Data were collected at a supermarket for the waiting time at the checkout during
ID: 3365517 • Letter: D
Question
Data were collected at a supermarket for the waiting time at the checkout during time periods where there were a constant number of checkout counters open. The total number of customers in the store and the waiting times (in minutes) were recorded. Based on a study of 30 time periods, the total number of customers in the store can be used to predict the waiting time at the checkout counter by the prediction line Yi=-0.432+0.153Xi For these data, SYX=0.378 , X=22.5 , and SSX=32.974.
Complete parts (a) through (c).
a. Construct a 90 % confidence interval estimate of the mean waiting time for all customers when there are 20 customers in the store.
b. Construct a 90 % prediction interval of the waiting time for an individual customer when there are 20 customers in the store.
c. Why is the interval in (a) narrower than the interval in (b)?
A. Since there is more variation in estimating a mean value than in predicting an individual value, the confidence interval is wider than the prediction interval.
B. Since there is more variation in predicting an individual value than in estimating a mean value, the prediction interval is narrower than the confidence interval.
C. Since there is more variation in predicting an individual value than in estimating a mean value, the prediction interval is wider than the confidence interval.
D. Since there is more variation in estimating a mean value than in predicting an individual value, the confidence interval is narrower than the prediction interval.
Explanation / Answer
Yi = -0.432 + 0.153 Xi
For these data, SYX=0.378 , x =22.5 , and SSX = 32.974
Here n = 30 (given)
90% confidence interval for the predicted value = y^ +- tcr se0
where dF = 30 -2= 28 and alpha = 0.10
tcr = t0.10,28 = 1.7011
se0 = se * sqrt [1/n + (x0 - x)2 / SSXX]
x = 22.5
SSXX = 32.974
y^(X = 20) = -0.432 + 0.153 * 20 = 2.628
90% confidenceinterval for the predicted value = y^ +- tcr se0
= 2.628 +- 1.7011 * 0.378 * sqrt [1/30 + (20 - 22.5)2 /32.974]
= 2.628 + 1.7011 * 0.378 * 0.4721
= (2.3244, 2.9316)
(b) 90% Prediction interval for the predicted value = y^ +- tcr sep
sep = se * sqrt [1 + 1/n + (x0 - x)2 / SSXX]
sep = 0.378 * sqrt [1 + 1/30 + (22.5 - 20)2 /32.974]
90% Prediction interval for the predicted value = y^ +- tcr sep
= 2.628 +- 1.7011 * 0.378 * sqrt [1 + 1/30 + (20 - 22.5)2 /32.974]
= 2.628 +- 0.643 * 1.1058
= (1.917, 3.339)
(c)
Since there is more variation in predicting an individual value than in estimating a mean value, the prediction interval is wider than the confidence interval. Option C is correct here.
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