Download Lecture 19, Nov 15.

Math/Stat 352
Lecture 19
Section 6.5 and 6.7
Large Sample Tests for the difference of two means and Small
Sample Tests for the difference of two means: independent
samples
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Independent and dependent samples
Two samples are independent if the sample values selected from
one population are not related to or somehow paired or
matched with the sample values selected from the other
population.
Examples: weights of students in different univ., test results of
students in different towns, yields on different fields, etc.
Two samples are dependent (or consist of matched pairs) if the
members of one sample can be used to determine the members
of the other sample.
Examples: Test results for students before and after a study
session, weight of a group of people before and after a weight
loss program, predicted and true max temps for several days in
a given month in Reno, etc.
Large Sample Tests for the Difference Between Two Means:
Independent samples

Now, we are interested in determining whether or not the means of two
populations are equal.

The data will consist of two samples, one from each population. The
samples are independent.

IDEA FOR THE TEST:
 We will compute the difference of the sample means.

If the difference is far from 0, we will conclude that the population
means are different.

If the difference is close to 0, we will conclude that the population
means might be the same, or in other words, not enough evidence to
conclude significant difference.
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COMPARING MEANS: INDEPENDENT LARGE SAMPLES
FRAMEWORK: 1ST sample: x1, x2, …, xm from population with mean μx;
2nd sample: y1, y2, …, yn from population with mean μy;
GOAL: Determine if μx –μy =>< 0 or if μx –μy =>< ∆
Test
Ho: μx – μy=≥≤ ∆
vs
Ha: μx - μy ≠ <>∆
∆ is the difference between two means we want to test for.
Example: Ho: μx – μy = 0, here ∆=0. Or
Ho: μx – μy ≥ 2, here ∆=2.
Some example questions: Are the mean heights of students in UNR and UNLV
the same?
Is the difference between the mean heights of Americans and Chinese larger
than 2 inches?
COMPARING MEANS: INDEPENDENT LARGE SAMPLES or σ 2
X
known
,σ
Case 1: Large samples or σ X2 , σ Y2 known.
STEP1. Ho: μx – μy=≥≤ ∆ vs Ha: μx - μy ≠ ><∆, significance level α.
STEP 2. Test statistic:
z=
( X − Y ) − ∆0
σ / n X + σ / nY
2
X
2
Y
.
Under the Ho, the test statistic has standard normal distribution.
STEP 3. Critical value? For one-sided test zα, for two-sided zα/2 .
STEP 4. DECISION-critical/rejection region(s) depends on Ha.
Ha: μx - μy ≠ ∆ Reject Ho if |z|> zα/2;
Ha: μx - μy > ∆ Reject Ho if z > zα;
Ha: μx - μy < ∆ Reject Ho if z < - zα.
STEP 5. Answer the question in the problem.
2
Y
COMPARING MEANS: INDEPENDENT LARGE SAMPLES or
known
σ ,σ
2
X
Note: If σx and σy are not known, substitute sx and sy in the formula for
the test statistic.
Use also when small sample, but normal population and
known.
σ X2 , σ Y2
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2
Y
COMPARING MEANS: INDEPENDENT SMALL SAMPLES
CASE 2: σx and σy not known, but assumed equal.
STEP 2. Test statistic:
where
𝒔𝒑 𝟐
𝒕 =
�−𝒚
� −∆
𝒙
𝒔𝒑
𝟏
𝟏
+
𝒏𝒙 𝒏𝒚
is a pooled estimate of the common variance
1
(m − 1) sx2 + (n − 1) s y2 } .
s
=
{
m+n−2
2
p
Under the Ho, the test statistic has t distribution with df = m+n-2.
STEP 3. Critical value? One-sided test tα, two-sided tα/2 .
STEP 4. DECISION-critical/rejection region(s) depends on Ha.
Ha: μx - μy ≠ ∆ Reject Ho if |t|> tα/2;
Ha: μx - μy > ∆ Reject Ho if t > tα;
Ha: μx - μy < ∆ Reject Ho if t < - tα.
COMPARING MEANS: INDEPENDENT SMALL SAMPLES
CASE 3: σx and σy not known, and may not be assumed equal.
STEP 2. Test statistic:
𝒕 =
�−𝒚
� −∆
𝒙
𝒔𝒙𝟐 𝒔𝒚𝟐
+
𝒏𝒙 𝒏𝒚
Under Ho, the degrees of freedom for the t distribution may be
approximated by
v=
( s X2
2
( s nX ) + ( s nY ) 
nX ) 2 (nX − 1)  + ( sY2 nY ) 2 (nY − 1) 
2
X
2
Y
STEP 3. Critical value? One-sided test tα, two-sided tα/2 .
STEP 4. DECISION-critical/rejection region(s) depends on Ha.
Ha: μx - μy ≠ ∆ Reject Ho if |t|> tα/2;
Ha: μx - μy > ∆ Reject Ho if t > tα;
Ha: μx - μy < ∆ Reject Ho if t < - tα.
EXAMPLE
A medication for blood pressure was administered to a group of 13
randomly selected patients with elevated blood pressure while a group
of 15 was given a placebo. At the end of 3 months, the following data
was obtained on their Systolic Blood Pressure.
Control group, x: n=15, sample mean = 180, s=50
Treated group, y: m=13, sample mean =150, s=30.
Test if the treatment has been effective. Assume the variances are the
same in both groups and use α=0.01.
Soln. Let μx= mean blood pressure for the control group;
μy= mean blood pressure for the treatment group.
