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Unformatted text preview: 90%
confidence interval estimate for the
difference between the 2 population
means? Specific Motors of Detroit has
developed a new automobile
known as the M car. 24 M cars
and 28 J cars (from Japan)
were road tested to compare
mpg performance. The sample
statistics are:
M Cars
Hypothesis Tests n
x
s J Cars 24 28 29.8 27.3 2.56 1.81 Difference Between 2
Population Means
DM–J = μM – μJ parameter
point
estimate d x −J= 5
=M x 2
. standard
error s 2 s2
s = M + J =06 5
.2
d
n
n
M
J margin of
error s 2 s2
M
+ J = .0 3
15
n
n
M
J t0.102 interval
estimate
degrees of df =
freedom s2 s22
1
+
nn
1
2 d ±tα 2 ( + ) ≈40
( )+ ( )
sM2
nM sM2
1
nM − nM
1 2 sJ 2
nJ 2 sJ 2
1
nJ − nJ
1 2 Example Given these samples, what is the 90%
confidence interval estimate for the
difference between the 2 population
means? Hypothesis Tests Specific Motors of Detroit has
developed a new automobile
known as the M car. 24 M cars
and 28 J cars (from Japan)
were road tested to compare
mpg performance. The sample
statistics are:
s12 s2 2
d ± Cars + J Cars
M tα 2
n1 n 2
2
n
24 s 2 s28
M
J
t 0.10 2
x d ± 29.8 n + 27.3
n
M s Difference Between 2
Population Means
DM–J = μM – μJ parameter
point
estimate d x −J= 5
=M x 2
. standard
error s 2 s2
s = M + J =06 5
.2
d
n
n
M
J margin of
error s 2 s2
M
+ J = .0 3
15
n
n
M
J t0.102 J 2 .5 ±
2.56 1.053 1.81
1.448 ≤ D0 ≤ 3.553 interval
estimate
degrees of df =
freedom s2 s22
1
+
nn
1
2 d ±tα 2 ( + ) ≈40
( )+ ( )
sM2
nM sM2
1
nM − nM
1 2 sJ 2
nJ 2 sJ 2
1
nJ − nJ
1 2 Example Given these samples, what is the 90%
confidence interval estimate for the
difference between the 2 population
means? Specific Motors of Detroit has
developed a new automobile
known as the M car. 24 M cars
and 28 J cars (from Japan)
were road tested to compare
mpg performance. The sample
statistics are: Hypothesis Tests We are 90% J Cars
confident
M Cars
that the mean 28
n
24
dxifference in mpg
29.8
27.3
between the 2 cars is
s
2.56
1.81
between 1.448 and
3.553 mpg. Difference Between 2
Population Means
DM–J = μM – μJ parameter
point
estimate d x −J= 5
=M x 2
. standard
error s 2 s2
s = M + J =06 5
.2
d
n
n
M
J margin of
error s 2 s2
M
+ J = .0 3
15
n
n
M
J t0.102 interval
estimate
degrees of df =
freedom s2 s22
1
+
nn
1
2 d ±tα 2 ( + ) ≈40
( )+ ( )
sM2
nM sM2
1
nM − nM
1 2 sJ 2
nJ 2 sJ 2
1
nJ − nJ
1 2 Hypothesis Tests about the Difference
between Two Population Means (σ1 and σ2
unknown) difference between 2 population means: D = μ1 – μ2 to
◦ ◦ make inferences about D, d x− 2
=1 x take independent simple random samples from 2
populations
use sampling distribution of Hypothesis Tests ◦ use s1 and s2 to estimate population standard deviations
use tdistribution for making inferences decision rules for hypothesis test based on either pvalue
or critical value approach calculate test statistic and decide based on decision rule Hypothesis Tests about the Difference
between Two Population Means (σ1 and
σ2 unknown)
Population Mean test statistic upper tail test tSTAT = x −µ
0
sn degrees of freedom = n –
1
H0: μ ≤ μ0
Ha: μ > μ0
reject H0 if tSTAT > tα or pvalue<α Hypothesis Tests lower tail test H0: μ ≥ μ0
Ha: μ < μ0
reject H0 if tSTAT <– tα or pvalue<α twotailed test H0: μ = μ0
Ha: μ ≠ μ0
reject H0 if tSTAT < –tα/2 or tSTAT >
tα/2 or if pvalue<α Recall from
chapter 9 Hypothesis Tests about the Difference
between Two Population Means (σ1 and...
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 Spring '13

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