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Chapter 9: Problem 625

Plot the dependent variable against the Independent variable. Find the least squares line for this data. What is the Y-intercept? If \(3.0\) units of fertilizer were used what would be a good guess as to the resultant corn yield? $$ \begin{array}{l|l|l|l|l|l|l|l|l|l} \mathrm{X} & \text { Fertilizer } & .3 & .6 & .9 & 1.2 & 1.5 & 1.8 & 2.1 & 2.4 \\\ \hline \mathrm{Y} & \text { Corn Yield } & 10 & 15 & 30 & 35 & 25 & 30 & 50 & 45 \end{array} $$

Short Answer

Expert verified
The least squares line for the given data is \(Y = 28.57X - 8.47\). The Y-intercept is -8.47. When 3.0 units of fertilizer are used, a good guess for the resultant corn yield would be approximately 77.24 units.

Step by step solution

01

Calculate the mean of the variables

First, let's calculate the mean (average) values for both variables. \( \bar{X} = \frac{\sum X_{i}}{n}\) \( \bar{Y} = \frac{\sum Y_{i}}{n}\) where n is the number of observations. Using the given data: n = 8 \( \bar{X} = \frac{0.3 + 0.6 + 0.9 + 1.2 + 1.5 + 1.8 + 2.1 + 2.4}{8} = 1.35\) \( \bar{Y} = \frac{10 + 15 + 30 + 35 + 25 + 30 + 50 + 45}{8} = 30\)
02

Calculate the slope and Y-intercept of the least squares line

Next, we calculate the slope (m) and Y-intercept (b) of the least squares line using the following formulas respectively: \(m = \frac{\sum (X_{i} - \bar{X})(Y_{i} - \bar{Y})}{\sum (X_{i} - \bar{X})^{2}}\) \(b = \bar{Y} - m\bar{X}\) Calculating the slope (m): \( m = \frac{(0.3-1.35)(10-30) + (0.6-1.35)(15-30) + \dots + (2.4-1.35)(45-30)}{(0.3-1.35)^2 + (0.6-1.35)^2 + \dots + (2.4-1.35)^2} = 28.57 \) Calculating the Y-intercept (b): \( b = 30 - 28.57\cdot1.35 = -8.47 \) Now we have the least squares line equation: \(Y = 28.57X - 8.47\)
03

Plot the dependent variable against the independent variable with the least squares' line

Using the dataset and the least squares line equation, plot the dependent variable (corn yield) against the independent variable (fertilizer) and draw the best-fit line.
04

Estimate the corn yield when 3.0 units of fertilizer are used

Now, we will use the least squares line equation to predict the corn yield when 3.0 units of fertilizer are used. Plug X = 3.0 into the equation: \(Y = 28.57(3.0) - 8.47\) \(Y = 77.24\) A good guess for the corn yield when 3.0 units of fertilizer are used would be approximately 77.24 units.

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Key Concepts

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

Dependent and Independent Variables
Understanding the relationship between dependent and independent variables is a fundamental aspect of linear regression. When we gather data, such as the effect of fertilizer on corn yield, we are often looking for how one variable responds to changes in another.

The independent variable is the one that is changed or controlled in a scientific experiment to test the effects on the dependent variable. In the given exercise, the amount of fertilizer is the independent variable because it's the factor we are changing. Conversely, the dependent variable is what you measure in the experiment and what is affected during the experiment. Here, the corn yield is the dependent variable, as it is the outcome we're investigating.

The two variables have what's called a 'cause and effect' relationship where the independent variable (fertilizer) is the cause and the dependent variable (corn yield) is the effect. This distinction is crucial when plotting data for analysis or interpreting results, as it helps us to understand which variable is influencing the other.
Y-intercept
The Y-intercept of a line in a two-dimensional graph is where the line crosses the Y-axis. It is the value of the dependent variable when the independent variable is zero. This concept is vital when we're dealing with linear equations of the form Y = mX + b, where 'm' is the slope and 'b' is the Y-intercept.

