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Chapter 12: Problem 15

The following data (Exercise 16, Section 12.2) 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. Estimate the expected strawberry texture for a coded storage temperature of \(x=-1\). Use a \(99 \%\) confidence interval. b. Predict the particular value of \(y\) when \(x=1\) with a \(99 \%\) prediction interval. c. At what value of \(x\) will the width of the prediction interval for a particular value of \(y\) be a minimum, assuming \(n\) remains fixed?

Short Answer

Expert verified
Based on the given step by step solution, answer the following question: Question: Estimate the expected strawberry texture for a coded storage temperature of \(x=-1\) with a \(99\%\) confidence interval. Answer: To estimate the expected strawberry texture for a coded storage temperature of \(x=-1\) with a \(99\%\) confidence interval, follow Step 3 in the provided solution. Use the regression equation found in Step 2, plug in \(x=-1\), and calculate the expected texture using a \(99\%\) confidence interval by conducting a \(t\)-test with the degrees of freedom \((n-2)\).

Step by step solution

01

Find mean and standard deviation

Calculate the mean and standard deviation for \(x\) and \(y\): $$ \bar{x} = \frac{(-2) + (-2) + 0 + 2+ 2}{5} = 0 \\ \bar{y} = \frac{4.0 + 3.5 + 2.0 + 0.5 + 0.0}{5} = 2 $$ $$ s_x = \sqrt{\frac{(-2-0)^2 + (-2-0)^2 + (0-0)^2 + (2-0)^2 + (2-0)^2}{4}} = 2 \\ s_y = \sqrt{\frac{(4-2)^2 + (3.5-2)^2 + (2-2)^2 + (0.5-2)^2 + (0-2)^2}{4}} = 1.5 $$ Now calculate the correlation coefficient (r) and regression equation.
02

Calculate correlation coefficient and regression equation

Calculate the correlation coefficient using the formula: $$ r = \frac{\sum ( x_i - \bar{x})( y_i - \bar{y} ) }{ n-1 } $$ Plug the values to get the correlation coefficient, and use the coefficient to find the regression equation (\( y = a + bx \)). a. Estimate the expected strawberry texture for a coded storage temperature of \(x=-1\) with a \(99\%\) confidence interval.
03

Estimate expected texture for \(x=-1\) with a \(99\%\) confidence interval

Using the regression equation found in step 2, put \(x=-1\) and calculate \(y\) with a \(99\%\) confidence interval. Use \(t\)-tests for finding the confidence interval, considering the degrees of freedom to be \((n-2)\). b. Predict the particular value of \(y\) when \(x=1\) with a \(99\%\) prediction interval.
04

Predict \(y\) for \(x=1\) with a \(99\%\) prediction interval

To predict the value of \(y\) when \(x=1\), use the regression equation found in step 2. Calculate the prediction interval using the standard errors and \(t\)-tests with a \(99\%\) prediction interval. c. At what value of \(x\) will the width of the prediction interval for a particular value of \(y\) be a minimum, assuming \(n\) remains fixed?
05

Minimize the width of the prediction interval

To find the value of \(x\) where the width of the prediction interval is minimized, derive the prediction interval formula with respect to \(x\) and set it to zero. After that, find the value of \(x\) that satisfies the condition.

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Key Concepts

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

Confidence Interval
Understanding confidence intervals is crucial when estimating the accuracy of statistical predictions. A confidence interval (CI) is a range of values, derived from the sample data, that is likely to contain the value of an unknown population parameter. To put it simply, if we were to take multiple samples and build a confidence interval for each sample, the true population parameter would lie within these intervals in 99% of the cases 鈥� given a 99% confidence interval.

Regarding the strawberry texture exercise, when we estimate the expected texture for a coded storage temperature with a 99% confidence interval, we are indicating that we are 99% confident that the actual value of the strawberry texture for that specific temperature falls within this range. Technically, it is calculated using the sample mean plus or minus a margin of error, which is influenced by the standard deviation and the level of confidence we desire (99% in the example). This margin of error is multiplied by the appropriate value from the t-distribution, determined by our sample size and the desired level of confidence.
Prediction Interval
In contrast to a confidence interval, a prediction interval gives a range within which we expect a single new observation to fall. This is inherently broader than a confidence interval, as it accounts for the variability between individual observations, not just the estimate of the mean. Thus, when we say that we have calculated a 99% prediction interval for the strawberry texture at a certain storage temperature, we mean that we predict with 99% confidence that the texture of any individual strawberry (not the mean texture) will fall within this range.

Applying Prediction Interval to Our Example

Using the regression equation to predict a particular value of strawberry texture gives us a point estimate. However, due to natural variability, we need to consider the prediction interval to account for the uncertainty around this point estimate for a single new observation. The 99% prediction interval includes a larger margin of error than the confidence interval due to this additional uncertainty.
Correlation Coefficient
The correlation coefficient (r) measures the strength and direction of the linear relationship between two variables. Its value ranges from -1 to 1. A value of 1 indicates a perfect positive linear correlation, -1 indicates a perfect negative linear correlation, and 0 indicates no linear correlation.

