Applied Statistics and Probability for Engineers

(Chris Devlin) #1
330 CHAPTER 10 STATISTICAL INFERENCE FOR TWO SAMPLES

EXAMPLE 10-1 A product developer is interested in reducing the drying time of a primer paint. Two formula-
tions of the paint are tested; formulation 1 is the standard chemistry, and formulation 2 has a
new drying ingredient that should reduce the drying time. From experience, it is known that
the standard deviation of drying time is 8 minutes, and this inherent variability should be un-
affected by the addition of the new ingredient. Ten specimens are painted with formulation 1,
and another 10 specimens are painted with formulation 2; the 20 specimens are painted in
random order. The two sample average drying times are minutes and
minutes, respectively. What conclusions can the product developer draw about the effective-
ness of the new ingredient, using 0.05?
We apply the eight-step procedure to this problem as follows:


  1. The quantity of interest is the difference in mean drying times,  1  2 , and  0 0.



  2. We want to reject H 0 if the new ingredient reduces mean drying time.

  3. 0.05

  4. The test statistic is


where ^21 ^22 64 and n 1 n 2 10.
6. Reject H 0 :  1  2 if z 0
1.645z0.05.


  1. Computations: Since minutes and minutes, the test statistic is


z 0 

121  112

B

1822
10

1822
10

2.52

x 1  121 x 2  112

1822

z ̨ 0 

x 1 x 2  0

B

^21
n 1

^22
n 2

H ̨ 1 :  1 
 2.

H ̨ 0 :  1  2 0, or H ̨ 0 : ̨  1  2.

x ̨ 1  121 x ̨ 2  112

distribution when H 0 is true, we would takez and z as the boundaries of the critical re-
gion just as we did in the single-sample hypothesis-testing problem of Section 9-2.1. This
would give a test with level of significance. Critical regions for the one-sided alternatives
would be located similarly. Formally, we summarize these results below.


2
2

Null hypothesis:

Test statistic: (10-2)

Alternative Hypotheses Rejection Criterion

H ̨ 1 :  1  2  0 z 0 z

H ̨ 1 :  1  2 
 0 z 0
z

H 1 ̨:  1  2  0 z 0
z
2 or z 0 z
2

Z 0 

X 1 X 2  0

B

^21
n 1

^22
n 2

H 0 :  1  2  0

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