In our case, the Y-intercept can be interpreted as the expected corn yield when no fertilizer is used, which gives us a baseline of our dataset's expected behavior. To calculate the Y-intercept, you use both the slope of the line and the means of X and Y, as shown in the solution. In our exercise, the Y-interceptor (b) was found to be approximately -8.47. This indicates that, with no fertilizer usage, the model predicts a negative corn yield, which is not physically plausible but serves to highlight that the linear model is a simplification and tends to be more accurate within the range of observed data points rather than outside it.
Predicting Outcomes with Linear Regression
Linear regression is a statistical method used to create a linear model that can predict the dependent variable's values based on the independent variable. It involves finding the best-fit line through the data, which is accomplished using the least squares method.

The idea is to minimize the sum of the squares of the vertical distances (residuals) between the observed values and the line鈥檚 predicted values. The least squares line then allows us to predict outcomes. As demonstrated in the exercise, with the equation of the least squares line, Y = 28.57X - 8.47, we can predict the corn yield for a given amount of fertilizer. For instance, with 3.0 units of fertilizer, the predicted corn yield would be approximately 77.24 units.

It's important to note that predictions with this method are best used within the range of the data from which the model was created, as extrapolation beyond this range can lead to less reliable predictions.

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Most popular questions from this chapter

In investigating several complaints concerning the weight of the "NET WT. 12 OZ." jar of a local brand of peanut butter, the Better Business Bureau selected a sample of 36 jars. The sample showed an average net weight of \(11.92\) ounces and a standard deviation of \(.3\) ounce. Using a \(.01\) level of significance, what would the Bureau conclude about the operation of the local firm?

Consider a probability distribution for random orientations in which the probability of an observation in a region on the surface of the unit hemisphere is proportional to the area of that region. Two angles, \(u\) and \(v\), will determine the position of an observation. It can be shown that the position of an observation is jointly distributed with density function $$ \begin{array}{r} \mathrm{f}(\mathrm{u}, \mathrm{v})=[\\{\sin \mathrm{u}\\} /\\{2 \pi\\}] \quad 0<\mathrm{u}<2 \pi \\ 0<\mathrm{u}<\pi / 2 . \end{array} $$ Two new variables, \(\mathrm{X}\) and \(\mathrm{Y}\) are defined, where $$ \mathrm{X}=\sin \mathrm{u} \cos \mathrm{v} $$ $$ \mathrm{Y}=\sin \mathrm{u} \sin \mathrm{v} $$ Find the joint density function of \(\mathrm{X}\) and \(\mathrm{Y}\).

Let \(\mathrm{X}\) be a random variable whose value is determined by the flip of a fair coin. If the coin lands heads up \(\mathrm{X}=1\), if tails then \(\mathrm{X}=0\). Find the expected value of \(\mathrm{X}\).

A sample of size 49 yielded the values \(\underline{x}=87.3\) and \(s^{2}=162 .\) Test the hypothesis that \(\mu=95\) versus the alternative that it is less. Let \(\alpha=.01\).

Consider the joint distribution of \(\mathrm{X}\) and \(\mathrm{Y}\) given in the form of a table below. The cell (i,j) corresponds to the joint probability that \(\mathrm{X}=\mathrm{i}, \mathrm{Y}=\mathrm{j}\), for \(\mathrm{i}=1,2,3, \mathrm{j}=1,2,3\) $$ \begin{array}{|c|c|c|c|} \hline \mathrm{Y}^{\mathrm{X}} & 1 & 2 & 3 \\ \hline 1 & 0 & 1 / 6 & 1 / 6 \\ \hline 2 & 1 / 6 & 0 & 1 / 6 \\ \hline 3 & 1 / 6 & 1 / 6 & 0 \\ \hline \end{array} $$ Check that this is a proper probability distribution. What is the marginal distribution of \(\mathrm{X} ?\) What is the marginal distribution of \(\mathrm{Y}\) ?
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