With H
0
:
0
2
1
=

μ
vs. H
a
:
0
2
1
≠

, we will reject H
0
if
α
<

value
p
.
(
29
(
29
(
29
6
8
.
6
4
5
2
5
240
.
2
6
164
.
2
5
240
.
6
164
.
2
2
2
2
≈
=
+
+
=
ν
, and the test statistic
17
.
6
1265
.
78
.
95
.
21
73
.
22
5
240
.
6
164
.
2
2
=
=
+

=
t
leads to a pvalue of 2[ P(t > 6.17)] < 2(.0005) =.001, which is less than most reasonable
s
'
, so we reject H
0
and conclude that
there is a difference in the densities of the two brick types./
We want a 95% confidence interval for
2
1

.
262
.
2
9
,
025
.
=
t
, so the interval is
(
29
(
29
80
.
6
,
40
.
20
007
.
3
262
.
2
6
.
13


=
±

.
Because the interval is
so wide, it does not appear that precise information is available./
Let
=
1
the true average strength for wirebrushing preparation and let
=
2
the average strength for handchisel preparation.
Since we are concerned about any possible difference
between the two means, a twosided test is appropriate.
We test
0
:
2
1
0
=

H
vs.
0
:
2
1
≠

a
H
. We need the degrees of freedom to find the rejection region:
(
29
(
29
(
29
33
.
14
1632
.
0039
.
3964
.
2
11
11
2
5
01
.
4
2
12
58
.
1
2
12
01
.
4
12
58
.
1
2
2
2
2
=
+
=
+
+
=
, which we round down to 14, so we reject H
0
if
145
.
2
14
,
025
.
=
≥
t
t
.
The test statistic is
(
29
159
.
3
2442
.
1
93
.
3
13
.
23
20
.
19
12
01
.
4
12
58
.
1
2
2

=

=
+

=
t
, which is
145
.
2

≤
, so we reject H
0
and conclude that there does appear to be a difference between the two population
average strengths.
/
a.
95% upper confidence bound:
x
+
t
.05,651
SE
= 13.4 + 1.671(2.05) = 16.83 seconds
Comment
:
The above is constructed using a
onesample
CI
b.
Let
μ
1
and
μ
2
represent the true average time spent by blackbirds at the experimental and natural locations, respectively. We wish to test H
0
:
μ
1
–
μ
2
= 0 v. H
a
:
μ
1
–
μ
2
> 0. The
relevant test statistic is
2
2
76
.
1
05
.
2
7
.
9
4
.
13
+

=
t
= 1.37, with estimated df =
49
76
.
1
64
05
.
2
)
76
.
1
05
.
2
(
4
4
2
2
2
+
+
≈ 112.9. Rounding to
t
= 1.4 and df = 120, the tabulated
P

value is very roughly .082. Hence, at the 5% significance level, we fail to reject the null hypothesis. The true average time spent by blackbirds at the experimental location is
not statistically significantly higher than at the natural location.
c.
95% CI for silvereyes’ average time – blackbirds’ average time at the natural location: (38.4 – 9.7) ± (2.00)
2
2
06
.
5
76
.
1
+
= (17.96 sec, 39.44 sec). The
t
value 2.00
is based on estimated df = 55.
Comment:
Notice
how standard error of the mean (not standard deviation of individual measurements) was
reported in the data and how calculations in the problem were
different because of this./
a.
Following the usual format for most confidence intervals:
statistic
±
(critical value)(standard error),
a pooled variance confidence interval for the difference between two
means is
(
29
n
m
p
n
m
s
t
y
x
1
1
2
,
2
/
+
⋅
±


+
.
b.
The sample means and standard deviations of the two samples are
90
.
13
=
x
,
225
.
1
1
=
s
,
20
.
12
=
y
,
010
.
1
2
=
s
.
The pooled variance
estimate is
=
2
p
s
(
29
(
29
2
2
2
2
2
1
010
.
1
2
4
4
1
4
225
.
1
2
4
4
1
4
2
1
2
1

+

+

+

=

+

+

+

s
n
m
n
s
n
m
m
260
.
1
=
, so
1227
.
1
=
p
s
.
With df = m+n2
= 6 for this interval,
447
.
2
6
,
025
.