decrease in plant density as altitude increases.
R-squared Value
The \( R^2 \) value, or coefficient of determination, tells us how much of the dependent variable's variation is explained by the independent variable in a regression model. In our example, \( R^2 = 0.61 \), meaning 61% of the variation in plant densities is explained by altitude.
This high \( R^2 \) value suggests a strong relationship in the model, where changes in altitude do a good job of predicting changes in species density. However, remember that \( R^2 \) does not indicate causation or that the model explains all variations possible, as there are still 39% of variations unexplained by altitude alone.
t-test
A t-test is a statistical method used to determine if there is a significant difference between the means of two groups. In linear regression, a t-test helps examine if the slope of the regression line significantly differs from zero.
In this exercise, when testing whether the slope (-7.79) is significantly different from zero, we use the formula \( t = \frac{b - 0}{SE_b} \), where \( b \) is the estimated slope, and \( SE_b \) is its standard error. Without \( SE_b \), you cannot compute \( t \) or the p-value exactly.
In practical scenarios, access to statistical software can provide \( SE_b \), leading to a calculated \( t \)-value and p-value, helping decide if the slope is significant—here indicating if density changes relate to altitude.
Hypothesis Testing
Hypothesis testing is a statistical process used to make inferences or draw conclusions about a population, based on sample data. It involves framing a null hypothesis (\( H_0 \)) and an alternative hypothesis (\( H_a \)).
For our case, \( H_0: \beta = 0 \) posits there is no relationship between altitude and plant density, while \( H_a: \beta < 0 \) suggests a negative relationship.
The significance level, \( \alpha \), is set at 0.05. If your data's p-value is less than \( \alpha \), you reject \( H_0 \). Note that a negative slope and \( r \) hint at a meaningful relation even when a precise p-value isn't available.
This process assists in determining if empirical data supports or contradicts the hypothesis, thus influencing conclusions drawn from statistical models.

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Most popular questions from this chapter

Suppose that we have postulated the model $$Y_{i}=\beta_{1} x_{i}+\varepsilon_{i} \quad i=1,2, \dots, n$$ where the \(\varepsilon_{i}\) 's are independent and identically distributed random variables with \(E\left(\varepsilon_{i}\right)=0 .\) Then \(\hat{y}_{i}=\widehat{\beta}_{1} x_{i}\) is the predicted value of \(y\) when \(x=x_{i}\) and \(\mathrm{SSE}=\sum_{i=1}^{n}\left[y_{i}-\widehat{\beta}_{1} x_{i}\right]^{2} .\) Find the least- squares estimator of \(\beta_{1}\). (Notice that the equation \(y=\beta x\) describes a straight line passing through the origin. The model just described often is called the no-intercept model.)

If \(\widehat{\beta}_{0}\) and \(\widehat{\beta}_{1}\) are the least-squares estimates for the intercept and slope in a simple linear regression model, show that the least-squares equation \(\hat{y}=\hat{\beta}_{0}+\hat{\beta}_{1} x\) always goes through the point \((\bar{x}, \bar{y})\). [Hint: Substitute \(\bar{x}\) for \(x\) in the least-squares equation and use the fact that \(\left.\widehat{\beta}_{0}=\bar{y}-\widehat{\beta}_{1} \bar{x} .\right]\)

Information about eight four-cylinder automobiles judged to be among the most fuel efficient in 2006 is given in the following table. Engine sizes are in total cylinder volume, measured in liters (L). $$\begin{array}{lcc}\text { Car } & \text { Cylinder Volume }(x) & \text { Horsepower }(y) \\\\\hline \text { Honda Civic } & 1.8 & 51 \\ \text { Toyota Prius } & 1.5 & 51 \\\\\text { WW Golf } & 2.0 & 115 \\\\\text { WW Beetle } & 2.5 & 150 \\\\\text { Toyota Corolla } & 1.8 & 126 \\ \text { WW Jetta } & 2.5 & 150 \\\\\text { Mini Cooper } & 1.6 & 118 \\\\\text { Toyota Yaris } & 1.5 & 106\end{array}$$ a. Plot the data points on graph paper. b. Find the least-squares line for the data. c. Graph the least-squares line to see how well it fits the data. d. Use the least-squares line to estimate the mean horsepower rating for a fuel-efficient automobile with cylinder volume \(1.9 \mathrm{L}\).

The correlation coefficient for the heights and weights of ten offensive backfield football players was determined to be \(r=.8261\) a. What percentage of the variation in weights was explained by the heights of the players? b. What percentage of the variation in heights was explained by the weights of the players? c. Is there sufficient evidence at the \(\alpha=.01\) level to claim that heights and weights are positively correlated? d. What is the attained significance level associated with the test performed in part (c)?

Utility companies, which must plan the operation and expansion of electricity generation, are vitally interested in predicting customer demand over both short and long periods of time. A short-term study was conducted to investigate the effect of each month's mean daily temperature \(x_{1}\) and of cost per kilowatt-hour, \(x_{2}\) on the mean daily consumption (in \(\mathrm{kWh}\) ) per household. The company officials expected the demand for electricity to rise in cold weather (due to heating), fall when the weather was moderate, and rise again when the temperature rose and there was a need for air conditioning. They expected demand to decrease as the cost per kilowatt-hour increased, reflecting greater attention to conservation. Data were available for 2 years, a period during which the cost per kilowatt-hour \(x_{2}\) increased due to the increasing costs of fuel. The company officials fitted the model $$Y=\beta_{0}+\beta_{1} x_{1}+\beta_{2} x_{1}^{2}+\beta_{3} x_{2}+\beta_{4} x_{1} x_{2}+\beta_{5} x_{1}^{2} x_{2}+\varepsilon$$ to the data in the following table and obtained \(\hat{y}=325.606-11.383 x_{1}+.113 x_{1}^{2}-21.699 x_{2}+.873 x_{1} x_{2}-.009 x_{1}^{2} x_{2}\) with \(\mathrm{SSE}=152.177\) When the model \(Y=\beta_{0}-\beta_{1} x_{1}+\beta_{2} x_{1}^{2}+\varepsilon\) was fit, the prediction equation was \(\hat{y}=130.009-3.302 x_{1}+.033 x_{1}^{2}\) with \(\mathrm{SSE}=465.134 .\) Test whether the terms involving \(x_{2}\left(x_{2}, x_{1} x_{2}, x_{1}^{2} x_{2}\right)\) contribute to a significantly better fit of the model to the data. Give bounds for the attained significance level.

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