Download COMPARING MEANS: INDEPENDENT SAMPLES

COMPARING MEANS: INDEPENDENT SAMPLES
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 based on the two samples.
Test
Ho: µx = µy
vs
Ha: µx ≠ µy or
Ha: µx > µy or Ha: µx < µy
Procedure depends on what can we assume about variability of the
populations: σx and σy.
CASE1. σx and σy are known.
CASE2. σx and σy are not known, but may be assumed equal σx=σy
CASE3. σx and σy are not known, and can not be assumed equal.
Test statistics are developed for each of the 3 cases.
COMPARING MEANS: INDEPENDENT SAMPLES
CASE 1: σx and σy known
Test on significance level α.
STEP1. Ho: µx = µy
STEP 2. Test statistic:
vs
Ha: µx ≠ µy or
z=
x−y
σ x2
m
+
σ
2
y
Ha: µx > µy
.
n
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: µ ≠ µo Reject Ho if |z|> zα/2;
Ha: µ > µo Reject Ho if z > zα;
Ha: µ < µo
Reject Ho if z < - zα.
STEP 5. Answer the question in the problem.
COMPARING MEANS: INDEPENDENT SAMPLES
CASE 2: σx and σy not known, but assumed equal.
STEP 2. Test statistic:
where
s 2p
t=
x−y
,
1 1
+
sp
m n
is a pooled estimate of the common variance
1
2
2
s =
(
m
1)
s
(
n
1)
s
−
+
−
{
x
y }.
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: µ ≠ µo Reject Ho if |t|> tα/2;
Ha: µ > µo Reject Ho if t > tα;
Ha: µ < µo
Reject Ho if t < - tα.
COMPARING MEANS: INDEPENDENT SAMPLES
CASE 3: σx and σy not known, and may not be assumed equal.
STEP 2. Test statistic:
t=
x−y
2
y
.
sx2 s
+
m n
Under Ho, the degrees of freedom for the t distribution may be
approximated by df=min(m-1, n-1).
STEP 3. Critical value? One-sided test tα, two-sided tα/2 .
STEP 4. DECISION-critical/rejection region(s) depends on Ha.
Ha: µ ≠ µo Reject Ho if |t|> tα/2;
Ha: µ > µo Reject Ho if t > tα;
Ha: µ < µo
Reject Ho if t < - tα.
EXAMPLE1
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,
equality of variances/st.dev. σx=σy
y
=150, sy =30. Assumed
EXAMPLE1 contd.
STEP1. Ho: µx = µy (medicine not effective) vs
Ha: µx > µy (med. effective)
STEP 2. Pooled variance:
s 2p =
(m − 1) sx2 + (n − 1) s y2
m+n−2
(15 − 1)502 + (13 − 1)302
=
= 1761.54.
15 + 13 − 2
Standard deviation s p = s 2p = 1761.54 = 41.97
Test statistic:
t=
x−y
180 − 150
=
= 1.8863.
1 1
1 1
41.97
+
+
sp
15 13
m n
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.
(1-α)100% CONFIDENCE INTERVAL FOR (µx- µy)
CASE 1: σx and σy known
( x − y ) ± zα /2
σ x2
m
+
σ y2
n
.
CASE 2: σx and σy not known, but assumed equal.
( x − y ) ± tα /2 s p
1 1
+ .
m n
Use t distribution with df=m+n-2.
CASE 3: σx and σy not known, and may not be assumed equal.
( x − y ) ± tα / 2
Use t distribution with df=min(m-1, n-1).
2
sx2 s y
+ .
m n
EXAMPLE1 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).
13 15
NOTE: The interval contains zero. Intuitively, that conforms our
decision that there is no difference in means between the medicine
and the placebo.
MINITAB EXERCISE
Is there a significant difference between
test scores on the 1st and 2nd test? Use
data in example176.xls (see class web
site).
MINITAB EXAMPLE contd.
Two Sample T-Test and Confidence Interval
Two sample T for exm1 vs exm2
Boxplots of exm1 and exm2
(means are indicated by solid circles)
N
Mean StDev SE Mean
exm1 30
77.2
10.5
1.9
exm2 30
69.0
21.9
4.0
100
90
80
70
60
50
40
30
20
exm1
95% CI for mu exm1 - mu exm2: ( -0.8, 17.1)
T-Test mu exm1 = mu exm2 (vs not =): T = 1.85 P = 0.072 DF = 41
exm2