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Chapter 9: Problem 65

FIBER IN CEREALS AS A PREDICTOR OF CALORIES In Example 9.10 on page \(592,\) we look at a model to predict the number of calories in a cup of breakfast cereal using the number of grams of sugars. In Exercises 9.64 and 9.65 , we give computer output with two regression intervals and information about a specific amount of sugar. Interpret each of the intervals in the context of this data situation. (a) The \(95 \%\) confidence interval for the mean response (b) The \(95 \%\) prediction interval for the response The intervals given are for cereals with 16 grams of sugars: Sugars 95 \(\mathrm{P}\) \(\begin{array}{rrr}\text { rs Fit } & \text { SE Fit } \\\ 6 & 157.88 & 7.10 & \text { (143.3 }\end{array}\) \(95 \% \mathrm{Cl}\) 35,172.42) \(9 \%\) \(\begin{array}{lllll}16 & 15788 & 7.10 & (143.35,172.42) & (101.46\end{array}\) 214.31)

Short Answer

Expert verified
The 95% confidence interval indicates that the true average calories for cereals with 16 grams of sugars is likely between 143.35 and 172.42. The 95% prediction interval allows us to predict that the calories in a single box of cereal with 16 grams of sugars will be between 101.46 and 214.31.

Step by step solution

01

Interpret the Confidence Interval

The 95% confidence interval given is (143.35, 172.42). This interval suggests that we can be 95% confident that the true average caloric content of all cereals with 16 grams of sugars lies within this range.
02

Interpret the Prediction Interval

The 95% prediction interval given is (101.46, 214.31). This means that we can predict with 95% certainty that the caloric content of an individual cereal box with 16 grams of sugars will fall within this range.
03

Understand The Difference Between Them

Note the difference between these two intervals. The confidence interval is about estimating the true population mean, while the prediction interval is about estimating individual data points.

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Key Concepts

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

prediction interval
A prediction interval provides a range in which we expect a future observation to fall. Unlike a confidence interval, which tells us where the mean of a population likely lies, a prediction interval accounts for more variability, which is why it tends to be wider.
In the cereal example, the 95% prediction interval (101.46 to 214.31) estimates the range of calorie content for a specific cereal box with 16 grams of sugar. This interval is broader because it has to consider both the variability of the average calories and the variability of individual cereal boxes.
Why the wider span? It acknowledges the possibility of differences from the predicted average due to:
  • Measurement errors
  • Inherent natural variance in calorie content among different boxes
  • Any unique characteristics not accounted for in the model
Understanding prediction intervals is particularly important in practical settings where individual outcomes matter.
linear regression
Linear regression is a statistical tool used to establish the relationship between two or more variables. It allows us to predict a dependent variable based on the values of one or more independent variables.
In the fiber and cereal example, linear regression is used to predict the calories in cereals by looking at the grams of sugar they contain. The linear regression model creates an equation of a line, usually in the format:\[ y = mx + b \]
Here, \(y\) is the dependent variable (calories), \(x\) is the independent variable (sugar content), \(m\) is the slope, and \(b\) is the y-intercept.
Key elements of linear regression:
  • Slope (m): Indicates how much \(y\) changes with a one-unit change in \(x\).
  • Intercept (b): The expected mean value of \(y\) when \(x\) is zero.
  • Fit of the model: How well the model explains the variation in data, often assessed by R-squared.
This model is foundational for making informed predictions and assessing decision-making processes in various fields.
statistical analysis
Statistical analysis is an umbrella term that encompasses various methods to analyze data and extract meaningful insights. It's all about applying mathematical theories to grow our understanding of data without being overwhelmed by its complexity.
In our context, statistical analysis helps in understanding the relationship between sugar content and calorie count in cereals. This involves using methods like regression to fit the best line through our data and establishing intervals to predict outcomes.
Why is statistical analysis important?
  • It allows for hypothesis testing, enabling researchers to confirm or refute assumptions.
  • Data-driven decisions can be made more accurately.
  • Statistical models, like those in regression analysis, help discern patterns or trends not immediately visible.
Statistical analysis bridges the gap between raw data and actionable insights, especially when dealing with large datasets.

