Consider Case Study 8.2 on page 238. Suppose 18 specimens were used for each type of paint in an experiment and XA -XB, the actual difference in mean drying time, turned out to be 1.0. (a) Does this seem to be a reasonable result if the two population mean drying times truly are equal? Make use of the result in the solution to Case Study 8.2. (b) If someone did the experiment 10,000 times under the condition that μA = μB, in how many of those 10,000 experiments would there be a difference XA - XB that was as large as (or larger than) 1.0? Case Study 8.2: Paint Drying Time: Two independent experiments are run in which two different types of paint are compared. Eighteen specimens are painted using type A, and the drying time, in hours, is recorded for each. The same is done with type B. The population standard deviations are both known to be 1.0. Assuming that the mean drying time is equal for the two types of paint, find P(XA-XB> 1.0), where XA and XB are average drying times for samples of size nA = nB = 18. Solution: From the sampling distribution of XA - XB, we know that the distribution is approximately normal with mean and variance SO PXA-XBPA-PB=0 OXA-XB Z= + TLA TB HA-HB=0 1 18 + 1 18 1 9° 0X₁-X₂=√1/9 1-(HA-HB) 1-0 √1/9 √1/9 Figure 8.5: Area for Case Study 8.2. The desired probability is given by the shaded region in Figure 8.5. Corre- sponding to the value XA - XB = 1.0, we have 1.0 XA-XB = 3.0; P(Z >3.0) = 1- P(Z <3.0)=1-0.9987= 0.0013.
Consider Case Study 8.2 on page 238. Suppose 18 specimens were used for each type of paint in an experiment and XA -XB, the actual difference in mean drying time, turned out to be 1.0. (a) Does this seem to be a reasonable result if the two population mean drying times truly are equal? Make use of the result in the solution to Case Study 8.2. (b) If someone did the experiment 10,000 times under the condition that μA = μB, in how many of those 10,000 experiments would there be a difference XA - XB that was as large as (or larger than) 1.0? Case Study 8.2: Paint Drying Time: Two independent experiments are run in which two different types of paint are compared. Eighteen specimens are painted using type A, and the drying time, in hours, is recorded for each. The same is done with type B. The population standard deviations are both known to be 1.0. Assuming that the mean drying time is equal for the two types of paint, find P(XA-XB> 1.0), where XA and XB are average drying times for samples of size nA = nB = 18. Solution: From the sampling distribution of XA - XB, we know that the distribution is approximately normal with mean and variance SO PXA-XBPA-PB=0 OXA-XB Z= + TLA TB HA-HB=0 1 18 + 1 18 1 9° 0X₁-X₂=√1/9 1-(HA-HB) 1-0 √1/9 √1/9 Figure 8.5: Area for Case Study 8.2. The desired probability is given by the shaded region in Figure 8.5. Corre- sponding to the value XA - XB = 1.0, we have 1.0 XA-XB = 3.0; P(Z >3.0) = 1- P(Z <3.0)=1-0.9987= 0.0013.
Linear Algebra: A Modern Introduction
4th Edition
ISBN:9781285463247
Author:David Poole
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Chapter7: Distance And Approximation
Section7.3: Least Squares Approximation
Problem 31EQ
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Transcribed Image Text:Consider Case Study 8.2 on page 238. Suppose 18 specimens were used for each type of paint in
an experiment and XA -XB, the actual difference in mean drying time, turned out to be 1.0.
(a) Does this seem to be a reasonable result if the two population mean drying times truly are
equal? Make use of the result in the solution to Case Study 8.2.
(b) If someone did the experiment 10,000 times under the condition that µÃ = µ³, in how many of
those 10,000 experiments would there be a difference XA XB that was as large as (or larger
than) 1.0?
Case Study 8.2: Paint Drying Time: Two independent experiments are run in which two different
types of paint are compared. Eighteen specimens are painted using type A, and
the drying time, in hours, is recorded for each. The same is done with type B.
The population standard deviations are both known to be 1.0.
Assuming that the mean drying time is equal for the two types of paint, find
P(XA-XB> 1.0), where XA and XB are average drying times for samples of size
nA = nB = 18.
Solution: From the sampling distribution of XA - XB, we know that the distribution is
approximately normal with mean
and variance
PXA-XB = PA-PB=0
SO
OXA-X
-XB
0² 0²3 1 1 1
+
nA nB 18 18
= + =
9
0X₁-X₁=√1/9
1-(HA-HB)
√1/9
RA-UB=0
Figure 8.5: Area for Case Study 8.2.
The desired probability is given by the shaded region in Figure 8.5. Corre-
sponding to the value XA - XB = 1.0, we have
1.0
1-0
√1/9
XA-XB
= 3.0;
P(Z >3.0) = 1- P(Z <3.0) = 1-0.9987= 0.0013.
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