Question

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)= β₀ + β₁x 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.

Answer 1

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 )

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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