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Most popular questions from this chapter

Data 9.1 on page 577 introduces the dataset InkjetPrinters, which includes information on all-in-one printers. Two of the variables are Price (the price of the printer in dollars) and CostColor (average cost per page in cents for printing in color). Computer output for predicting the price from the cost of printing in color is shown: $$ \begin{aligned} &\text { The regression equation is Price }=378-18.6 \text { CostColor }\\\ &\begin{array}{lrrrrr} \text { Analysis of Variance } & & & & \\ \text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { P } \\ \text { Regression } & 1 & 57604 & 57604 & 13.19 & 0.002 \\ \text { Residual Error } & 18 & 78633 & 4369 & & \\ \text { Total } & 19 & 136237 & & & \end{array} \end{aligned} $$ (a) What is the predicted price of a printer that costs 10 cents a page for color printing? (b) According to the model, does it tend to cost more or less (per page) to do color printing on a cheaper printer? (c) Use the information in the ANOVA table to determine the number of printers included in the dataset. (d) Use the information in the ANOVA table to compute and interpret \(R^{2}\). (e) Is the linear model effective at predicting the price of a printer? Use information from the computer output and state the conclusion in context.

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 \(\begin{array}{lrrrrr}\text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { P } \\ \text { Regression } & 1 & 7376.1 & 7376.1 & 7.44 & 0.011 \\ \text { Residual Error } & 28 & 27774.1 & 991.9 & & \\\ \text { Total } & 29 & 35150.2 & & & \end{array}\)

Exercise 2.143 on page 102 introduces a study examining years playing football, brain size, and percentile score on a cognitive skills test. We show computer output below for a model to predict Cognition score based on Years playing football. (The scatterplot given in Exercise 2.143 allows us to proceed without serious concerns about the conditions.) Pearson correlation of Years and Cognition \(=-0.366\) P-Value \(=0.015\) Regression Equation Cognition \(=102.3-3.34\) Years Coefficients \(\begin{array}{lrrrr}\text { Term } & \text { Coef } & \text { SE Coef } & \text { T-Value } & \text { P-Value } \\ \text { Constant } & 102.3 & 15.6 & 6.56 & 0.000 \\ \text { Years } & -3.34 & 1.31 & -2.55 & 0.015 \\ & & & & \\ & \text { S } & \text { R-sq } & \text { R-sq(adj) } & \text { R-sq(pred) } \\ 25.4993 & 13.39 \% & 11.33 \% & 5.75 \%\end{array}\) Analysis of Variance \(\begin{array}{lrrrrr}\text { Source } & \text { DF } & \text { Adj SS } & \text { Adj MS } & \text { F-Value } & \text { P-Value } \\\ \text { Regression } & 1 & 4223 & 4223.2 & 6.50 & 0.015 \\ \text { Error } & 42 & 27309 & 650.2 & & \\ \text { Total } & 43 & 31532 & & & \\ & \-- & & & \end{array}\) (a) What is the correlation between these two variables? What is the p-value for testing the correlation? (b) What is the slope of the regression line to predict cognition score based on years playing football? What is the t-statistic for testing the slope? What is the p-value for the test? (c) The ANOVA table is given for testing the effectiveness of this model. What is the F-statistic for the test? What is the p-value? (d) What do you notice about the three p-values for the three tests in parts \((\mathrm{a}),(\mathrm{b}),\) and \((\mathrm{c}) ?\) (e) In every case, at a \(5 \%\) level, what is the conclusion of the test in terms of football and cognition?

Use the computer output (from different computer packages) to estimate the intercept \(\beta_{0},\) the slope \(\beta_{1},\) and to give the equation for the least squares line for the sample. Assume the response variable is \(Y\) in each case. $$ \begin{aligned} &\text { The regression equation is }\\\ &Y=808-3.66 \mathrm{~A}\\\ &\begin{array}{lrrrr} \text { Predictor } & \text { Coef } & \text { SE Coef } & \text { T } & \text { P } \\ \text { Constant } & 807.79 & 87.78 & 9.20 & 0.000 \\ \text { A } & -3.659 & 1.199 & -3.05 & 0.006 \end{array} \end{aligned} $$

Show some computer output for fitting simple linear models. State the value of the sample slope for each model and give the null and alternative hypotheses for testing if the slope in the population is different from zero. Identify the p-value and use it (and a \(5 \%\) significance level) to make a clear conclusion about the effectiveness of the model.$$ \begin{array}{lrrrr} \text { Coefficients: } & \text { Estimate } & \text { Std.Error } & \mathrm{t} \text { value } & \operatorname{Pr}(>|\mathrm{t}|) \\ \text { (Intercept) } & 7.277 & 1.167 & 6.24 & 0.000 \\ \text { Dose } & -0.3560 & 0.2007 & -1.77 & 0.087 \end{array} $$
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