1. (25 points) Establishing the properties of materials is an important problem

Question

1. (25 points) Establishing the properties of materials is an important problem in identifying a suitable substitute for biodegradable materials in the fast food packaging industry. Consider the following data on product density (g/cm3) and thermal conductivity K-factor. Thermal Conductivity (y) 0.0480 0.0525 0.0540 0.0535 0.0570 0.0610 Product Density (x) 0.1750 0.2200 0.2250 0.2260 0.250 0.2765 For the data above perform the following using Minitab or the formulas: a. (4 points) Estimate the intercept and the slope and write the regression line b. (4 points) Compute SSE and estimate the variance c. (4 points) Find the Standard error of the slope and intercept coefficients d. (4 points) Find the coefficient of determination R2 e. (3 points) Use a t-test to test the significance of the slope coefficients at -0.05 and comment on your results of this Cl and your finding in parts (e) provides an adequate fit? f· (3 points) Construct a 95% CI on the slope. Comment on the relationship g. (3 points) Perform model adequacy checks. Do you believe the model

Explanation / Answer

Solution:

Here, we have to use Minitab for estimating the regression model by using given data.

Required Minitab output is given as below:

Welcome to Minitab, press F1 for help.

Regression Analysis: Thermal Conductivity (Y) versus Product Density (X)

The regression equation is

Thermal Conductivity (Y) = 0.0249 + 0.129 Product Density (X)

Predictor        Coef     SE Coef          T        P

Constant     0.024934    0.001786      13.96    0.000

Product      0.128522    0.007738      16.61    0.000

S = 0.0005852   R-Sq = 98.6%     R-Sq(adj) = 98.2%

Analysis of Variance

Source            DF          SS          MS         F        P

Regression         1 0.000094464 0.000094464    275.84    0.000

Residual Error     4 0.000001370 0.000000342

Total              5 0.000095833

Part a

The value for the intercept is given as 0.024934 and the slope for the regression equation is given as 0.128522. The regression equation is given as below:

Thermal Conductivity (Y) = 0.0249 + 0.129 Product Density (X)

Part b

The value of SSE by using Minitab output is given as below:

SSE = 0.000001370

Estimate for standard deviation is given as 0.0005852, so estimate for variance is given as 0.0005852*0.0005852 = 0.0000003425.

Part c

Standard error of the slope is given as 0.007738.

Standard error of the intercept is given as 0.001786.

Part d

The coefficient of determination or the value of R square is given as 98.6% or 0.986.

Part e

The p-value for the t test for significance of slope coefficient is given as 0.00 which is less than = 0.05, so we reject the null hypothesis that the slope is not statistically significant. There is sufficient evidence to conclude that the slope for regression equation is statistically significant.

Part f

CI = slope -/+ t*SE

Critical value t = 2.57 by using t-table with df = 6 – 1 = 5.

We are given sample size n = 6.

CI = 0.128522 -/+ 2.57*0.007738

CI = 0.128522 -/+ 0.01988666

Lower limit = 0.128522 - 0.01988666 = 0.10863534

Upper limit = 0.128522 + 0.01988666 = 0.14840866

The value 0 does not lies within this confidence interval so we reject the null hypothesis that the slope is not statistically significant. There is sufficient evidence to conclude that the slope for regression equation is statistically significant.

Part g

The p-value for overall model is given as 0.00 which is less than level of significance or alpha value 0.05, so we reject the null hypothesis that the given regression model is not statistically significant. This means, there is sufficient evidence to conclude that the regression model is statistically significant.

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.