Question

The following data were obtained in an experiment relating the dependent variable, \(y\) (texture of strawberries), with \(x\) (coded storage temperature). $$ \begin{array}{l|rrrrr} x & -2 & -2 & 0 & 2 & 2 \\ \hline y & 4.0 & 3.5 & 2.0 & 0.5 & 0.0 \end{array} $$ a. Find the least-squares line for the data. b. Plot the data points and graph the least-squares line as a check on your calculations. c. Construct the ANOVA table.

Short answer

Based on the given dataset, the least-squares line was calculated as \(y = -0.9x + 2\). After plotting the datapoints and the least-squares line, it was observed that the line is a reasonable fit. In the ANOVA table, the total sum of squares (SST) is 13.5, the regression sum of squares (SSR) is 9.68, and the error sum of squares (SSE) is 3.82. The F statistic is found to be 10.07.

Step-by-step solution

Write down the data points

First, let's write down the given data points in a compact way: $$ \begin{array}{l|rrrrr} x & -2 & -2 & 0 & 2 & 2 \\\ \hline y & 4.0 & 3.5 & 2.0 & 0.5 & 0.0 \\ \end{array} $$

Calculate the means of \(x\) and \(y\)

To find the means, we add up the values in each row and divide by the number of values: $$ \bar{x} = \frac{(-2) + (-2) + 0 + 2 + 2}{5} = 0 \\ \bar{y} = \frac{4.0 + 3.5 + 2.0 + 0.5 + 0.0}{5} = 2 $$

Compute the covariance and variance of \(x\) and \(y\)

The covariance and variance are calculated as follows: $$ cov(x, y) = \frac{\sum_{i=1}^{5} (x_i - \bar{x})(y_i - \bar{y})}{5-1} \\ var(x) = \frac{\sum_{i=1}^{5} (x_i - \bar{x})^2}{5-1} $$ By plugging in the values, we get: $$ cov(x, y) = \frac{(-2)(2) + (-2)(1.5) + (0)(0) + (2)(-1.5) + (2)(-2)}{4} = \frac{-18}{4} = -4.5 \\ var(x) = \frac{(-2)^2 + (-2)^2 + (0)^2 + (2)^2 + (2)^2}{4} = \frac{20}{4} = 5 $$

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

Now we can calculate the slope and y-intercept of the least-squares line: $$ b = \frac{cov(x, y)}{var(x)} = \frac{-4.5}{5} = -0.9 \\ a = \bar{y} - b\bar{x} = 2 - (-0.9)(0) = 2 $$ So, the least-squares line is given by: $$ y = -0.9x + 2 $$

Plot the data points and graph the least-squares line

Now we can plot the data points and the least-squares line to visually check our calculations. You can use graphing software or graph paper to do this. Plot the data points given: $$ (-2, 4.0), \, (-2, 3.5), \, (0, 2.0), \, (2, 0.5), \, (2, 0.0) $$ And then draw the line: $$ y = -0.9x + 2 $$ You should see that the line goes through the middle of the data points and is a reasonable fit.

Compute the SST, SSR, and SSE for the ANOVA table

Next, we need to compute the SST, SSR, and SSE for the ANOVA table: $$ SST = \sum_{i=1}^{5}(y_i - \bar{y})^2 \\ SSR = \sum_{i=1}^{5}(\hat{y}_i - \bar{y})^2 \\ SSE = \sum_{i=1}^{5}(y_i - \hat{y}_i)^2 $$ Where \(\hat{y}_i\) are the predicted values of \(y_i\) from the least-squares line. We can compute these values as follows: $$ SST = (4.0-2)^2+(3.5-2)^2+(2.0-2)^2+(0.5-2)^2+(0.0-2)^2=13.5 \\ $$ $$ \hat{y}_1 = -0.9(-2) + 2 = 4.8 \\ \hat{y}_2 = -0.9(-2) + 2 = 4.8 \\ \hat{y}_3 = -0.9(0) + 2 = 2.0 \\ \hat{y}_4 = -0.9(2) + 2 = 0.2 \\ \hat{y}_5 = -0.9(2) + 2 = 0.2 \\ $$ $$ SSR = (4.8-2)^2+(4.8-2)^2+(2.0-2)^2+(0.2-2)^2+(0.2-2)^2=9.68 \\ $$ $$ SSE = (4.0-4.8)^2+(3.5-4.8)^2+(2.0-2.0)^2+(0.5-0.2)^2+(0.0-0.2)^2=3.82 \\ $$

