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An experiment was conducted to observe the effect of an increase in temperature on the potency of an antibiotic. Three 1 -ounce portions of the antibiotic were stored for equal lengths of time at each of these temperatures: \(30^{\circ}, 50^{\circ}, 70^{\circ},\) and \(90^{\circ} .\) The potency readings observed at each temperature of the experimental period are listed here: $$ \begin{array}{l|l|l|l|l} \text { Potency Readings, } y & 38,43,29 & 32,26,33 & 19,27,23 & 14,19,21 \\ \hline \text { Temperature, } x & 30^{\circ} & 50^{\circ} & 70^{\circ} & 90^{\circ} \end{array} $$ Use an appropriate computer program to answer these questions: a. Find the least-squares line appropriate for these data. b. Plot the points and graph the line as a check on your calculations. c. Construct the ANOVA table for linear regression. d. If they are available, examine the diagnostic plots to check the validity of the regression assumptions. e. Estimate the change in potency for a 1 -unit change in temperature. Use a \(95 \%\) confidence interval. f. Estimate the average potency corresponding to a temperature of \(50^{\circ} .\) Use a \(95 \%\) confidence interval. g. Suppose that a batch of the antibiotic was stored at \(50^{\circ}\) for the same length of time as the experimental period. Predict the potency of the batch at the end of the storage period. Use a \(95 \%\) prediction interval.

Short Answer

Expert verified
Answer: To estimate the change in potency for a 1-unit change in temperature, we need to find the slope (b) of the least-squares line obtained from the linear regression analysis. The value of b represents the average change in potency per unit change in temperature. We can then find the 95% confidence interval for the slope to determine the range of change in potency with a given level of confidence.

Step by step solution

01

Organize the data

Before starting the analysis, we need to organize the data properly. We have 4 temperatures and their corresponding potency readings: Temperature: 30°, 50°, 70°, 90° Potency Readings: (38,43,29), (32,26,33), (19,27,23), (14,19,21) We'll later use a software or calculator to perform the necessary calculations but you need to input these data.
02

Find the least-squares line

Using a calculator or software (such as Excel, R, or Python), apply a linear regression analysis on our dataset. This will give us the least-squares line that best fits the data. You will get the equation of the line in the form \(y = a + b x\), where \(a\) is the intercept, and \(b\) is the slope.
03

Plot the points and graph the line

Now, using the same calculator or software, plot the original potency readings along with the temperatures on a graph. Afterward, plot the least-squares line found in Step 2 on the same graph. This will give you a visual representation of the relationship between the temperature and potency readings.
04

Construct the ANOVA table for linear regression

Using the software or calculator, construct the ANOVA (Analysis of Variance) table for linear regression. This table displays the information that helps us assess the goodness of fit of the regression line and the statistical significance of the predictors (in our case, temperature).
05

Examine diagnostic plots

If the calculator or software you are using provides diagnostic plots, examine these plots to check the validity of the regression assumptions. These plots typically include residual plot, QQ-plot, and scale-location plot, which helps you assess the assumptions of linearity, normality, and homoscedasticity of the residuals.
06

Estimate the change in potency for a 1-unit change in temperature

Using the software or calculator, find the 95% confidence interval for the slope of the least-squares line (the value of \(b\)). This will give you an estimate of the change in potency for a 1-unit change in temperature along with the given confidence level.
07

Estimate the average potency corresponding to a temperature of 50°

To estimate the average potency at a temperature of 50°, substitute \(x=50\) into the least-squares line equation \(y = a + b x\). Then, using the software or calculator, find the 95% confidence interval for this estimate.
08

Predict the potency of the batch at the end of the storage period

Now, suppose that a batch of the antibiotic was stored at 50° for the same length of time as the experimental period. To predict the potency of the batch at the end of the storage period, use the same least-squares line equation and input \(x=50\). Find the 95% prediction interval for this estimate using the software or calculator. After completing all these steps, you will have fully analyzed the given exercise, found the least-squares line, plotted the data, constructed the ANOVA table, checked the validity of the regression assumptions, estimated the change in potency for a 1-unit change in temperature, estimated the average potency at a specific temperature, and predicted the potency of a batch at the end of storage.

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Key Concepts

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

Least Squares Method
When it comes to finding the best fit line for a set of data points, the Least Squares Method is incredibly useful. This method helps us to determine the regression line equation, which is typically represented as \( y = a + bx \). Here, \( a \) represents the y-intercept, and \( b \) represents the slope of the line. By applying this technique, we minimize the sum of the squares of the differences between the observed values and the values predicted by the line.

