A chemical engineer is investigating the effect ofprocess operating temperature on product yield.The study results inthe following data:
| Temperature | | Yield |
| 100 | | 61.07 |
| 110 | | 66.01 |
| 120 | | 79.28 |
| 130 | | 75.04 |
| 140 | | 80.30 |
| 150 | | 97.95 |
| 160 | | 98.17 |
| 170 | | 110.07 |
| 180 | | 121.28 |
| 190 | | 118.21 |
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 = 0.9763).
1. what is the mean temperature?
130
155
140
145
2. what is the mean yield?
89.6555
90.7380
91.6321
91.6321
3. what is the standard deviation of temperature?
900.1573
30.2765
21.4653
101.5487
4. what is the standard deviation of yield?
460.7586
601.5487
21.4653
30.2765
5. The slope of the fitted regression line is closest to:
-9.6310
191.1070
82.1912
0.6922
207.8088
6. The intercept of the fitted regression line is closest to:
191.1070
82.1912
207.8088
-9.6310
7. The yield predicted by the regression model for a temperature of150 degrees is closest to:
-1443.9578
80.355
90.738
94.199
97.66
8. The residual error for a temperature of 150 degrees is closestto:
-97.95
99
-3.7510
1
3.7510
97.95
9. If the yield were measured in ounces instead of grams (note that1 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 (notethat 1 gram is 0.35274 ounces), the correlation coefficient wouldincrease by a factor of:
0.35274
1/0.35274
would not change
None of the above
