91Ó°ÊÓ

The article "Analysis of the Modeling Methodologies for Predicting the Strength of Air-Jet Spun Yarns" (Textile Res. \(J ., 1997: 39-44\) ) reported on a study carried out to relate yarn tenacity ( \(y\), in \(\mathrm{g} /\) tex \()\) to yarn count \(\left(x_{1}\right.\), in tex), percentage polyester \(\left(x_{2}\right)\), first nozzle pressure \(\left(x_{3}\right.\), in \(\left.\mathrm{kg} / \mathrm{cm}^{2}\right)\), and second nozzle pressure \(\left(x_{4}\right.\), in \(\mathrm{kg} /\) \(\left.\mathrm{cm}^{2}\right)\). The estimate of the constant term in the corresponding multiple regression equation was \(6.121\). The estimated coefficients for the four predictors were \(-.082\), \(.113, .256\), and \(-.219\), respectively, and the coefficient of multiple determination was 946 . a. Assuming that the sample size was \(n=25\), state and test the appropriate hypotheses to decide whether the fitted model specifies a useful linear relationship between the dependent variable and at least one of the four model predictors. b. Again using \(n=25\), calculate the value of adjusted \(R^{2}\). c. Calculate a \(99 \%\) confidence interval for true mean yarn tenacity when yarn count is \(16.5\), yarn contains \(50 \%\) polyester, first nozzle pressure is 3 , and second nozzle pressure is 5 if the estimated standard deviation of predicted tenacity under these circumstances is . 350 .

Step by step solution

01

Formulate Hypotheses for the F-Test

To test if the linear model is useful, we need to test the null hypothesis that all regression coefficients are zero against the alternative that at least one is non-zero. That is:- Null Hypothesis, \( H_0: \beta_1 = \beta_2 = \beta_3 = \beta_4 = 0 \)- Alternative Hypothesis, \( H_1: \) at least one \( \beta_i eq 0 \).

02

Calculate F-Statistic for Overall Significance

The F-statistic is calculated using the formula:\[ F = \frac{(R^2 / k)}{((1 - R^2) / (n - k - 1))} \]where \( R^2 = 0.946 \), \( k = 4 \), and \( n = 25 \).Let's plug in the numbers:\[ F = \frac{(0.946 / 4)}{((1 - 0.946) / (25 - 4 - 1))} = \frac{0.2365}{0.00284} \approx 83.3 \]

03

Determine Critical Value and Decision for F-Test

For a significance level \( \alpha = 0.05 \), degrees of freedom for the numerator \( k = 4 \), and denominator \( n - k - 1 = 20 \), the critical F-value is approximately 2.87 (from F-distribution tables).Since the calculated F-statistic \( 83.3 \) is greater than 2.87, we reject \( H_0 \). Thus, the model is useful for predicting yarn tenacity.

04

Calculate Adjusted R-Squared

The formula for adjusted \( R^2 \) is:\[ R^2_{adj} = 1 - \left(\frac{1-R^2}{n-k-1} \times (n-1)\right) \]Substitute values:\[ R^2_{adj} = 1 - \left(\frac{1-0.946}{25-4-1} \times (25-1)\right) \approx 1 - 0.0131 \times 24 \approx 0.937 \]

05

Set Up Confidence Interval Formula

For a 99% confidence interval of mean response \( \mu_y \), use the formula:\[ \hat{y} \pm t^* \times s_e \]where \( \hat{y} \) is predicted tenacity, \( s_e = 0.35 \), and \( t^* \) is the critical t-value from the t-distribution with \( n-k-1 = 20 \) degrees of freedom.

06

Calculate Predicted Yarn Tenacity

Substitute given values into the regression equation to calculate \( \hat{y} \):\[ \hat{y} = 6.121 - 0.082(16.5) + 0.113(50) + 0.256(3) - 0.219(5) \]\[ \hat{y} = 6.121 - 1.353 + 5.65 + 0.768 - 1.095 \approx 10.091 \]

07

Compute Critical t-value for 99% Confidence Interval

The critical \( t^* \) for 99% confidence level with \( df = 20 \) is approximately 2.845 (from t-distribution tables).

08

Calculate 99% Confidence Interval

Using \( \hat{y} = 10.091 \) and \( s_e = 0.35 \), the confidence interval is:\[ 10.091 \pm 2.845 \times 0.35 \]\[ \approx 10.091 \pm 0.996 \]\[ \approx (9.095, 11.087) \]

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Key Concepts

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

Yarn Tenacity Prediction

Predicting yarn tenacity involves understanding how various factors contribute to the strength of yarns produced by the air-jet spinning method. Yarn tenacity, measured in grams per tex, is influenced by several predictors:

• Yarn count, which refers to the thickness of the yarn.
• Percentage of polyester content in the yarn.
• First nozzle pressure in kg/cm².
• Second nozzle pressure in kg/cm².

All these variables interplay in a multiple regression analysis to forecast the yarn's strength. This statistical tool helps in establishing a relationship between the dependent variable (yarn tenacity) and the independent variables (predictors). The estimated regression equation pieces together these variables with specific coefficients, providing a mathematical framework to predict tenacity based on given input values.

Hypothesis Testing

Hypothesis testing in the context of multiple regression helps determine if the independent variables collectively have a meaningful relationship with the dependent variable. Here, a central task is to evaluate whether the regression model significantly predicts yarn tenacity.

• The null hypothesis (\( H_0 \)) assumes that none of the predictor variables have a relationship with the dependent variable, i.e., all regression coefficients are zero.
• The alternative hypothesis (\( H_1 \)) suggests that at least one predictor variable significantly affects yarn tenacity, implying at least one coefficient is non-zero.

The test involves computing the F-statistic, which measures the model's overall significance. If the F-statistic exceeds the critical F-value from statistical tables, we can reject \( H_0 \), indicating a useful model exists for predicting yarn tenacity.

Coefficient of Determination

The coefficient of determination, represented as \( R^2 \), indicates how well the regression model explains the variability of the dependent variable. An \( R^2 \) value close to 1 suggests a highly effective model. In this exercise, the given \( R^2 \) is 0.946, signifying that approximately 94.6% of the variation in yarn tenacity can be explained by the predictors in the model.
Adjusted \( R^2 \) provides a more accurate measure, especially when multiple predictors are involved. It accounts for the number of predictors in the model, preventing overestimation of the model's predictive power. The formula for adjusted \( R^2 \) helps refine the value to 0.937, reflecting a well-fitted model beyond simple \( R^2 \).

Confidence Intervals

Confidence intervals offer a range within which we expect the true mean yarn tenacity to fall, considering the model's predictions and variability. A 99% confidence interval gives very high assurance in the results.

• To compute this interval, the predicted yarn tenacity (\( \hat{y} \)) is calculated based on the regression equation by substituting real values for yarn count, polyester percentage, and nozzle pressures.
• The standard error (\( s_e \)), captured from predictions, is also used.
• Critical t-value (\( t^* \)) from the t-distribution table allows calculating the interval's margin of error.

The interval calculation is \( \hat{y} \pm t^* \times s_e \), providing a confidence range. For example, using a predicted tenacity of 10.091 and a standard error of 0.35, the 99% interval is (9.095, 11.087), indicating that with 99% confidence, the true mean yarn tenacity under specified conditions lies within this range.