Stat 104 – Lecture 27
Two Independent Samples
• Do males and females at
I.S.U. spend the same amount
of time, on average, at the
Lied Recreation Athletic
Center?
1
Populations
random
selection
2. Male
Inference
1. Female
Samples
random
selection
2
Time (minutes)
1. Females
63, 32, 86, 53, 49
73, 39, 56, 45, 67
49, 51, 65, 54, 56
2. Males
52, 75, 74, 68, 93
77, 41, 87, 72, 53
84, 65, 66, 69, 62
3
1
Stat 104 – Lecture 27
Time (minutes)
Sex=F
Mean
Std
Dev
n
Sex=M
55.87 Mean
13.527 Std
Dev
15 n
69.20
13.790
15
4
Comment
• This sample of I.S.U.
females spends, on average,
13.33 minutes less time at the
Lied Recreation Athletic
Center than this sample of
I.S.U. males.
5
Confidence Interval: μ
1
( y1 − y2 ) ± t
*
− μ2
s12 s22
+
n1 n2
t * from Table T,
df = nasty formula
6
2
Stat 104 – Lecture 27
Finding t*
• Use Table T.
• Confidence Level in last row.
• df = 27.99 or 28.
7
Table T
80%
Confidence Level
90% 95% 98%
99%
df
1
2M
3
4
2.048
28
8
s12 s22
+
=
n1 n2
(13.527)2 + (13.792)2
15
15
= 24.88 = 4.988
9
3
Stat 104 – Lecture 27
Confidence Interval:μ
1
( y1 − y2 ) ± t *
− μ2
s12 s22
+
n1 n2
(55.87 − 69.2) ± 2.048(4.988)
− 13.33 ± 10.22
− 23.55 to − 3.11
10
Interpretation
• We are 95% confident that
I.S.U. females spend, on
average, from 3.11 to 23.55
minutes less time at the Lied
Recreation Athletic Center
than I.S.U. males do.
11
Test of Hypothesis:μ
1
− μ2
• Step 1: Assumptions.
–Quantitative response for two
groups.
–Independent random samples.
–Approximately normal
distribution for both groups.
12
4
Stat 104 – Lecture 27
Test of Hypothesis:μ
1
− μ2
• Step 2: Hypotheses.
H 0 : μ1 = μ2 or H 0 : μ1 − μ2 = 0
H A : μ1 ≠ μ 2 or H A : μ1 − μ2 ≠ 0
13
Test of Hypothesis:μ
1
− μ2
• Step 3: Test Statistic.
t=
( y1 − y 2 ) − 0
s12 s 22
+
n1 n 2
14
Time (minutes)
Sex=F
Mean
Std
Dev
n
Sex=M
55.87 Mean
13.527 Std
Dev
15 n
69.20
13.790
15
15
5
Stat 104 – Lecture 27
(13.527)2 + (13.792)2
s12 s22
+
=
n1 n2
15
15
= 24.88 = 4.988
16
Test of Hypothesis:μ
1
− μ2
• Step 3: Test Statistic.
t=
( y1 − y2 ) − 0 = (55.87 − 69.20)
s12 s22
+
n1 n2
4.988
t = −2.672
17
Table T
0.100
Right-Tail probability
0.050 0.025 0.010 Prob
0.005
df
1
2
3
4
M
28
1.313
1.701 2.048
2.467
2.672
2.763
18
6
Stat 104 – Lecture 27
Test of Hypothesis:μ
1
− μ2
• Step 4: Probability value
–The P-value is between
2(0.005) = 0.010 and 2(0.010)
= 0.020.
19
Test of Hypothesis: μ
1
− μ2
• Step 5: Results.
– Reject the null hypothesis because
the P-value is smaller than α = 0.05
– The difference in mean times is not
zero. Therefore, on average, females
and males at I.S.U. spend different
amounts of time at the Lied
Recreation Athletic Center.
20
Comment
• This conclusion agrees with the
results of the confidence interval.
• Zero is not contained in the 95%
confidence interval (–23.55 mins to
–3.11 mins), therefore the
difference in population mean
times is not zero.
21
7
Stat 104 – Lecture 27
JMP
• Data in two columns.
–Response variable:
• Numeric – Continuous
–Explanatory variable:
• Character – Nominal
22
JMP Starter
• Basic – Two-Sample t-Test
–Y, Response: Time
–X, Grouping: Sex
23
Oneway Analysis of Time By Sex
100
90
Time
80
70
60
50
40
30
F
M
Sex
Means and Std Deviations
Level
F
M
Number
15
15
Mean
55.8667
69.2000
Std Dev
13.5270
13.7903
Std Err
Mean Lower 95%
3.4927
48.376
3.5606
61.563
Upper 95%
63.358
76.837
t Test
M-F
Assuming unequal variances
Difference
13.3333 t Ratio
Std Err Dif
4.9877 DF
Upper CL Dif
23.5503 Prob > |t|
Lower CL Dif
3.1164 Prob > t
Confidence
0.95 Prob < t
2.67326
27.98961
0.0124 *
0.0062 *
0.9938 -15
-10
-5
0
5
10
15
24
8