In the Realm of Strawberries

In our case, the correlation coefficient provides insight into how well the storage temperature predicts the texture of strawberries. A high absolute value of the correlation coefficient would suggest that temperature is a good predictor of texture. When solving for the correlation coefficient within the exercise, knowing this coefficient helps us build a more accurate regression model, which in turn allows us to make better predictions and understand the relationship between storage temperature and strawberry texture.

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Most popular questions from this chapter

Use the data in Exercises \(7-8\) to calculate the coefficient of determination, \(r^{2} .\) What information does this value give about the usefulness of the linear model? $$ \begin{array}{r|rrrrr} x & -2 & -1 & 0 & 1 & 2 \\ \hline y & 1 & 1 & 3 & 5 & 5 \end{array} $$

You can monitor every step you take, your speed, your pace, or some other aspect of your daily activity. The data that follows lists the overall rating scores for 14 fitness trackers and their prices. \({ }^{13}\) $$\begin{array}{lcc}\hline \text { Fitness Trackers } & \text { Score } & \text { Price (\$) } \\\\\hline \text { Fitbit Surge } & 87 & 250 \\\\\text { TomTom Spark 3 } & 85 & 250 \\\\\text { Garmin Forerunner 38 } & 85 & 200 \\\\\text { TomTom Spark } & 84 & 200 \\\\\text { Fitbit Charge 2 } & 83 & 150 \\\\\text { Garmin Vivosmart HR } & 83 & 120 \\\\\text { Fitbit Blaze } & 82 & 200 \\\\\text { Huawei Fit } & 82 & 130 \\\\\text { Garmin Vivosmart HR+ } & 79 & 180 \\\\\text { Withings Steel HR } & 79 & 145 \\\\\text { Fitbit Alta } & 78 & 130 \\\\\text { Garmin Vivoactive HR } & 77 & 250 \\\\\text { Samsung Gear Fit 2 } & 76 & 180 \\\\\text { Under Armour Band } & 74 & 80 \\\\\hline\end{array}$$ a. Use a scatterplot of the data to check for a relationship between the rating scores and prices for the fitness trackers. b. Calculate the sample coefficient of correlation \(r\) and interpret its value. c. By what percentage was the sum of squares of deviations reduced by using the least-squares predictor \(\hat{y}=a+b x\) rather than \(\bar{y}\) as a predictor of \(y ?\)

10\. Recidivism Recidivism refers to the return to prison of a prisoner who has been released or paroled. The data that follow reports the group median age at which a prisoner was released from a federal prison and the percentage of those arrested for another crime. \({ }^{7}\) Use the MS Excel printout to answer the questions that follow. $$ \begin{array}{l|lllllll} \text { Group Median Age }(x) & 22 & 27 & 32 & 37 & 42 & 47 & 52 \\ \hline \text { \% Arrested }(y) & 64.7 & 59.3 & 52.9 & 48.6 & 44.5 & 37.7 & 23.5 \end{array} $$ $$ \begin{aligned} &\text { SUMMARY OUTPUT }\\\ &\begin{array}{ll} \hline \text { Regression Statistics } & \\ \hline \text { Multiple R } & 0.9779 \\ \text { R Square } & 0.9564 \\ \text { Adjusted R Square } & 0.9477 \\ \text { Standard Error } & 3.1622 \\ \text { Observations } & 7.0000 \\ \hline \end{array} \end{aligned} $$ $$ \begin{aligned} &\text { ANOVA }\\\ &\begin{array}{llrrr} \hline & & & & {\text { Significance }} \\ & \text { df } & \text { SS } & \text { MS } & \text { F } & \text { F } \\ & & & & & \\ \hline \text { Regression } & 1 & 1096.251 & 1096.251 & 109.631 & 0.000 \\ \text { Residual } & 5 & 49.997 & 9.999 & & \\ \text { Total } & 6 & 1146.249 & & & \\ \hline \end{array} \end{aligned} $$ $$ \begin{array}{lrrrrrr} \hline& {\text { Coeffi- Standard }} \\ & \text { cients } & \text { Error } & \text { tStat } & \text { P-value } & \text { Lower } 95 \% & \text { Upper } 95 \% \\ \hline \text { Intercept } & 93.617 & 4.581 & 20.436 & 0.000 & 81.842 & 105.393 \\ \mathrm{x} & -1.251 & 0.120 & -10.471 & 0.000 & -1.559 & \- \\ \hline \end{array} $$ a. Find the least-squares line relating the percentage arrested to the group median age. b. Do the data provide sufficient evidence to indicate that \(x\) and \(y\) are linearly related? Test using the \(t\) statistic at the \(5 \%\) level of significance. c. Construct a \(95 \%\) confidence interval for the slope of the line. d. Find the coefficient of determination and interpret its significance.

Use the data given in Exercises 5-6 (Exercises 17-18, Section 12.1). Do the data provide sufficient evidence to indicate that \(y\) and \(x\) are linearly related? Test using the \(t\) statistic at the 1\% level of significance. Construct a \(99 \%\) confidence interval for the slope of the line. What does the phrase "99\% confident" mean? $$ \begin{array}{r|rrrrr} x & -2 & -1 & 0 & 1 & 2 \\ \hline y & 1 & 1 & 3 & 5 & 5 \end{array} $$

What diagnostic plot can you use to determine whether the incorrect model has been used? What should the plot look like if the correct model has been used?
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