Question

Six points have these coordinates: $$ \begin{array}{l|llllll} x & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline y & 5.6 & 4.6 & 4.5 & 3.7 & 3.2 & 2.7 \end{array} $$ a. Find the least-squares line for the data. b. Plot the six points and graph the line. Does the line appear to provide a good fit to the data points? c. Use the least-squares line to predict the value of \(y\) when \(x=3.5\) d. Fill in the missing entries in the MINITAB analysis of variance table. (Table)

Short answer

Answer: The predicted value of \(y\) when \(x = 3.5\) is approximately \(4.05\).

Step-by-step solution

Calculate the slope and y-intercept for the least-squares line

First, we need to calculate the means of \(x\) and \(y\) values. _mean of x-values:_ \(\bar{x} = \frac{1+2+3+4+5+6}{6} = \frac{21}{6} = 3.5\) _mean of y-values:_ \(\bar{y} = \frac{5.6+4.6+4.5+3.7+3.2+2.7}{6} = \frac{24.3}{6} = 4.05\) Now, we'll calculate slope \(m\) and y-intercept \(b\) of the line using the least-squares method. _slope \(m\):_ \(m = \frac{\sum (x_i - \bar{x})(y_i - \bar{y})}{\sum (x_i - \bar{x})^2}\) First, we calculate \((x_i - \bar{x})(y_i - \bar{y})\) and \((x_i - \bar{x})^2\) for each data point: $$ \begin{array}{c|c|c|c|c} \text{} & (x_{i} - \bar{x}) & (y_{i} - \bar{y}) & (x_{i} - \bar{x})(y_{i} - \bar{y}) & (x_{i} - \bar{x})^{2} \\ \hline 1, 5.6 & -2.5 & 1.55 & -3.875 & 6.25 \\ 2, 4.6 & -1.5 & 0.55 & -0.825 & 2.25 \\ 3, 4.5 & -0.5 & 0.45 & -0.225 & 0.25 \\ 4, 3.7 & 0.5 & -0.35 & -0.175 & 0.25 \\ 5, 3.2 & 1.5 & -0.85 & 1.275 & 2.25 \\ 6, 2.7 & 2.5 & -1.35 & 3.375 & 6.25 \\ \end{array} $$ Now, we can sum up the last two columns and calculate slope \(m\): $$m= \frac{ (-3.875 - 0.825 - 0.225 - 0.175 + 1.275 + 3.375)}{(6.25 + 2.25 + 0.25 + 0.25 + 2.25 + 6.25)} = \frac{-0.45}{17.5} = -0.0257$$ _y-intercept \(b\):_ \(b = \bar{y} - m\bar{x} = 4.05 - (-0.0257)(3.5) \approx 4.14\) The least-squares line equation is: \(y = -0.0257x + 4.14\)

Plot the points and graph the line

Use graphing software or graph paper to plot the given data points and draw the line \(y = -0.0257x + 4.14\). Check if the line appears to fit the data points well. If the line passes close to most of the points, then the fit is generally considered good.

Predict the value of \(y\) when \(x = 3.5\)

To predict the value of \(y\) when \(x = 3.5\), substitute \(x\) with \(3.5\) in the least-squares line equation: $$y = -0.0257(3.5) + 4.14 \approx 4.05$$ Thus, the predicted value of \(y\) when \(x = 3.5\) is approximately \(4.05\).

Complete the missing entries in the MINITAB analysis of variance table

The MINITAB analysis of variance table is not provided, so we cannot fill in the missing entries.

Key concepts

Slope Calculation

When you're working with Least Squares Regression, one of the key steps is to calculate the slope of the line. The slope, often represented by the letter \(m\), dictates how steep the line is. In layman's terms, it's a measure of how much \(y\) (the dependent variable) changes for a unit change in \(x\) (the independent variable). To calculate the slope, we use the formula: \(m = \frac{\sum (x_i - \bar{x})(y_i - \bar{y})}{\sum (x_i - \bar{x})^2}\). Here's a step-by-step breakdown of how this works:

• Compute the mean or average of the \(x\) values, which is denoted as \(\bar{x}\).
• Similarly, find the mean of the \(y\) values, denoted as \(\bar{y}\).
• For each data point, subtract \(\bar{x}\) from the \(x_i\) value and \(\bar{y}\) from the \(y_i\) value.
• Multiply the result of each \((x_i - \bar{x})\) by \((y_i - \bar{y})\) to get \((x_i - \bar{x})(y_i - \bar{y})\).
• Sum up all of these products to get the numerator for the slope calculation.
• For each \((x_i)\), square \((x_i - \bar{x})\) and then sum them all up to get the denominator of the slope formula.

In the example provided, this step produced a slope \(m = -0.0257\), which tells us that for each additional unit increase in \(x\), \(y\) decreases by -0.0257 units.

Y-intercept Calculation

The y-intercept of a line, represented by \(b\), is the point where the line crosses the y-axis. To find it, you need the slope \(m\) and the mean values of \(x\) and \(y\). The formula to calculate \(b\) is: \(b = \bar{y} - m\bar{x}\). Let's break this down:

• You already have \(\bar{y}\), which is the average of the \(y\) values.
• Multiply the slope \(m\) by \(\bar{x}\), the average of the \(x\) values.
• Subtract the product obtained from \(\bar{y}\) to get \(b\).

In our example, with \(m\) calculated as \(-0.0257\) and \(\bar{x} = 3.5\), the y-intercept is calculated as \(b = 4.05 - (-0.0257)(3.5)\) which approximates to \(4.14\). This value means that our line will intersect the y-axis at \(y = 4.14\). The full equation of the line thus becomes \(y = -0.0257x + 4.14\), a simple model we can now use in further analysis.

Data Plotting

Data plotting is a crucial visual step in understanding how well your least-squares line fits the dataset. Think of it as putting together a puzzle; each point on the graph is a piece. Here's how to create and interpret a plot of your data and line:

• First, lay out a graph with x-values on the horizontal axis and y-values on the vertical axis.
• Plot each point based on its \((x, y)\) coordinate; for example, \((1, 5.6)\), \((2, 4.6)\), etc.
• Next, draw the least-squares line using the equation \(y = -0.0257x + 4.14\).

Once plotted, look at how close the data points are to this line. If most points are near or on the line, it means you have a good fit. However, points further from the line indicate that the model may not be capturing all data trends effectively. In our example, observation shows that the line follows the decline in y-values realistically, pointing to a decent fit.

Predictive Modeling

Predictive modeling is basically the practical application of the least-squares line, allowing you to estimate unknown values. In this exercise, the task is to predict the y-value when \(x\) is \(3.5\). Using our line's equation \(y = -0.0257x + 4.14\), here's how to make a prediction:

• Substitute \(3.5\) for \(x\) in the equation.
• Compute: \(y = -0.0257 \times 3.5 + 4.14\).
• The math here gives you \(4.05\) as the predicted value for \(y\).

This technique is a basic yet powerful method for making forecasts based on a linear relationship. It is widely used in various decision-making scenarios where a quick estimate can be useful, provided the linear assumption holds true.