Assignment 2
Answers
1
(a)
The null and the alternative hypothesis can be represented as follows;
0
:
0
:
1
0
≠

=

girls
boys
girls
boys
H
H
μ
μ
μ
μ
(b)
The difference in height and the standard error of the difference are as follows;
inches
Y
Y
SE
inches
Y
Y
girls
boys
girls
boys
77
.
0
57
)
2
.
4
(
55
)
9
.
3
(
)
(
6
.
0
2
2
_
_
_
_
=
+
=


=

(c)
95% confidence interval for the difference in height:
).
10
.
2
,
90
.
0
(
77
.
0
96
.
1
6
.
0

=
×
±

(d)
78
.
0
77
.
0
6
.
0
)
(
)
(

=

=


=





girls
boys
girls
boys
Y
Y
SE
Y
Y
stat
t
Based on the above, the 0.78<2.58. The 2.58 represents the critical value at the 1%
level. Hence you cannot reject the null hypothesis. The critical value for the onesided
hypothesis would have been 2.33. Assuming a onesided hypothesis implies that you
have some information about the problem at hand, and, as a result, can be more easily
convinced than if you had no prior expectation.
2
(a)
The confidence interval for the mean weekly earnings is;
]
69
.
452
,
29
.
416
[
20
.
18
49
.
434
1744
67
.
294
58
.
2
49
.
434
=
±
=
×
±
Based on the sample at hand, the best guess for the population mean is $434.49.
However, because of random sampling error, this guess is likely to be wrong. Instead, the
interval estimate for the average earnings lies between $416.29 and $452.69. Committing
to such and interval repeatedly implies that the resulting statement is incorrect 1 out of
100 times. For a 90% confidence interval is smaller. A smaller interval implies, given the
same average earnings and the standard deviation, that the statement will be false more
often. The larger the confidence interval, the more likely it is to contain the population
value.
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 Spring '07
 Francisco
 Statistics, Null hypothesis, Statistical hypothesis testing, critical value

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