SECTION 6.1
133
Chapter 6
Section 6.1
1. (a)
X
783,
s
120,
n
73. The null and alternate hypotheses are
H
0
:
μ
750 versus
H
1
:
μ
750.
z
783
750
120
73
2 35. Since the alternate hypothesis is of the form
μ
μ
0
, the
P
value is the
area to the right of
z
2 35.
Thus
P
0 0094.
(b) The
P
value is 0.0094, so if
H
0
is true then the sample is in the most extreme 0.94% of its distribution.
3. (a)
X
1 90,
s
21 20,
n
160. The null and alternate hypotheses are
H
0
:
μ
0 versus
H
1
:
μ
0.
z
1 90
0
21 20
160
1 13. Since the alternate hypothesis is of the form
μ
μ
0
, the
P
value is the
sum of the areas to the right of
z
1 13 and to the left of
z
1 13.
Thus
P
0 1292
0 1292
0 2584.
(b) The
P
value is 0.2584, so if
H
0
is true then the sample is in the most extreme 25.84% of its distribution.
5. (a)
X
51 2,
s
4 0,
n
110. The null and alternate hypotheses are
H
0
:
μ
50 versus
H
1
:
μ
50.
z
51 2
50
4 0
110
3 15. Since the alternate hypothesis is of the form
μ
μ
0
, the
P
value is area
to the right of
z
3 15.
Thus
P
0 0008.
(b) If the mean tensile strength were 50 psi, the probability of observing a sample mean as large as the value of 51.2
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 Fall '11
 MattGilg
 Statistics, Null hypothesis, Harshad number, alternate hypotheses

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