/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 43 The shelf life of packaged food ... [FREE SOLUTION] | 91Ó°ÊÓ

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The shelf life of packaged food depends on many factors. Dry cereal is considered to be a moisturesensitive 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 (days stored at \(73^{\circ} \mathrm{F}\) and \(50 \%\) relative humidity) and \(y=\) moisture content \((\%)\) were recorded. The resulting data are from. \(\begin{array}{rrrrrrrr}x & 0 & 3 & 6 & 8 & 10 & 13 & 16 \\ y & 2.8 & 3.0 & 3.1 & 3.2 & 3.4 & 3.4 & 3.5 \\ x & 20 & 24 & 27 & 30 & 34 & 37 & 41 \\ y & 3.1 & 3.8 & 4.0 & 4.1 & 4.3 & 4.4 & 4.9\end{array}\) a. Summary quantities are \(\sum x=269 \quad \sum y=51 \quad \sum x y=1081.5\) $$ \sum y^{2}=190.78 \quad \sum x^{2}=7745 $$ 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 artide, 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

Expert verified
a. The estimated regression line equation will be calculated from provided quantities. b. The linear regression model's usefulness will be determined from a significance test. c. A 95% prediction interval for a 30-day old cereal will be calculated. d. If the upper limit of the 95% prediction interval is below or equivalent to 4.1%, the 30-day old cereal box is acceptable. Otherwise, it is unacceptable.

Step by step solution

01

Calculate the estimated regression line

Use summary quantities to estimate the regression line parameters 'a' (intercept) and 'b' (slope). The formula for 'b' (slope) is: \[ b = \frac{(n \sum xy - \sum x \sum y)}{(n \sum x^2 - (\sum x)^2)} \], and for 'a' (intercept) is \[ a = \bar{y} - b \bar{x} \]. Here, 'n' is the total number of observations, \(\sum xy\) is the sum of the product of x and y, \(\sum x\) and \(\sum y\) are the sum of x and y observations respectively, while \( \bar{x} \) and \( \bar{y} \) represent the mean of x and y observations respectively. Using the given summary quantities, calculate 'a' and 'b'.
02

Evaluate the usefulness of the model

To assess whether the model is useful for predicting moisture content based on shelf time, conduct a significance test. The null hypothesis (H0) is that the shelf time does not affect the moisture content (b=0), while the alternative hypothesis (H1) is that it does (b≠0). Calculate the t-statistic and p-value and interpret the results. If the p-value is small (typically ≤ 0.05), reject the null hypothesis and conclude that the model is useful.
03

Calculate the prediction interval

Find a 95% prediction interval for the moisture content of cereal that has been on the shelf for 30 days. The formula for the prediction interval is \[ \hat{y} ± t*\sqrt{MSE*(1+1/n+(x-\bar{x})^2/\sum(x-\bar{x})^2)} \], where \(\hat{y}\) is the predicted value for a given x, t is the t-value for the desired level of confidence, n is the total number of observations, x is the value for which the prediction is being done, and MSE is the mean squared error. Here, assume that MSE is constant and can be estimated from the data.
04

Explain the acceptability of the cereal box

Based on the 95% prediction interval calculated in step 3, evaluate whether a cereal box with a shelf life of 30 days will have acceptable moisture content. If the upper limit of the prediction interval is below or equivalent to the threshold (4.1%), then the cereal box is acceptable. Otherwise, it is not.

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Key Concepts

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

Regression Line
A regression line is a fundamental concept in linear regression, which represents the relationship between two variables. In this case, we're trying to predict the moisture content (y) based on the number of days the cereal has been stored (x).
This is known as the 'best-fit' line, and it's derived from the equation of a straight line: \[ y = a + bx \] where 'a' is the intercept, and 'b' is the slope of the line.
The slope 'b' tells us how much y changes for a one-unit change in x. To calculate these values, we use a set of summary statistics derived from the data.
  • To find the slope 'b', the formula used is: \[ b = \frac{(n \sum xy - \sum x \sum y)}{(n \sum x^2 - (\sum x)^2)} \]
  • For the intercept 'a', the formula is: \[ a = \bar{y} - b \bar{x} \]
With these calculations, we can plot the regression line, which helps in making predictions and understanding the trend of the data.
Prediction Interval
A prediction interval gives us a range of values for the dependent variable, in this case, moisture content, where we expect an individual new observation to fall with a specified probability, such as 95%.
This is particularly useful when you're making predictions for future data points and you want to understand how precise your predictions are.
The formula for a prediction interval is given by:\[ \hat{y} \pm t*\sqrt{MSE*(1+1/n+(x-\bar{x})^2/\sum(x-\bar{x})^2)} \] where:
  • \(\hat{y}\) is the predicted value at a specific value of x
  • \(t\) is the t-value correlating to the confidence level chosen
  • MSE is the mean squared error
  • \(n\) is the number of observations
This interval accounts for variability not only in the dataset but also in the prediction itself. In other words, it represents how confident we are in predicting the moisture content for a cereal box that has been on the shelf for a specified number of days.
Significance Test
A significance test in linear regression helps us decide whether the predictor variable, which is 'time on shelf' in this scenario, significantly impacts the response variable, 'moisture content'.
It essentially checks if there's a meaningful relationship between the two variables.
For this, we establish two hypotheses:
  • Null hypothesis (\(H_0\)): The slope \(b\) is zero, meaning the shelf time does not affect moisture content.
  • Alternative hypothesis (\(H_1\)): The slope \(b\) is not zero, indicating an effect of shelf time on moisture content.
We'll use a t-statistic to determine significance:\[ t = \frac{b}{SE_b} \] where \(SE_b\) is the standard error of the slope.
If the computed p-value from this t-statistic is less than a critical value (usually 0.05 for a 5% significance level), we reject the null hypothesis in favor of the alternative, suggesting the shelf time is a significant predictor for moisture content.
This allows us to be confident in using the regression model for predictions.

