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The accompanying data on \(x=\) frequency \((\mathrm{MHz})\) and \(y=\) output power (W) for a certain laser configuration was read from a graph in the article "Frequency Dependence in RF Discharge Excited Waveguide \(\mathrm{CO}_{2}\) Lasers" (IEEE J. of Quantum Electronics, 1984: 509-514). \begin{tabular}{r|rrrrrrrr} \(x\) & 60 & 63 & 77 & 100 & 125 & 157 & 186 & 222 \\ \hline\(y\) & 16 & 17 & 19 & 21 & 22 & 20 & 15 & 5 \end{tabular} A computer analysis yielded the following information for a quadratic regression model: \(\hat{\beta}_{0}=-1.5127\), \(\hat{\beta}_{1}=.391901, \quad \hat{\beta}_{2}=-.00163141, \quad s_{\hat{\beta}_{2}}=.00003391\), \(\mathrm{SSE}=.29, \mathrm{SST}=202.88\), and \(s_{\dot{Y}}=.1141\) when \(x=100\). a. Does the quadratic model appear to be suitable for explaining observed variation in output power by relating it to frequency? b. Would the simple linear regression model be nearly as satisfactory as the quadratic model? c. Do you think it would be worth considering a cubic model? d. Compute a \(95 \%\) CI for expected power output when frequency is 100 . e. Use a \(95 \%\) PI to predict the power from a single experimental run when frequency is 100 .

Step by step solution

01

Assess Quadratic Model Suitability

Calculate the coefficient of determination, \( R^2 \), to assess the fit of the quadratic model.\[R^2 = 1 - \frac{\text{SSE}}{\text{SST}} = 1 - \frac{0.29}{202.88} \approx 0.9986\]The \( R^2 \) value is very close to 1, suggesting that the quadratic model is suitable for explaining the variation in output power with respect to frequency.

02

Compare with Linear Model

Calculate and interpret the coefficient of determination for the linear model. For linear models:\[\text{SSR} + \text{SSE} = \text{SST}\]Assume the \( R^2 \) for the quadratic model to be very high (0.9986), which indicates a much better fit compared to most possible linear models. Since the quadratic model accounts for more variation than a linear model typically would, it would be significantly better than a linear model in this context.

03

Consider a Cubic Model

Since the quadratic model already explains almost all of the variation (\( R^2 \approx 0.9986 \)), adding more complexity with a cubic model is likely not necessary unless the data shows clear non-linear patterns unaccounted for by quadratic terms. Since the \( R^2 \) is very high with the quadratic model itself, it is unlikely a cubic model would improve this fit substantially.

04

Compute 95% Confidence Interval (CI)

Use the formula for the confidence interval for predictions in quadratic regression:\[\text{CI} = \hat{Y} \pm t_{\alpha/2} \cdot s_{\hat{Y}}\]Where \(\hat{Y}\) is the predicted power at \(x = 100\):\[\hat{Y} = -1.5127 + 0.391901(100) - 0.00163141(100^2) = 21\]The 95% CI with \(t\) value (drawn from t-tables for appropriate degrees of freedom, approximate, say 10) is:\[\text{CI} = 21 \pm t_{0.025} \cdot 0.1141 \]Using \(t_{0.025, 6} \approx 2.447\):\[\text{CI} = 21 \pm 2.447 \cdot 0.1141 \approx (20.72, 21.28)\]

05

Compute 95% Prediction Interval (PI)

Use the formula for prediction interval:\[\text{PI} = \hat{Y} \pm t_{\alpha/2} \cdot \sqrt{s_{\hat{Y}}^2 + s^2}\]Substitute \(s = \sqrt{\frac{\text{SSE}}{\text{df}}}\), approximate \(s = \sqrt{\frac{0.29}{5}} \approx 0.24\) where \(df\) is degrees of freedom.Thus, for PI (\(t_{0.025, 6} \approx 2.447\)):\[\text{PI} = 21 \pm 2.447 \cdot \sqrt{0.1141^2 + 0.24^2} \approx 21 \pm 2.447 \times 0.264\approx (20.35, 21.65)\]

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Key Concepts

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

Coefficient of Determination

The coefficient of determination, also known as \( R^2 \), is a statistical measure that assesses how well a model explains and predicts future outcomes. In the context of quadratic regression analysis, \( R^2 \) indicates the proportion of variability in the dependent variable that is predictable from the independent variable.For the quadratic model provided, the \( R^2 \) value is approximately 0.9986. This very high value suggests that the model nearly perfectly explains the variation in output power based on frequency. A coefficient of determination close to 1 means that the model's predictions are highly accurate, capturing almost all of the data's variability.In practical applications, a high \( R^2 \) is often desirable as it represents the model's ability to provide a consistent approximation of the dependent variable outcomes within the observed data range.

Predictive Modeling

Predictive modeling uses statistical techniques to predict outcomes based on data input. For the quadratic regression model, predictive modeling allows us to predict the laser's output power based on its operating frequency.The equations governing predictive modeling involve arranging the relationship between variables into a mathematical equation. The quadratic equation used here:\[\hat{Y} = \hat{\beta}_{0} + \hat{\beta}_{1}x + \hat{\beta}_{2}x^2\]Here, \( \hat{Y} \) is the predicted output power when given a frequency \( x \). By plugging the observed frequency value into this equation, one can obtain a predicted outcome.Predictive modeling is crucial in engineering and scientific studies where it helps forecast product performance, system behavior, or experimental outcomes. Insights gained from these models can be instrumental in decision-making processes.

Confidence Interval

A confidence interval (CI) provides a range of plausible values for a parameter in a statistical model, allowing users to infer about where the true parameter value might lie.In the quadratic regression model provided, a 95% CI was calculated for the expected power output at a frequency of 100 MHz. By using the formula:\[CI = \hat{Y} \pm t_{\alpha/2} \cdot s_{\hat{Y}}\]We can compute the interval around the predicted output power. The calculated interval is \((20.72, 21.28)\), indicating that with 95% confidence, the true mean output power at this frequency falls within this range.Confidence intervals play a critical role in statistics, as they offer a measure of the reliability of an estimate. When conducting experiments or predictions, establishing a CI helps interpret the stability and potential variability within the model outcomes.

Prediction Interval

While similar to a confidence interval, a prediction interval (PI) is used to predict the range within which a single new observation is expected to fall. This is particularly useful when considering variability from the normal operating conditions.For the given problem, a 95% PI was computed using the formula:\[PI = \hat{Y} \pm t_{\alpha/2} \cdot \sqrt{s_{\hat{Y}}^2 + s^2}\]This accounts for both the uncertainty in the mean prediction \( s_{\hat{Y}} \) and the variance of an individual output \( s \). The resulting interval was \((20.35, 21.65)\).Prediction intervals are pivotal in risk analysis and quality control. They allow researchers and engineers to anticipate the range in which an individual measurement might lie, offering a comprehensive view of potential deviations from expected outcomes.