Inference for 
1
 2
Who? Students at I.S.U.
 What? Time (minutes).
 When? Fall 2000.
 Where? Lied Recreation Athletic Center.
 How? Measure time from when student
arrives on 2nd floor until she/he leaves.
 Why? Part of a Stat 101 data collection
project.

1
Inference for 
1
 2
 Do
males and females at I.S.U.
spend the same amount of
time, on average, at the Lied
Recreation Athletic Center?
2
random
selection
Inference
1. Female
Populations
Samples
2. Male
random
selection
3
Time (minutes)
1. Females
2. Males
63, 32, 86, 53, 49
73, 39, 56, 45, 67
49, 51, 65, 54, 56
52, 75, 74, 68, 93
77, 41, 87, 72, 53
84, 65, 66, 69, 62
4
Time (minutes)
Sex=F
Mean
Sex=M
55.87 Mean
69.20
Std Dev
13.527 Std Dev
13.790
Std Err
Mean
N
3.4927 Std Err
3.5606
Mean
15 N
15
5
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.
6
Conditions & Assumptions
 Randomization
Condition
 10%
Condition
 Nearly Normal Condition
 Independent Groups
Assumption
–How were the data collected?
7
Conditions & Assumptions
 Randomization
Condition
–Random sample of males.
–Random sample of females.
 Independence
Assumption
–Two separate random samples.
 10%
Condition
8
.99
2
.95
.90
.75
Females
.50
1
0
.25
.10
.05
.01
Normal Quantile Plot
3
-1
-2
-3
5
3
2
Count
4
1
30
40
50
60
70
80
Time (min)
90 100
9
.99
2
.95
.90
.75
Males
.50
1
0
.25
.10
.05
.01
Normal Quantile Plot
3
-1
-2
-3
4
3
2
Count
5
1
30
40
50
60
70
80
Time (min)
90 100
10
Nearly Normal Condition
 The
female sample data could
have come from a population
with a normal model.
 The male sample data could
have come from a population
with a normal model.
11
Confidence Interval for 
1
y
1
 2
 y2   t SE y1  y2 
*
2
1
2
2
s s
SE y1  y2  

n1 n2
SE y1  y2  
SE y   SE y 
2
1
2
2
12
2
1
2
2
s s
SE y1  y2  

n1 n2
13.527   13.792 
2

15
2
15
 24.88  4.988
13
SE y1  y2  

SE y   SE y 
3.4927   3.5606 
2
1
2
2
2
2
 24.88  4.988
14
Finding
*
t
 Use
Table T.
 Confidence Level in last row.
 df = a really nasty formula (so
the value will be given to you).
–df = 28 for our example.
15
Table T
df
1
2
3
4

2.048
28
Confidence Levels 80%
90%
95%
98%
99%
16
Confidence Interval for 
1
 2
 y  y   t SE y  y 
55.87  69.2   2.0484.988 
*
1
2
1
2
 13.33  10.22
 23.55 to  3.11
17
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.
18
Inference for
1   2
 Do
males and females at I.S.U.
spend the same amount of
time, on average, at the Lied
Recreation Athletic Center?
 Could the difference between
the population mean times be
zero?
19
Test of Hypothesis for 
1
 2
 Step
1: Set up the null and
alternative hypotheses.
H 0 : 1  2 or H0 : 1  2  0
H A : 1  2 or HA : 1  2  0
20
Test of Hypothesis for 
1
 Step
 2
2: Check Conditions.
–Randomization Condition
• Two Independent Random
Samples
–10% Condition
–Nearly Normal Condition
21
.99
2
.95
.90
.75
Females
.50
1
0
.25
.10
.05
.01
Normal Quantile Plot
3
-1
-2
-3
5
3
2
Count
4
1
30
40
50
60
70
80
Time (min)
90 100
22
.99
2
.95
.90
.75
Males
.50
1
0
.25
.10
.05
.01
Normal Quantile Plot
3
-1
-2
-3
4
3
2
Count
5
1
30
40
50
60
70
80
Time (min)
90 100
23
Nearly Normal Condition
 The
female sample data could
have come from a population
with a normal model.
 The male sample data could
have come from a population
with a normal model.
24
Test of Hypothesis for 
1
 2
 Step
3: Compute the value of the
test statistic and find the P-value.

y
t
 y2   0
SE y1  y2 
1
2
1
2
2
s
s
SE y1  y2  

n1 n2
25
Time (minutes)
Sex=F
Mean
Sex=M
55.87 Mean
69.20
Std Dev
13.527 Std Dev
13.790
Std Err
Mean
N
3.4927 Std Err
3.5606
Mean
15 N
15
26
2
1
2
2
s s
SE y1  y2  

n1 n2
13.527   13.792 
2

15
2
15
 24.88  4.988
27
Test of Hypothesis for 
1
 2
 Step
3: Compute the value of the
test statistic and find the P-value.

y
t
 y2   0 55.87  69.20 

 2.672
SE y1  y2 
4.988
1
28
Table T
Two tail probability
0.20
0.10
0.05
0.02 P-value 0.01
df
1
2
3
4

28
1.313 1.701 2.048 2.467 2.672 2.763
29
Test of Hypothesis for 
1
 2
 Step
4: Use the P-value to
make a decision.
–Because the P-value is small (it is
between 0.01 and 0.02), we
should reject the null hypothesis.
30
Test of Hypothesis for 
1
 2
 Step
5: State a conclusion within
the context of the problem.
– 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.
31
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.
32
Alternatives
H 0 : 1  2
H A : 1  2 , One tail prob (Pr  t )
H A : 1  2 , One tail prob (Pr  t )
H A : 1  2 , Two tail prob (Pr  t )
33
JMP
 Data
in two columns.
–Response variable:
• Numeric – Continuous
–Explanatory variable:
• Character – Nominal
34
JMP Starter
 Basic
– Two-Sample t-Test
–Y, Response: Time
–X, Grouping: Sex
35
Oneway Analysis of Time By Se x
100
90
Time
80
70
60
50
40
30
F
M
Sex
M e ans and Std Deviations
Level
F
M
Number
15
15
Mean
55.8667
69.2000
Std Dev Std Err Mean Lower 95% Upper 95%
13.5270
3.4927
48.376
63.358
13.7903
3.5606
61.563
76.837
t Te st
F-M
Assuming unequal variances
Difference
-13.333 t Ratio
Std Err Dif
4.988 DF
Upper CL Dif
-3.116 Prob > |t|
Lower CL Dif
-23.550 Prob > t
Confidence
0.95 Prob < t
-2.67326
27.98961
0.0124
0.9938
-15 -10
0.0062
-5
0
5
10
15
36