Transcript 9-5

9-5
Day 1
Testing Paired Differences
(independent samples)
What do you do if samples are
independent?
Comparing period 5 Calc average on the
midterm and period 6 Calc average
midterm?
Get a random sample of seniors and compare
GPAs of girls to GPA of boys…
If the samples are independent then you
compare mean differences using a new
formula
σ Known or Unknown
H0 : μ1  μ2  0
μ1  μ2
HA : μ1  μ2  0
μ1  μ2
Left Tailed
HA : μ1  μ2  0
μ1  μ2
Right Tailed
HA : μ1  μ2  0
μ1  μ2
Customary…
Two Tailed
And the usuals…
x1 and x2 should have normal
distributions with mean μ1 and μ2. If
both n1 and n2 are larger than 30, then
the CLT assures you that the
distributions of the xbars are normal.
How do I do it?
So glad you asked!!
Using the same pattern
1. State null and alternate hypotheses
2. Compute your test statistic
3. Find the P-value that corresponds to
the sample test statistic
4. Conclude
5. State your conclusion
σ known
x1  x 2   μ1  μ2 

z
2
1
2
2
σ
σ

n1 n2
z
 x1  x 2 
2
1
2
2
σ
σ

n1 n2
σ unknown
x1  x 2   μ1  μ2 

t
2
1
2
2
s
s

n1 n2
t
 x1  x 2 
2
1
2
2
s
s

n1 n2
Example
A random sample of n1=12 winter days in
Denver gave a mean pollution index of 43.
For Englewood (a suburb of Denver) a
random sample of n2=14 winter days gave a
sample pollution index of 36. Assume that
pollution index is normally distributed, and
previous studies show σ1= 21 and σ2 = 15.
Does this information suggest that the mean
population pollution index of Englewood is
different from Denver in the winter?
Use 1% level of significance.
Example
A random sample of n1 =16 communities in western
Kansas gave an average rate of hay fever (per
1000, under 25 yrs of age) of 109.50 with a sample
standard deviation of 15.41. A random sample of n2
=14 regions in western Kansas gave an average
rate of hay fever (per 1000, over 50 yrs of age) of
99.36 with a sample standard deviation of 11.57.
Assuming that the hay fever rate of each group is
approximately normal, does the data suggest that
the 50+ has a lower rate of hay fever?
Use a 5% level of confidence.