This preview shows pages 1–2. Sign up to view the full content.
Solutions  Hypothesis Testing
9.4 Iowa GPA
:
0
H
:
μ= 2.80
a
H
:
μ
≠
2.80
In the above hypotheses,
0
H
: is the notation for the null hypothesis,
a
H
:
is the notation for the
alternative hypothesis, and μ is the parameter, the mean GPA of the population, about which
we’re hypothesizing.
9.8 Pvalue:
a) This Pvalue does not give strong evidence against the null hypothesis.
b) This extreme Pvalue does give strong evidence against the null hypothesis.
9.17 Another test of therapeutic touch
:
a)
p
= proportion of trials guessed correctly
0
H
:
p
= 0.50 and
a
H
:
p
> 0.50
b)
p
ˆ
= 53/130 = 0.4077
se
=
(
29
=

n
p
p
/
1
(
29
=

130
/
50
.
1
50
.
0.0439
z
= (0.4077 – 0.5)/0.0439 =  2.10; the sample proportion is a bit more than 2 standard errors
less than would be expected if the null hypothesis were true.
c) The Pvalue is 0.9821 (rounds to 0.98). Because the Pvalue is so much greater than the
significance level of 0.05, we do not reject the null hypothesis. The probability would be 0.98
of getting a test statistic at least as extreme as the value observed if the null hypothesis were
true, and the population proportion were 0.50.
d)
np
= (130)(0.5) = 65=
n
(1
p
); the sample size was large enough to make the inference in (c).
We also would need to assume randomly selected practitioners and subjects for this to apply
to all practitioners and subjects.
9.21 Garlic to repel ticks
:
a) The relevant variable is whether garlic or placebo is more effective, and the parameter is the
population proportion,
p
= those for whom garlic is more effective than placebo.
b)
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview. Sign up
to
access the rest of the document.
 '08
 BARROSO,JOAOR
 Statistics

Click to edit the document details