In: Math
A particular meat-processing plant slaughters steers and cuts and wraps the beef for its customers. Suppose a complaint has been filed with the Food and Drug Administration (FDA) against the processing plant. The complaint alleges that the consumer does not get all the beef from the steer he purchases. In particular, one consumer purchased a cut and wrapped beef. To settle the complaint, the FDA collected data on the live weights and dressed weights of nine steers processed by a reputable meat processing plant (not the firm in question). The results are listed in the table.
Live Weight |
Dressed Weight |
x, pounds |
y, pounds |
420 |
280 |
380 |
250 |
480 |
310 |
340 |
210 |
450 |
290 |
460 |
280 |
430 |
270 |
370 |
240 |
390 |
250 |
a. Fit the model E(y)= β_{0} + β_{1}x to the data
b. Construct a 95% prediction interval for the dressed weight y of a 300-pound steer.
c. Would you recommend that the FDA use the interval obtained in part b to determine whether the dressed weight of 150 pounds is a reasonable amount to receive from a 300-pound steer? Explain.
using excel add-in for regresssion, we get following output-
SUMMARY OUTPUT | ||||||
Regression Statistics | ||||||
Multiple R | 0.9660 | |||||
R Square | 0.9332 | |||||
Adjusted R Square | 0.9236 | |||||
Standard Error | 8.3034 | |||||
Observations | 9 | |||||
ANOVA | ||||||
df | SS | MS | F | Significance F | ||
Regression | 1 | 6739.60 | 6739.60 | 97.75 | 0.00 | |
Residual | 7 | 482.62 | 68.95 | |||
Total | 8 | 7222.22 | ||||
Coefficients | Standard Error | t Stat | P-value | Lower 95% | Upper 95% | |
Intercept | 5.7106 | 26.3152 | 0.2170 | 0.8344 | -56.5150 | 67.9362 |
x, pounds | 0.6260 | 0.0633 | 9.8869 | 0.0000 | 0.4763 | 0.7757 |
so,
y^= 5.7106 + 0.6260X
b)
Predicted interval
where
where S_{s} is standard error of predicted y-value of x regression
n=number of observation
SSx=sum of squares of deviations of X
=11.31786
t_{alpha/2,df} = t_{0.05/2,8}=2.306
predicted y at 300 is
y^= 5.7106 + 0.6260*300=193.5106
so, Predicted interval is
193.5106+/-2.306*11.31786
( 167.412,219.610 )
--------------
c)
As the above interval does not contain the value 150, it is not recommended that the FDA use the interval obtained to determine the dressed weight of 150 pounds to receive from a 300-pound steer