Construct the ANOVA table

Now we can construct the ANOVA table: | Source | df | Sum of squares | Mean square | F statistic | |-------------|----|----------------|-------------|-------------| | Regression | 1 | 9.68 | 9.68 | 10.07 | | Error | 3 | 3.82 | 1.27 | | | Total | 4 | 13.5 | | | The F statistic is calculated as \(F = \frac{MSR}{MSE} = \frac{9.68}{1.27} = 10.07\). The ANOVA table is now complete.

Key concepts

ANOVA Table

The ANOVA (Analysis of Variance) table is a crucial component in regression analysis, as it helps to determine whether there is a significant relationship between the dependent and independent variables. Let's simplify the ANOVA table using the example from our exercise.

Imagine the ANOVA table as a ledger that helps us account for the total variability in the dependent variable, which in this case is the texture of strawberries. It breaks down this variability into two parts: variability explained by the regression (SSR) and the unexplained variability or error (SSE).

As we constructed in the exercise, the ANOVA table includes columns for sources of variation (Regression and Error), degrees of freedom (df), sum of squares, mean squares, and the F statistic. The degrees of freedom associated with Regression is typically the number of independent variables, and for Error, it's the total number of observations minus the number of parameters estimated (including the intercept).

The Sum of Squares measures variability; SST is the total variability in y, SSR is attributed to the regression line's ability to predict y, and SSE is the error or residual. The Mean Square is the Sum of Squares divided by its respective degrees of freedom, which is used to calculate the F statistic. This statistic tells us the ratio of variance explained by the model to variance unexplained, which is a way to test the overall significance of the regression model. If the F value is larger than the F-critical value from F distribution tables, we reject the null hypothesis that the model with the independent variables is not better than a model with no independent variables.

Covariance and Variance

Understanding the relationship between two variables is essential in regression analysis, and here covariance is the statistic that measures this. Covariance indicates the direction of the linear relationship between variables. If both variables tend to increase or decrease together (positive covariance) or if one increases when the other decreases (negative covariance), this gives us an insight into their correlation.

In our exercise, we calculated the covariance between storage temperature (x) and texture of strawberries (y). A negative covariance of -4.5 indicates that an increase in coded storage temperature tends to be associated with a decrease in the texture rating of strawberries.

Variance, on the other hand, measures how much values of a single variable spread out from the mean. In the context of regression, the variance of the independent variable (x) contributes to the determination of the slope of the least-squares regression line. It's worth highlighting that the variance of x in the exercise was 5, which is strictly positive, affirming that there's variation in our independent variable, a necessity for regression analysis.

Both covariance and variance are building blocks in calculating the slope (b) of the least-squares regression line. By knowing how these two statistics affect our regression line, we can better understand the relationship our data is exhibiting and ensure the reliability of our regression model.

Hypothesis Testing

Hypothesis testing in regression analysis is a statistical method used to make inferences about the population parameters based on sample data. Specifically, we often want to test the significance of our regression coefficients to ensure they are not the result of random chance.

For instance, in the context of the least-squares regression line from our exercise, hypothesis testing can be used to determine whether the slope of the regression line is statistically significantly different from zero. The null hypothesis (H0) generally states there is no effect or no relationship, in our case, it would be that the slope is zero, meaning storage temperature does not affect the texture of strawberries.

To test this, we use the F statistic from the ANOVA table, which compares the variance explained by the model to the unexplained variance. If the F statistic is large, it provides evidence against the null hypothesis. In our exercise, the calculated F statistic of 10.07 would then be compared to a critical value from the F distribution. Exceeding this critical value would lead us to reject the null hypothesis, thus affirming the significance of the storage temperature in explaining changes in the texture of strawberries.

Hypothesis testing in regression not only allows us to infer the relevance of predictors but also validates the utility of the regression model itself, making it a fundamental tool in data analysis.