IE 300/GE 331 Homework 8 Solutions
[Type text]
Page 1
910
The heat evolved in calories per gram of a cement mixture
is approximately normally distributed.
The mean is thought to be 100 and the standard deviation is 2. We wish to test
g
G
: ± = 100
versus
g
²
:± ≠ 100
with a sample of n=9 specimens.
(a)
If the acceptance region is defined as 98.5
≤ ³´ ≤
101.5, find the type
µ
error probability
¶
.
(b)
Find
·
for the case where the true mean heat evolved is 103.
(c)
Find
·
for the case where the true mean heat evolved is 105. This value of
·
is smaller than the
one found in part (b) above. Why?
Solutions
(a)
¶ = ¸¹º
»
< 98.5¼ + ¸¹³´ > 101.5¼ = ¸ ½
¾
»
¿²GG
À/√Á
<
ÁÂ.Ã¿²GG
À/√Á
Ä + ¸ ½
¾
»
¿²GG
À/√Á
>
²G².Ã¿²GG
À/√Á
Ä
= ¸¹Å < −2.25¼ + ¸¹Å > 2.25¼ = 0.01222 + 1 − 0.98778 = 0.02444
(b)
· = ¸¹98.5 ≤ º
»
≤ 101.5 ÆℎÇÈ ± = 103¼ = ¸ ½
ÁÂ.Ã¿²GÉ
À/√Á
≤
¾
»
¿²GÉ
À/√Á
≤
²G².Ã¿²GÉ
À/√Á
Ä
= ¸¹−6.75 ≤ Å ≤ −2.25¼ = ¸¹Å ≤ −2.25¼ − ¸¹Å ≤ −6.75¼ = 0.01222 − 0 = 0.01222
(c)
· = ¸¹98.5 ≤ º
»
≤ 101.5 ÆℎÇÈ ± = 105¼ = ¸ ½
ÁÂ.Ã¿²GÃ
À/√Á
≤
¾
»
¿²GÃ
À/√Á
≤
²G².Ã¿²GÃ
À/√Á
Ä
= ¸¹−9.75 ≤ Å ≤ −5.25¼ = ¸¹Å ≤ −5.25¼ − ¸¹Å ≤ −9.75¼ = 0 − 0 = 0
The probability of accepting the null hypothesis when it is actually false is smaller in part (c)
since the true mean,
± = 105
, is further from the acceptance region. A larger difference exists.
940
The mean water temperature downstream from a power plant cooling tower discharge pipe should
be no more than 100F. Past experience has indicated that the standard deviation of temperature is 2F.
The water temperature is measured on nine randomly chosen days, and the average temperature is
found to be 98F. (Assume that the data are normally distributed)
(a)
Is there evidence that he water temperature is acceptable at
¶ = 0.05
?
(b)
What is the Pvalue for this test?
(c)
What is the probability of accepting the null hypothesis at
¶ = 0.05
if the water has a true
mean temperature of 104F?
Solutions
(a)
(1) The parameter of interest is the true mean water temperature,
±
(2)
g
G
: ± = 100
(3)
g
²
:± < 100
(4)
¶ = 0.05
(5)
Å
G
=
Ê´¿Ë
Ì/√Í
(6) ) Reject
g
G
if
Î
G
< −Î
Ï
,where
−Î
G.GÃ
= −1.645
(7)
³´
=98,
σ
=2,n=9,
Î
G
=
ÁÂ¿²GG
À/√Á
=3
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 Spring '09
 Zafarani
 Normal Distribution, Statistical hypothesis testing, Type text

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