statistics 300: elementary statistics section 9-3
TRANSCRIPT
Statistics 300:Elementary Statistics
Section 9-3
Section 9-3 concernsConfidence Intervals
andHypothesis tests forthe difference of two
means, (1 – 2)
222yxyx
What is the variance of
? )( 21 xx • Apply the new concept here also:
Application: Use sample values
y
y
x
xyx
yxyx
yxyx
n
s
n
ss
sss22
2
222
222
Alternative approach when two samples come from
populations with equal variances
[ “homogeneous variances” or “homoscedastic” ]
22
21
When samples come from populations with equal variances:
222
221
222
21
of estimatean is
of estimatean is
common
common
common
s
s
and
When samples come from populations with equal variances:
21
222
2112
2
: is
of estimatebest the
dfdf
sdfsdfsp
common
Alternative notation for thepreceding formula:
11
11
21
222
2112
nn
snsnsp
When doing confidence intervalsof hypothesis tests involving
(m1 – m2) one must first decide whether the variances are
different or the same,heterogeneous or homogeneous.
If the variances are different(heterogeneous), then use thefollowing expression as part
of the confidence interval formulaor the test statistic:
2
22
1
21
n
s
n
s
If the variances are the same(homogeneous), then use thisalternative expression as part
of the confidence interval formulaor the test statistic:
pooled"" means
"" where2
2
1
2
p
n
s
n
s pp
When the variances are pooledto estimate the common
(homogeneous) variance, then the degrees of freedom for both
samples are also pooled by adding them together.
Application to CI(1 - 2)
212/
2121 )()(
xxstE
ExxCI
Application to CI(1 - 2)when variances are different
2
22
1
21
2/
2121 )()(
n
s
n
stE
ExxCI
In this case, the degrees of freedom for “t” will bethe smaller of the two sample degrees of freedom.
Application to CI(1 - 2)when variances are the same
2
2
1
2
2/
2121 )()(
n
s
n
stE
ExxCI
pp
In this case, the degrees of freedom for “t” will bethe sum of the two sample degrees of freedom.
Tests concerning (1 - 2)Test Statistic
)(
02121
21
)()(
xxs
xx
Test statistic for (1 - 2)when variances are different
2
22
1
21
02121 )()(
ns
ns
xx
In this case, the degrees of freedom for “t” will bethe smaller of the two sample degrees of freedom.
Test statistic for (1 - 2)when variances are the same
2
2
1
2
02121 )()(
n
s
n
s
xx
pp
In this case, the degrees of freedom for “t” will bethe ssum of the two sample degrees of freedom.
Section 9-3Handling Claims / Hypotheses
• Write the claim in a symbolic expression as naturally as you can
• Then rearrange the expression to have the difference between the two means on one side of the relational operator (< > = …)
Section 9-3Handling Claims / Hypotheses
• Statement: Mean #2 is less than four units more than Mean #1
• So:
• Rearrange:
• H0:
• H1:
412 412
412 412
Section 9-3Handling Claims / Hypotheses
• Statement: On average, treatment A produces 18 more than treatment B
• So:
• Rearrange:
• H0:
• H1:
18 BA 18 BA
18 BA
18 BA