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Weeds among the corn Lamb's-quarter is a common weed that interferes with the growth of corn. An agriculture researcher planted corn at the same rate in 16 small plots of ground and then weeded the plots by hand to allow a fixed number of lamb'squarter plants to grow in each meter of corn row. The decision of how many of these plants to leave in each plot was made at random. No other weeds were allowed to grow. Here are the yields of corn (bushels per acre) in each of the plots: Some computer output from a least-squares regression analysis on these data is shown below. $$ \begin{array}{lllll} \text { Predictor } & \text { Coef } & \text { SE Coef } & \text { T } & \text { P } \\ \text { Constant } & 166.483 & 2.725 & 61.11 & 0.000 \\ \begin{array}{l} \text { Weeds per } \\ \text { meter } \end{array} & -1.0987 & 0.5712 & -1.92 & 0.075 \\ \mathrm{~S}=7.97665 & \mathrm{R}-\mathrm{Sq}=20.9 \% & \mathrm{R}-\mathrm{Sq}(\mathrm{adj}) & =15.3 \% \end{array} $$ (a) What is the equation of the least-squares regression line for predicting corn yield from the number of lamb's quarter plants per meter? Interpret the slope and \(y\) intercept of the regression line in context. (b) Explain what the value of \(s\) means in this setting. (c) Do these data provide convincing evidence at the \(\alpha=0.05\) level that more weeds reduce corn yield? Assume that the conditions for performing inference are met.

Step by step solution

01

Writing the Regression Equation

The least-squares regression line can be expressed as \( \hat{y} = a + bx \), where \( a \) is the intercept and \( b \) is the slope. From the output: \[ a = 166.483 \]\[ b = -1.0987 \] Thus, the regression equation is:\[ \hat{y} = 166.483 - 1.0987x \]The slope \(-1.0987\) indicates that for each additional lamb's-quarter plant per meter, the corn yield decreases by about 1.0987 bushels per acre. The intercept \(166.483\) represents the estimated corn yield when there are no weeds per meter.

02

Understanding the Standard Deviation of Residuals

The value \( s = 7.97665 \) is the standard deviation of the residuals, which measures the average distance that the observed values fall from the regression line. In this context, it indicates that the actual corn yields typically deviate from the predicted yield by about 7.97665 bushels per acre.

03

Hypothesis Testing for the Slope

The null hypothesis \( H_0 \) is that the slope \( b \) of the population regression line is zero, meaning that there is no relationship between the number of weeds and the corn yield. The alternative hypothesis \( H_a \) is that the slope is less than zero, suggesting that more weeds reduce the corn yield. The computer output gives a t-statistic of -1.92 and a p-value of 0.075. Since the p-value (0.075) is greater than the significance level \( \alpha = 0.05 \), we do not have enough evidence to reject \( H_0 \). Therefore, there is not sufficient evidence at the 0.05 level to conclude that more weeds reduce corn yield.

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Key Concepts

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

Least-Squares Regression

Least-squares regression is a statistical method used to determine the best-fitting line through a set of points. In our context, it helps predict how the number of weeds affects corn yield. The regression equation we derived is \( \hat{y} = 166.483 - 1.0987x \). Here, \( \hat{y} \) is the predicted corn yield, and \( x \) is the number of lamb's-quarter plants per meter.

The slope, \(-1.0987\), shows that with each additional weed per meter, the corn yield decreases by approximately 1.0987 bushels per acre. This negative slope suggests an inverse relationship between weeds and corn yield.

The intercept, \(166.483\), is the estimated corn yield when there are no weeds. Thus, even in an ideal situation with no weeds, the yield would be around 166.483 bushels per acre.

Hypothesis Testing

Hypothesis testing in regression analysis helps us understand whether there is a statistically significant relationship between the variables. Here, our null hypothesis \( H_0 \) assumes no relationship, meaning the slope \( b \) equals zero. The alternative hypothesis \( H_a \) suggests that the slope is less than zero, implying more weeds reduce corn yield.

Using the computer output, we have a t-statistic of -1.92 and a p-value of 0.075. By comparing this p-value to our significance level \( \alpha = 0.05 \), we see that 0.075 is greater, which means we fail to reject the null hypothesis. Therefore, we don't have strong enough evidence to say that the number of weeds significantly reduces the corn yield at this significance level.

Standard Deviation of Residuals

The standard deviation of residuals, denoted as \( s \), provides insight into how closely the actual data points cluster around the regression line. For our analysis, \( s = 7.97665 \). This tells us the typical deviation of observed corn yields from predicted yields based on our regression model.

A smaller \( s \) would mean that our model's predictions are very close to the actual yields, while a larger \( s \) suggests more variability in how closely predicted and observed values match. In our case, the deviation of nearly 8 bushels per acre indicates reasonable precision of predictions, though some variability remains.

Regression Analysis

Regression analysis is a powerful tool for investigating relationships between dependent and independent variables. It allows us to quantify the strength and direction of the relationship and make predictive insights. In this exercise, it helped quantify the impact of lamb's-quarter plants on corn yield.

Key output from this analysis includes the coefficients, t-statistic, p-value, and the standard deviation of residuals. Together, these outputs give us a comprehensive view of the relationship and its statistical significance.

Understanding such metrics can guide agricultural planning and weed management, helping maximize yield by revealing the extent to which weeds might affect the productivity of crops.