STAT 101  Agresti
Homework 4 Solutions
10/8/10
Chapter 6
6.1. (a) null hypothesis (b) alternative hypothesis (c) alternative hypothesis (d)
0
:
0.50
H
π
=
;
:
0.24
a
H
π
<
;
:
100
a
H
μ
.
6.2. (a) Let
µ
= the population mean ideal number of children.
0
:
2
H
μ
=
and
:
2
a
H
μ
≠
. (b) The test
statistic is
t
= 20.80. This test statistic was obtained with the following:
0
2.49
2
20.80
0.850
1302
y
t
se
μ


=
=
=
. (c) The
P
value is the probability, assuming the null hypothesis is
true, that the test statistic equals the observed value or a value even more extreme in the direction
predicted by the alternative hypothesis. In this case, the
P
value = 0.0000 (rounded to 4 decimal places),
which means that if the true mean were 2, we would see results as extreme as or more extreme than we
did almost never.
6.3. (a)
(
29
(
29
(
29
2
1.04
2
1.04
2 0.1492
0.30
P
P t
P z
=
≈
=
=
. If the true mean were 0, we would see
results at least as extreme as we did about 30% of the time. (b) Now,
(
29
(
29
(
29
2
2.50
2
2.50
2 0.0062
0.012
P
P t
P z
=
< 
≈
< 
=
=
. Since the
P
value is smaller than the
P

value calculated in part (a), we have stronger evidence against the null hypothesis. (c) (i) 0.15 (ii) 0.85.
6.6. “Upper” = 11.0; “tvalue” = 4.19; “df” = 16.
6.9. (a) The null hypothesis is
0
:
0
H
μ
=
, and the alternative hypothesis is
:
0
a
H
μ
≠
. (b)
0
0.052
0
1.31
0.0397
y
t
se
μ



=
=
= 
. The
P
value is
P
= 0.19. Since
P
> 0.05, we fail to reject the null
hypothesis. There is not enough evidence to conclude that the mean score differs from the neutral value of
zero. (c) We cannot accept
0
:
0
H
μ
=
, since failing to reject the null hypothesis does not prove the null
hypothesis to be true. We just did not have enough evidence to reject the null hypothesis. (d) A 95%
confidence interval is
(
29
0.052
1.96
0.0397
0.13 to 0.03
s
y
t
n
±
= 
±
= 
. Note that zero falls within
our confidence interval. When the null hypothesis value falls within the confidence interval, we fail to
reject the null hypothesis; when the null hypothesis value falls outside of the confidence interval, we
reject the null hypothesis (with Type I error probability = the error probability for the CI).
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6.11. Results of 99% confidence intervals for means are consistent with results of twosided tests at
α
= 1
– 0.99 = 0.01 level.
6.14. (a)
(
29
0
0.50 0.50
0.0158
1000
se
=
=
(b)
0
0
ˆ
0.55
0.50
3.16
0.0158
z
se
π
π


=
=
=
The sample proportion
falls 3.16 standard errors in absolute value away from the null hypothesis value of the proportion. (c)
(
29
(
29
2
3.16
2 0.0008
0.0016
P
P z
=
=
=
If the population proportion were 0.50, then the probability
equals 0.0016 that a sample proportion of
n
= 1000 subjects would fall as extreme as or more extreme
than 0.55. We reject the null hypothesis. There is enough evidence to conclude that the population
proportion of Canadian adults who believe the bill should stand differs from 0.50.
6.15. (a)
0
:
0.50
H
π
=
and
:
0.50
a
H
π
≠
. (b)
z
= –3.91 The sample proportion falls 3.91 standard
errors below the null hypothesis value of the proportion. (c)
P
= 0.000 (rounded).
If the null hypothesis
were true, then there is almost no chance that we would see results as extreme as or more extreme than
our sample results.
We would conclude that the proportion of Americans who would be willing to pay
higher taxes in order to protect the environment differs from 0.50. (d) The confidence interval is entirely
below 0.50, which leads us to believe that a minority of Americans would be willing to pay higher taxes
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 Fall '10
 ALAN
 Statistics, Null hypothesis, Statistical hypothesis testing

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