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Most popular questions from this chapter

Occasionally an investigator may wish to compute a confidence interval for \(\alpha\), the \(y\) intercept of the true regression line, or test hypotheses about \(\alpha\). The estimated \(y\) intercept is simply the height of the estimated line when \(x=0\), since \(a+b(0)=a\). This implies that \(s_{0}\) the estimated standard deviation of the statistic \(a\), results from substituting \(x^{*}=0\) in the formula for \(s_{a+b \alpha}\). The desired confidence interval is then \(a \pm(t\) critical value \() s_{a}\) and a test statistic is $$ t=\frac{a-\text { hypothesized value }}{s_{a}} $$ a. The article used the simple linear regression model to relate surface temperature as measured by a satellite \((y)\) to actual air temperature \((x)\) as determined from a thermocouple placed on a traversing vehicle. Selected data are given (read from a scatterplot in the article). $$ \begin{array}{rrrrrrrr} x & -2 & -1 & 0 & 1 & 2 & 3 & 4 \\ y & -3.9 & -2.1 & -2.0 & -1.2 & 0.0 & 1.9 & 0.6 \end{array} $$ \(\begin{array}{llll}x & 5 & 6 & 7\end{array}\) \(\begin{array}{llll}y & 2.1 & 1.2 & 3.0\end{array}\) Estimate the population regression line. b. Compute the estimated standard deviation \(s_{a r}\). Carry out a test at level of significance \(.05\) to see whether the \(y\) intercept of the population regression line differs from zero. c. Compute a \(95 \%\) confidence interval for \(\alpha\). Does the result indicate that \(\alpha=0\) is plausible? Explain.

The authors of the paper studied a number of variables they thought might be related to bone mineral density (BMD). The accompanying data on \(x=\) weight at age 13 and \(y=\) bone mineral density at age 27 are consistent with summary quantities for women given in the paper. A simple linear regression model was used to describe the relationship between weight at age 13 and \(\mathrm{BMD}\) at age 27\. For this data: $$ \begin{array}{lll} a=0.558 & b=0.009 & n=15 \\ \mathrm{SSTo}=0.356 & \text { SSResid }=0.313 & \end{array} $$ a. What percentage of observed variation in \(\mathrm{BMD}\) at age 27 can be explained by the simple linear regression model? b. Give a point estimate of \(\sigma\) and interpret this estimate. c. Give an estimate of the average change in BMD associated with a \(1 \mathrm{~kg}\) increase in weight at age 13 . d. Compute a point estimate of the mean BMD at age 27 for women whose age 13 weight was \(60 \mathrm{~kg}\).

A sample of \(n=61\) penguin burrows was selected, and values of both \(y=\) trail length \((\mathrm{m})\) and \(x=\) soil hardness (force required to penetrate the substrate to a depth of \(12 \mathrm{~cm}\) with a certain gauge, in \(\mathrm{kg}\) ) were determined for each one. The equation of the least- squares line was \(\hat{y}=11.607-1.4187 x\), and \(r^{2}=.386\). a. Does the relationship between soil hardness and trail length appear to be linear, with shorter trails associated with harder soil (as the article asserted)? Carry out an appropriate test of hypotheses. b. Using \(s_{e}=2.35, \bar{x}=4.5\), and \(\sum(x-\bar{x})^{2}=250\), predict trail length when soil hardness is \(6.0\) in a way that conveys information about the reliability and precision of the prediction. c. Would you use the simple linear regression model to predict trail length when hardness is \(10.0\) ? Explain your reasoning.

A subset of data read from a graph that appeared in the paper "Decreased Brain Volume in Adults with Childhood Lead Exposure" (Public Library of Science Medicine [May 27, 2008]: ell2) was used to produce the following Minitab output, where \(x=\) mean childhood blood lead level \((\mu \mathrm{g} / \mathrm{dL})\) and \(y=\) brain volume change (percentage). (See Exercise \(13.19\) for a more complete description of the study described in this paper) Regression Analysis: Response versus Mean Blood Lead Level The regression equation is Response \(=-0.00179-0.00210\) Mean Blood Lead Level \(\begin{array}{lrrrr}\text { Predictor } & \text { Coef } & \text { SE Coef } & \text { T } & \text { P } \\ \text { Constant } & -0.001790 & 0.008303 & -0.22 & 0.830 \\ \text { Mean Blood Lead Level } & -0.0021007 & 0.0005743 & -3.66 & 0.000\end{array}\) a. What is the equation of the estimated regression line? b. For this dataset, \(n=100, \bar{x}=11.5, s_{e}=0.032\), and \(S_{x x}=1764 .\) Estimate the mean brain volume change for people with a childhood blood lead level of \(20 \mu \mathrm{g} / \mathrm{dL}\), using a \(90 \%\) confidence interval. c. Construct a \(90 \%\) prediction interval for brain volume change for a person with a childhood blood lead level of \(20 \mu \mathrm{g} / \mathrm{dL}\). d. Explain the difference in interpretation of the intervals computed in Parts (b) and (c).

If the sample correlation coefficient is equal to 1, is it necessarily true that \(\rho=1\) ? If \(\rho=1\), is it necessarily true that \(r=1 ?\)

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