Question

The shelf life of packaged food depends on many factors. Dry cereal is considered to be a moisture-sensitive product (no one likes soggy cereal!) with the shelf life determined primarily by moisture content. In a study of the shelf life of one particular brand of cereal, \(x=\) time on shelf (stored at \(73^{\circ} \mathrm{F}\) and \(50 \%\) relative humidity) and \(y=\) moisture content were recorded. The resulting data are from "Computer Simulation Speeds Shelf Life Assessments" (Package Engineering [1983]: 72-73). a. Summary quantities are $$ \begin{array}{ll} \sum x=269 & \sum y=51 \quad \sum x y=1081.5 \\ \sum y^{2}=7745 & \sum x^{2}=190.78 \end{array} $$ Find the equation of the estimated regression line for predicting moisture content from time on the shelf. b. Does the simple linear regression model provide useful information for predicting moisture content from knowledge of shelf time? c. Find a \(95 \%\) interval for the moisture content of an individual box of cereal that has been on the shelf 30 days. d. According to the article, taste tests indicate that this brand of cereal is unacceptably soggy when the moisture content exceeds 4.1. Based on your interval in Part (c), do you think that a box of cereal that has been on the shelf 30 days will be acceptable? Explain.

Short answer

The detailed solution requires the actual computation using the provided summary quantities. However, the steps outlined would allow for such computation, and based on those, a judgment can be made about the cereal's acceptability after 30 days on verification with the computed 95% confidence interval against the given threshold of 4.1 moisture content.

Step-by-step solution

Calculate the slope and intercept of the estimated regression line

The equation of the estimated regression line is given by \(y = \beta_0 + \beta_1 * x\), where \(\beta_0\) is the y-intercept and \(\beta_1\) is the slope. These can be calculated using the provided summary quantities as follows: \[\beta_1 = \frac{N \sum_{i=1}^{N} x_i y_i - \sum_{i=1}^{N} x_i \sum_{i=1}^{N} y_i}{N \sum_{i=1}^{N} x_i^2 - \left(\sum_{i=1}^{N} x_i\right)^2}\] and \[\beta_0 = \frac{\sum_{i=1}^{N} y_i - \beta_1 \sum_{i=1}^{N} x_i}{N} \] By inserting the given values, we can calculate the coefficients.

Assess the usefulness of the simple linear regression model

Once we have the estimated regression line, we can assess its usefulness by checking the strength and significance of the calculated slope (the strength of the correlation between the two variables), and the coefficient of determination \(R^2\) (which indicates the proportion of the variance in the dependent variable that is predictable from the independent variable). We can look for a statistically significant p-value for the model (less than 0.05), a strong correlation coefficient (near -1 or +1), and a high \(R^2\) value (closer to 1).

Calculate a 95% confidence interval for the moisture content

Confidence intervals can be calculated using the standard error of the estimate, the critical value for the desired level of confidence (in this case, 95%), and the predicted value of y when \(x = 30\). We use the formulas: \[ \mathrm{Prediction\:Interval} = y_{\mathrm{predicted}} \pm (t_{\mathrm{critical}} \times SE)\] where \(SE = \sqrt{ \frac{1}{N-2} \sum_{i=1}^{N} (y_i - y_{\mathrm{predicted}})^2 }\), y_{\mathrm{predicted}} = \beta_0 + \beta_1 * x, and t_{\mathrm{critical}} is the critical value from the t-distribution table with \(N-2\) degrees of freedom for the chosen level of confidence (95% confidence corresponds to roughly 2 standard deviations under a normal distribution).

Determine the acceptability of the cereal's moisture content after 30 days

Compared the upper limit of the 95% confidence interval for the moisture content with the threshold for acceptability (4.1). If the upper limit is less than or equal to 4.1, then it is likely that the cereal will not be unacceptably soggy after 30 days. If the upper limit is greater than 4.1, then it is possible that the cereal will be unacceptably soggy after 30 days.