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A chemical engineer is investigating the effect of processoperating temperature on product yield.The study results in thefollowing data:

Temperature Yield

100 59.63

110 72.28

120 70.98

130 74.35

140 88.23

150 100.28

160 106.41

170 113.01

180 113.71

190 124.07

You can use Minitab to answer the following questions.However, you should be able to calculate the slope and intercept ofthe least squares regression model by hand, which requires only themeans and standard deviations of X and Y, and the correlationcoefficient (here r = 0.9839).

1. what is the mean temperature?

155

140

145

130

2. what is the mean yield?

92.2950

91.2125

92.4613

85.324

3. what is the standard deviation of temperature?

22.1631

30.2765

101.5487

900.1573

4. what is the standard deviation of yield?

601.5487

30.2765

491.2040

22.1631

5. The slope of the fitted regression line is closestto:

0.7202

211.4709

78.5291

-12.1340

196.7240

6. The intercept of the fitted regression line is closestto:

196.7240

78.5291

-12.1340

211.4709

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

81.492

95.896

92.295

-1819.3798

99.497

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

-4.3840

-100.28

6

102

4.3840

100.28

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

0.35274

1/0.35274

would not change

None of the above

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

0.35274

1/0.35274

would not change

None of the above