x
Then, n=15,
= 180, sx=50, m=13,
of variances/st.dev. σx=σy
y
=150, sy =30. Assumed equality
EXAMPLE contd.
STEP1. Ho: μx = μy (medicine not effective) vs
Ha: μx > μy (med. effective)
STEP 2. Pooled variance:
2
2
2
2
m
−
s
+
n
−
s
(
1)
(
1)
(15
1)50
(13
1)30
−
+
−
x
y
=
s 2p =
= 1761.54.
15 + 13 − 2
m+n−2
Standard deviation
=
sp
=
s 2p
1761.54
= 41.97
Test statistic:
x−y
180 − 150
=
t =
= 1.8863.
1 1
1 1
sp
+
41.97
+
m n
15 13
STEP 3. Critical value=t0.01=2.479, df=26.
STEP 4. t=1.8863 not > 2.479, do not reject Ho.
STEP 5. Not enough evidence to conclude that the medicine is effective.
Example
Sample statistics are shown for the distances of the home runs hit
in record-setting seasons by Mark McGwire and Barry Bonds. Use
a 0.05 significance level to test the claim that the distances come
from populations with different means.
McGwire
Bonds
n
70
73
x
418.5
403.7
s
45.5
30.6
Soln. Let μx= mean distance for McGwire;
μy= mean distance for Bonds.
CASE3. σx and σy are not known, and can not be assumed equal.
EXAMPLE contd.
STEP1. Ho: μx = μy (same mean distances) vs Ha: μx ≠ μy
(different mean distances)
Test statistic:
=
t
x−y
=
2
2
s
sx
+ y
m n
418.5 − 403.7
= 2.273.
2
45.5 30.62
+
70
73
STEP 3. Critical value= t0.025 = 1.994 , df = 120
STEP 4. t=2.273 > 1.994, reject Ho.
STEP 5. There is enough evidence to conclude that the mean
distances of the home runs for the two players are different.
NOTE: Since s. sizes are large, we could use z-test. Then
zα/2=z0.025 = 1.96, same conclusion.
Blood pressure EXAMPLE contd.
Construct a 95% CI for the difference in the means of blood
pressures for the two groups (μx - μy).
Soln. We already know n=15,
sy =30, sp=41.97.
x
= 180, sx=50, m=13, y
=150,
CASE 2. 95% CI, so α=0.05, so α/2=0.025, t(26)0.025 = 2.056.
95% CI is: (180 − 150) ± (2.056)(41.97)
1 1
+
=
(−2.7, 62.7).
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NOTE: The interval contains zero. Intuitively, that confirms our
decision that there is no difference in mean effects between the
medicine and the placebo.
Example
An article compares properties of welds made using carbon dioxide as a
shielding gas with those of welds made using a mixture of argon and
carbon dioxide. One property studied was the diameter of inclusions,
which are particles embedded in the weld. A sample of 544 inclusions in
welds made using argon shielding averaged 0.37µm in diameter, with a
standard deviation of 0.25 µm. A sample of 581 inclusions in welds made
using carbon dioxide shielding averaged 0.40 µm in diameter, with a
standard deviation of 0.26 µm. Can you conclude that the mean
diameters of inclusions differ between the two shielding gases?
MINITAB: Two-Sample T-Test and CI
Sample N Mean StDev SE Mean
1
10 32.30 8.56
2.7
2
10 44.1 10.1
3.2
Difference = mu (1) - mu (2)
Estimate for difference: -11.80
95% upper bound for difference: -4.52
T-Test of difference = 0 (vs <): T-Value = -2.82 P-Value = 0.006 DF = 17
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Example
Good website design can make Web navigation easier. An article presents a
comparison of item recognition between two designs. A sample of 10 users
using a conventional Web design averaged 32.3 items identified, with a
standard deviation of 8.56. A sample of 10 users using a new structured
Web design averaged 44.1 items identified, with a standard deviation of
10.09. Can we conclude that the mean number of items identified is greater
with the new structured design?
Two-Sample T-Test and CI
Sample N Mean StDev SE Mean
1
544 0.370 0.250 0.011
2
581 0.400 0.260 0.011
Difference = mu (1) - mu (2)
Estimate for difference: -0.0300
95% CI for difference: (-0.0598, -0.0002)
T-Test of difference = 0 (vs not =): T-Value = -1.97 P-Value = 0.049 DF = 1122
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Example
Two methods have been developed to determine the nickel content of steel. In a
sample of five replications of the first method, X, on a certain kind of steel, the average
measurement (in percent) was 3.16 with a standard deviation of 0.042. The average of
seven replications of the second method, Y, was 3.24, and the standard deviation was
0.048. Assume that it is known that the population variances are nearly equal. Can we
conclude that there is a difference in the mean measurements between the two
methods?
Minitab: Two-Sample T-Test and CI
Sample N Mean StDev SE Mean
1
5 3.1600 0.0420 0.019
2
7 3.2400 0.0480 0.018
Difference = mu (1) - mu (2)
Estimate for difference: -0.0800
95% CI for difference: (-0.1396, -0.0204)
T-Test of difference = 0 (vs not =): T-Value = -2.99 P-Value = 0.014 DF = 10
Both use Pooled StDev = 0.0457
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