Homework Solutions
Chapter 9
Section 9.1
2.
The data is summarized in the table.
a.
The critical value is z = 1.96.
The standard error for a difference in means is:
The confidence interval for
X
–
Y
is
b.
The research question is; does the mean height of women aged 2039 exceed
the mean height of women 60 and older by
at least one inch?
The following hypotheses take opposite sides of the question.
H
0
:

= 1 (it does not)
versus
H
a
:

> 1 (it does).
The large sample test statistic is
The rejection region for a onesided test
on the right is
, into which the test statistic clearly falls.
Therefore, we reject the null hypothesis.
The
data strongly indicate that the mean height of the younger women exceeds the mean height of the older women by more
than one inch.
(They haven’t had to raise teenage daughters yet!)
c.
Using the Excel NORMSDIST function, the pvalue for this test is P(Z > 5.64) = 1
–
5.64) = 1
–
0.999 999 9915
= .000 000 0085.
Unless one believes in 117,612,352 to 1 odds, one would reject the null hypothesis.
d. H
0
:

=
–
1
versus
H
a
:

<
–
1
5.
The data is summarized in the table.
Since the population data is normal with
known standard deviations, a “large sample” procedure is appropriate.
The
respective standard errors are
and
.
a. The hypotheses are
H
0
:

=
–
1.0 (=
0
) versus
H
a
:

<
–
1.0.
The alternate hypothesis says that the mean heat output for non
sufferers of Reynaud’s syndrome exceeds that of
sufferers by more than one cal/cm
2
/min.
The test statistic is
.
At the
= .01 level of significance, the critical value for this
onesided test on the left is z
.99
= 2.33.
Since 2.90 < 2.33, we reject the null hypothesis and conclude that the mean
heat output of Reynaud’s syndrome suffers is more than one cal/cm
2
/min than the mean for nonsuffers.
b.
The pvalue = P(Z < 2.90) = .0019.
c.
In this problem,
’ represents the difference in means under the alternate hypothesis.
That is,
’

=
–
1.2.
Now
d.
Using m = n, set
So the goal will be accomplished if n = 66.
8.
The data is summarized in the table.
a. The hypotheses are
H
0
:
Y

X
= 10
versus
H
a
:
Y

X
> 10
The test statistic is
.
The pvalue is essentially zero, so at any level of
significance, we reject H
0
.
The data overwhelmingly suggest that the mean tensile strength of the 1078 grade exceeds
the mean of the 1064 grade by more than 10 kg/mm
2
.
b.
A 95% confidence interval for
Y

X
will provide information on precision and reliability.
The confidence interval is given
by
.
We are 95%
confident that the mean tensile strength of all 1078 grade wire rods exceeds the mean tensile strength of the 1064 grade
wire rods by 15.59 to 16.41 kg/mm
2
.
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 Spring '04
 Chung
 Normal Distribution, Standard Error, 4 week, 0.1%, actual size, 1 min, 5 degrees

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