Question

In Data 9.2 on page 592 , we introduce the dataset Cereal, which has nutrition information on 30 breakfast cereals. Computer output is shown for a linear model to predict Calories in one cup of cereal based on the number of grams of Fiber. Is the linear model effective at predicting the number of calories in a cup of cereal? Give the F-statistic from the ANOVA table, the p-value, and state the conclusion in context. The regression equation is Calories \(=119+8.48\) Fiber Analysis of Variance \(\begin{array}{lrrrrr}\text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { P } \\ \text { Regression } & 1 & 7376.1 & 7376.1 & 7.44 & 0.011 \\ \text { Residual Error } & 28 & 27774.1 & 991.9 & & \\\ \text { Total } & 29 & 35150.2 & & & \end{array}\)

Short answer

Yes, the linear model is effective at predicting the number of calories in a cup of cereal. The F-statistic is 7.44; this is significantly greater than 1 and indicates that the regression model fits the data to some extent. The p-value is 0.011, which is less than 0.05, suggesting that the relationship between fiber and calories is statistically significant.

Step-by-step solution

Understanding The Regression Equation

First, let's understand the regression equation provided - Calories = 119 + 8.48 Fiber. This equation suggests that for each gram of fiber, the caloric content increases by approximately 8.48 calories, starting from a base of 119 calories.

Identifying The F-statistic And P-value

Next, from the given ANOVA table the F-statistic is 7.44 and the P-value is 0.011.

Drawing The Inference And Conclusion

The F-statistic measures how significant the fit of the linear model is. If it is significantly greater than 1, it indicates that the regression model has some validity. In this case, given that the F-statistic is 7.44, it is significantly greater than 1 which indicates that the model has some validity. The p-value is used to determine the significance of the model. In our case, the p-value is 0.011, less than the typically used significance level threshold of 0.05. This entails that the number of grams of fiber is a significant predictor of the calorie count in a cup of cereal.

Key concepts

ANOVA

Analysis of Variance (ANOVA) is a statistical technique used to compare the means of three or more samples to see if at least one sample mean is significantly different from the others. In the context of regression analysis, like the problem we're examining, ANOVA helps us to understand whether there is a statistically significant relationship between the independent variables (in this case, Fiber) and the dependent variable (Calories).
ANOVA breaks down the total variation in the data into two parts: variation due to the regression and the residual error variation. In the given exercise, these components are presented as Sum of Squares in the ANOVA table under 'SS'. The Regression SS shows the variation explained by the model, while the Residual Error SS shows the variation that the model fails to explain. The presented Degrees of Freedom (DF) help in calculating the Mean Square values (MS), which are used along with the F-statistic to determine the model's validity.

F-statistic

The F-statistic is a ratio that compares the model's explained variance to the unexplained variance, essentially measuring how well the model fits the data. It is calculated by dividing the Mean Square due to Regression (MSR) by the Mean Square due to Residual Error (MSE). In the linear regression model exercise, the calculated F-statistic is 7.44, which indicates the ratio of explained to unexplained variance.
An F-statistic significantly greater than 1 suggests that the predictor has an association with the response variable. In our case, with an F-statistic of 7.44, we infer that there is a valid relationship between the amount of fiber and the number of calories. This higher F-statistic signals that the variation captured by the regression is not merely due to chance.

p-value

The p-value is a fundamental concept in hypothesis testing used to measure the probability that the observed data would occur by random chance if there were no true effect or relationship. In simpler terms, it tells us how surprising the data is under a null hypothesis which assumes no effect.
The exercise quotes a p-value of 0.011, which means there is a 1.1% probability that fiber and calorie content would be this closely related if fiber actually had no effect on calories. Since this p-value is below the common alpha level of 0.05, we can reject the null hypothesis, concluding that the relationship between fiber and calories is statistically significant and not due to a random fluctuation in the data.

Statistical Significance

Statistical significance is the likelihood that a result or relationship is caused by something other than mere random chance. Statistical significance is quantified by the p-value. In most social science research, a threshold of 0.05 (or 5%) is used to judge whether an effect is statistically significant.
In this exercise, since the p-value (0.011) is less than the significance level of 0.05, we conclude that the regression model is statistically significant. This indicates that there is only a 1.1% chance that the relationship between fiber and calories in a cup of cereal could be an outcome of random variation, reinforcing the belief that the linear regression model is effective in predicting calorie content based on fiber.