A chemical engineer is investigating the effect of process operating temperature

Question

A chemical engineer is investigating the effect of process operating temperature on product yield.The study results in the following data:

Temperature Yield

100 62.95

110 68.07

120 73.94

130 83.95

140 82.70

150 96.68

160 109.66

170 109.76

180 108.30

190 125.30

You can use Minitab to answer the following questions. However, you should be able to calculate the slope and intercept of the least squares regression model by hand, which requires only the means and standard deviations of X and Y, and the correlation coefficient (here r = 0.9788).

1. what is the mean temperature?

2. what is the mean yield?

3. what is the standard deviation of temperature?

4. what is the standard deviation of yield?

5. The slope of the fitted regression line is closest to:

6. The intercept of the fitted regression line is closest to:

7. The yield predicted by the regression model for a temperature of 150 degrees is closest to:

8. The residual error for a temperature of 150 degrees is closest to:

9. If the yield were measured in ounces instead of grams (note that 1 gram is 0.35274 ounces), the slope would change by a factor of:

10. If the yield were measured in ounces instead of grams (note that 1 gram is 0.35274 ounces), the correlation coefficient would increase by a factor of:  

Explanation / Answer

Solution

Back-up Theory

Since study is investigating the effect of process operating temperature on product yield, process operating temperature is the independent variable, X and product yield is the dependent variable, Y.

Linear regression of Y on X is estimated as: y = a + bx, where b = r.SD(Y)/SD(X) and a = Ybar – b.Xbar; r is the correlation coefficient, SD(Y) = standard deviation of Y, SD(X) = standard deviation of X, Ybar = mean of Y and Xbar = mean of X.

Now, to work out solution,

From the given data all the above are computed and presented against the respective parts of the question below.

Part (a)

Mean temperature = Xbar = 145 ANSWER

Part (b)

Mean yield = Ybar = 92.131 ANSWER

Part (c)

Standard deviation of temperature = SD(X) = 28.723 ANSWER

Part (d)

Standard deviation of yield = SD(Y) = 19.793 ANSWER

Part (e)

Slope of the fitted regression line is = b = 0.6744 ANSWER

Part (f)

Intercept of the fitted regression line is = a = - 5.6705 ANSWER

Part (g)

The yield predicted by the regression model for a temperature of 150 degrees is a + 150b = 95.4895 ANSWER

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