The primary goal is to reduce the discrepancies between actual data points and the predictions made by our linear model. This method is mostly executed using computational tools or software, such as Excel, R, or Python, which make the calculations easier to handle, especially for larger datasets.
  • Intercept \(a\): This is the point where the line crosses the y-axis, indicating the potency level when the temperature is zero.
  • Slope \(b\): It shows the rate of change in potency readings for each one-degree increase in temperature.
Executing the least squares method allows us to visualize the relationship between variables and make informed predictions.
ANOVA in Regression
ANOVA, or Analysis of Variance, in the context of regression is a technique used to evaluate the model's effectiveness. It helps us understand whether our regression model is a good fit for the data by comparing the model's variance against the residual variance.

The primary components in an ANOVA table typically include - Sum of Squares for Regression (SSR), Sum of Squares for Error (SSE), and Total Sum of Squares (SST). Each part plays a critical role.
  • SSR (Sum of Squares for Regression): This shows the variation explained by the regression model.
  • SSE (Sum of Squares for Error): This part of the table indicates the variation in data that the model fails to explain.
  • SST (Total Sum of Squares): It represents the total variation present in the response variable \( y \).
A low SSE in comparison to SSR means that our model adequately describes the relationship between the variables. Evaluating the ANOVA table also involves checking the F-statistic, which helps determine if the regression model terms are significant.
Confidence Interval
Confidence intervals provide us with a range to estimate a parameter with a certain level of confidence, often 95% in many cases. In regression analysis, confidence intervals are used to estimate the range within which the true regression coefficient or the mean of the dependent variable lies, given certain values of the independent variable.

For example, in our exercise, we assess the confidence interval of the slope \( b \) to determine the change in potency for a one-unit change in temperature. This range provides an interval of values that, with a certain degree of confidence, includes the true parameter.
  • 95% Confidence Level: Indicates that if we were to select numerous samples and compute a confidence interval for each, 95% of these would contain the true parameter value.
In practice, computing confidence intervals helps us assess the reliability of our estimates and make informed decisions based on statistical evidence.
Prediction Interval
Unlike confidence intervals, prediction intervals provide an estimate range for individual new observations. In the context of our example, this means predicting the potency of a new batch of antibiotics stored at a given temperature and determining a range likely to contain the actual potency reading.

Prediction intervals account for both the mean response and the variability around that response, thus they are typically wider than confidence intervals.
  • 95% Prediction Interval: Implies a 95% probability that the actual potency value for a new batch lies within this range.
It is essential to note that while confidence intervals handle the average response, prediction intervals are more applicable for forecasting specific future outcomes.

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Most popular questions from this chapter

G. W. Marino investigated the variables related to a hockey player's ability to make a fast start from a stopped position. \({ }^{11}\) In the experiment, each skater started from a stopped position and attempted to move as rapidly as possible over a 6-meter distance. The correlation coefficient \(r\) between a skater's stride rate (number of strides per second) and the length of time to cover the 6 -meter distance for the sample of 69 skaters was -.37 . a. Do the data provide sufficient evidence to indicate a correlation between stride rate and time to cover the distance? Test using \(\alpha=.05 .\) b. Find the approximate \(p\) -value for the test. c. What are the practical implications of the test in part a?

What is the difference between deterministic and probabilistic mathematical models?

The following data (Exercises 12.16 and 12.24 ) 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?

Graph the line corresponding to the equation \(y=-2 x+1\) by graphing the points corresponding to \(x=0,1,\) and 2 . Give the \(y\) -intercept and slope for the line. How is this line related to the line \(y=2 x+1\) of Exercise \(12.1 ?\)

Athletes and others suffering the same type of injury to the knee often require anterior and posterior ligament reconstruction. In order to determine the proper length of bone-patellar tendonbone grafts, experiments were done using three imaging techniques to determine the required length of the grafts, and these results were compared to the actual length required. A summary of the results of a simple linear regression analysis for each of these three methods is given in the following table. \({ }^{15}\) $$ \begin{array}{llrcc} \text { Imaging Technique } & \text {Coeffcient of Determination, } r^{2} & \text { Intercept } & \text { Slope } & p \text { -value } \\ \hline \text { Radiographs } & 0.80 & -3.75 & 1.031 & <0.0001 \\ \text { Standard MRI } & 0.43 & 20.29 & 0.497 & 0.011 \\ \text { 3-dimensional MRI } & 0.65 & 1.80 & 0.977 & <0.0001 \end{array} $$ a. What can you say about the significance of each of the three regression analyses? b. How would you rank the effectiveness of the three regression analyses? What is the basis of your decision? c. How do the values of \(r^{2}\) and the \(p\) -values compare in determining the best predictor of actual graft lengths of ligament required?

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