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STAT 350 Assignment 8 Solutions
(60 points)
1.
(5 points = 0.5 each)
(a) Yes,
> 100 is a statement about a population standard deviation, i.e., a statement about a population
parameter.
(b)
No, this is a statement about the statistic
x
, not a statement about a population parameter.
(c)
Yes, this is a statement about the population median
~
.
(d)
No, this is a statement about the statistic s (s is the sample
, not population, standard deviation.
(e)
Yes, the parameter here is the ratio of two other parameters; i.e,
1
/
2
describes some aspect of the
populations being sampled, so it is a parameter, not a statistic.
(f)
No, saying that the difference between two samples means is 5.0 is a statement about sample results, not
about population parameters.
(g)
Yes, this is a statement about the parameter
of an exponential population.
(h)
Yes, this is a statement about the proportion
of successes in a population.
(i) Yes, this is a legitimate hypothesis because we can make a hypothesis about the population distribution [see
(4) at the beginning of this section].
(j) Yes, this is a legitimate hypothesis.
We can make a hypothesis about the population parameters [see (3) at the
beginning of this section].
10.
(3 points = 0.5 each)
(a) H
0
should not
be rejected, since pvalue = .084 >
= .05.
(b)
H
0
should not
be rejected, since pvalue = .003 >
= .001.
(c)
H
0
should be rejected, since pvalue = .048 <
= .05.
(d)
H
0
should be rejected, since pvalue = .084 <
= .10.
(e)
H
0
should not
be rejected, since pvalue = .039 >
= .01.
(f)
H
0
should be rejected, since pvalue = .017 <
= .10.
12.
(4 points = 1 each)
(a)
87
.
2
50
06
.
1
34
43
.
34
z
So, pvalue = 2P(z > 2.87) = 2(.0021) = .0042
[Note:
Since H
a
is a twotailed test, the pvalue is the sum of the area in the two tails.
That is, pvalue = P(z
< 2.87) + P(z > 2.87) = 2P(z > 2.87)]
(b)
87
.
2
50
06
.
1
34
57
.
33
z
So, pvalue = 2P(z > 2.87) = 2(.0021) = .0042
(c)
24
.
2
32
89
.
1
34
25
.
33
z
So, pvalue = 2P(z > 2.24) = 2(.0125) = .025
(d)
57
.
1
36
53
.
2
34
66
.
33
z
So, pvalue = 2P(z > 1.57) = 2(.0582) = .1164
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 Spring '08
 Staff
 Statistics, Standard Deviation

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