91Ó°ÊÓ

The following data on \(y=\) glucose concentration (g/L) and \(x=\) fermentation time (days) for a particular blend of malt liquor was read from a scatter plot in the article "Improving Fermentation Productivity with Reverse Osmosis" (Food Tech., 1984: 92-96): $$ \begin{array}{l|cccccccc} x & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ \hline y & 74 & 54 & 52 & 51 & 52 & 53 & 58 & 71 \end{array} $$ a. Verify that a scatter plot of the data is consistent with the choice of a quadratic regression model. b. The estimated quadratic regression equation is \(y=84.482-15.875 x+1.7679 x^{2}\). Predict the value of glucose concentration for a fermentation time of 6 days, and compute the corresponding residual. c. Using SSE \(=61.77\), what proportion of observed variation can be attributed to the quadratic regression relationship? d. The \(n=8\) standardized residuals based on the quadratic model are \(1.91,-1.95,-.25\), \(.58, .90, .04,-.66\), and .20. Construct a plot of the standardized residuals versus \(x\) and a normal probability plot. Do the plots exhibit any troublesome features? e. The estimated standard deviation of \(\hat{\mu}_{Y \cdot 6}\)-that is, \(\hat{\beta}_{0}+\hat{\beta}_{1}(6)+\hat{\beta}_{2}(36)-\) is 1.69. Compute a \(95 \%\) CI for \(\mu_{Y \cdot 6}\). f. Compute a \(95 \%\) PI for a glucose concentration observation made after 6 days of fermentation time.

Step by step solution

01

Plot the Data Points

First, we plot the data points given by the exercise to form a scatter plot using the pairs \((x, y)\) listed, which represent fermentation time and glucose concentration respectively. By plotting the points \((1, 74)\), \((2, 54)\), \((3, 52)\), \((4, 51)\), \((5, 52)\), \((6, 53)\), \((7, 58)\), \((8, 71)\), the pattern should resemble the shape of a quadratic function.

02

Verify Quadratic Fit

Examine the plotted points to see if they form a parabolic curve that opens upwards, which would justify using a quadratic model. The data indeed shows a non-linear pattern with a minimum point suggesting a better fit to a quadratic model than a linear one.

03

Predict Glucose Concentration for 6 Days

Use the given quadratic regression equation \(y = 84.482 - 15.875x + 1.7679x^2\) and substitute \(x = 6\). Calculate: \(y = 84.482 - 15.875(6) + 1.7679(6^2)\). Simplifying, \(y = 84.482 - 95.25 + 63.6844 = 52.9164\). This is the predicted glucose concentration for 6 days.

04

Calculate the Residual

Compare the predicted value with the actual observed value from the table for 6 days which is 53. The residual is computed as: \(Residual = Observed - Predicted = 53 - 52.9164 = 0.0836\).

05

Calculate the Proportion of Variation Explained

The explained variation by the model is given by the formula \(1 - \frac{SSE}{SST}\). Find SST using the formula for total sum of squares. SST = \( \sum (y_i - \bar{y})^2 \), where \( \bar{y} = \frac{383}{8} = 47.875 \). Substituting in the observed values, calculate SST and solve for \(1 - \frac{61.77}{SST}\).

06

Visualize and Analyze Residuals

Plot standardized residuals \(1.91, -1.95, -0.25, 0.58, 0.90, 0.04, -0.66, 0.20\) against \(x\) values to see if they are randomly scattered around zero, indicating no clear pattern. Also create a normal probability plot to check for normal distribution of residuals; they should fall along a straight diagonal line.

07

Compute the 95% Confidence Interval

The 95% Confidence Interval for \(\mu_{Y\cdot6}\) is given by \(\hat{y} \pm t(1.69)\), where \(t\) is the t-value for 95% confidence level with \(n-3\) degrees of freedom. Use the predicted value \(52.9164\) and standard deviation \(1.69\). Look up the t-value for \(n - 3 = 5\).

08

Compute the 95% Prediction Interval

The 95% Prediction Interval is given by \(\hat{y} \pm t \times \sqrt{s^2 + (1.69)^2}\), where \(t\) is the same as Step 7 and \(s^2\) is the variance derived from SSE and \(n - 3\). Use simplified form \([52.9164 - 3.3642, 52.9164 + 3.3642]\) for final prediction bounds.

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Key Concepts

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

Scatter Plot Analysis

Scatter plot analysis is an essential first step in visualizing the relationship between two variables. In this exercise, the variables are fermentation time (\(x\)) and glucose concentration (\(y\)). By plotting these data points as \((x, y)\) pairs, you can examine how the glucose concentration changes over different fermentation times. The plotted points in this example:

• \((1, 74)\)
• \((2, 54)\)
• \((3, 52)\)
• \((4, 51)\)
• \((5, 52)\)
• \((6, 53)\)
• \((7, 58)\)
• \((8, 71)\)

suggest a non-linear trend. At a glance, this scatter plot resembles a parabola, indicating that a quadratic model may be suitable. This shape occurs because the data reaches a minimum point and then increases, supporting the quadratic nature of the relationship.

Standardized Residuals

Standardized residuals help in assessing the fit of a regression model. They are calculated by taking the residual (the difference between observed and predicted values) and dividing it by an estimate of the standard deviation of the residuals. The standardized residuals for this quadratic regression model are:

• 1.91
• -1.95
• -0.25
• 0.58
• 0.90
• 0.04
• -0.66
• 0.20

By plotting these residuals against the \(x\) values, you can visually inspect for any patterns or outliers. Ideally, standardized residuals should scatter randomly around zero, indicating a good model fit. Additionally, by plotting them on a normal probability plot, you are checking for a normal distribution of residuals. Residuals falling close to a straight diagonal line suggest a normal distribution, which further validates the model.

Confidence Interval

A confidence interval (CI) estimates the range within which a population parameter will fall, with a certain degree of confidence. In this exercise, we are interested in a 95% CI for the mean glucose concentration after 6days of fermentation. The formula for a CI in this context is:\[\hat{y} \pm t \cdot (standard\ deviation) \]where:

• \(\hat{y}\) is the predicted mean using the regression equation
• \(t\) is the t-value associated with a 95% confidence interval and the degrees of freedom
• The given standard deviation is 1.69

By applying the values (\(\hat{y} = 52.9164\)), obtain the CI endpoints using an appropriate \(t\) value. These endpoints determine the range where you can be 95% confident that the true mean glucose concentration lies.

Prediction Interval

A prediction interval (PI) provides a range within which a future observation is expected to fall, with a certain confidence level. In contrast to a CI, which estimates a population parameter, a PI considers wider uncertainty as it predicts a single value. For a 95% PI after 6days of fermentation, the formula is:\[\hat{y} \pm t \cdot \sqrt{s^2 + (standard\ deviation)^2}\]Here:

• \(\hat{y}\) is the predicted concentration (52.9164)
• \(s^2\) is the known variance derived from SSE
• The standard deviation remains 1.69
• \(t\) is the critical value for 95%PI

By using the calculated variance and given standard deviation, you can find the interval endpoints. These show the range where future glucose concentration measurements after 6days are expected, with 95% confidence.