Inference for 
 Who?
Young adults.
 What? Heart rate (beats per minute).
 When?
 Where? In a physiology lab.
 How? Take pulse at wrist for one
minute.
 Why? Part of an evaluation of
general health.
1
Inference for 
 What
is the mean heart rate for
all young adults?
 Use the sample mean heart
rate, y , to make inferences
about the population mean
heart rate,  .
2
Inference for 
 Sampling
distribution of y
–Shape: Approximately normal
–Center: Mean, 
–Spread: Standard Deviation,
SD y  

n
3
Problem
population standard 
deviation, is unknown.

 Therefore, SD y  
is
n
unknown as well.
 The
4
Solution
 Use
the sample standard
deviation, s to get the standard
error of y
s
SE y  
n
5
Problem
 The
distribution of the
standardized sample mean
y
SE y 
does not follow a normal model.
6
Solution
 The
distribution of the
standardized sample mean
y
SE y 
does follow a Student’s t-model
with df = n – 1.
7
8
9
Inference for 
 Do
NOT use Table Z!
Table Z
 Use
Table T instead!
10
11
Conditions
 Randomization
condition.
 10%
condition.
 Nearly normal condition.
12
Randomization Condition
 Data
arise from a random
sample from some population.
 Data arise from a randomized
experiment.
13
10% Condition
 The
sample is no more than
10% of the population.
 Not as critical for means as it is
for proportions.
14
Nearly Normal Condition
 The
data come from a
population whose shape is
unimodal and symmetric.
–Look at the distribution of the
sample.
–Could the sample have come
from a normal model?
15
Confidence Interval for 
y  t SE y  to y  t SE y 
*
n 1
t
*
n 1
*
n 1
is from Table T
s
SE y  
n
16
Table T
df
1
2
3
4

t
n–1
Confidence Levels 80%
90%
*
n 1
95%
98%
99%
17
Inference for 
 What
is the mean heart rate for
all young adults?
 Use the sample mean heart
rate, y , to make inferences
about the population mean
heart rate,  .
18
Sample Data
 Random
sample of n = 25 young
adults.
 Heart rate – beats per minute
70, 74, 75, 78, 74, 64, 70, 78, 81, 73
82, 75, 71, 79, 73, 79, 85, 79, 71, 65
70, 69, 76, 77, 66
19
Summary of Data
n = 25
 y=
74.16 beats
s = 5.375 beats

s
SE y  
= 1.075 beats
n
20
Conditions
 Randomization
condition:
–Met because we have a random
sample of 25 young adults.
 10%
condition:
–Met because 25 is less than 10%
of all young adults that could
have been sampled.
21
Conditions
 Nearly
Normal Condition
–Could the sample have come
from a population described by a
normal model?
22
Normal Quantile Plot
3
.99
2
.95
.90
.75
.50
1
0
.25
.10
.05
.01
-1
-2
-3
6
4
Count
8
2
60
65
70
75
Heart rate
80
85
90
23
Nearly Normal Condition
quantile plot – data
follows straight line for a normal
model.
 Box plot – symmetric.
 Histogram – unimodal and
symmetric.
 Normal
24
Confidence Interval for 
y  t SE y  to y  t SE y 
*
n 1
t
*
n 1
*
n 1
is from Table T
s
SE y  
n
25
Table T
df
1
2
3
4

2.064
24
Confidence Levels 80%
90%
95%
98%
99%
26
Confidence Interval for 
y  t SE y  to y  t SE y 
*
n 1
74.16  2.064 1.075 
*
n 1
74.16  2.22 to 74.16  2.22
71.94 beats to 76.38 beats
27
Interpretation
 We
are 95% confident that the
population mean heart rate of
young adults is between
71.94 beats/min and 76.38 beats/min
28
Interpretation
 Plausible
values for the
population mean.
 95% of intervals produced using
random samples will contain the
population mean.
29
JMP:Analyze – Distribution
Mean
Std Dev
Std Err Mean
Upper 95% Mean
Lower 95% Mean
N
74.16
5.375
1.075
76.38
71.94
25
30
Test of Hypothesis for 
 Could
the population mean
heart rate of young adults be 70
beats per minute or is it
something higher?
31
Test of Hypothesis for 
 Step
1: State your null and
alternative hypotheses.
H 0 :   70
H A :   70
32
Test of Hypothesis for 
 Step
2: Check conditions.
–Randomization condition, met.
–10% condition, met.
–Nearly normal condition, met.
33
Test of Hypothesis for 
 Step
3: Calculate the test statistic
and convert to a P-value.
y  0
t
SE y 
s
SE y  
n
34
Summary of Data
n = 25
 y=
74.16 beats
s = 5.375 beats

s
SE y  
= 1.075 beats
n
35
Value of Test Statistic
y  0 74.16  70
t

SE y 
1.075
t  3.87
Use Table T to find the P-value.
36
Table T
One tail probability 0.10
df
1
2
3
4

24
0.05
0.025
2.064
0.01
2.492
0.005 P-value
2.797 3.87
The P-value is less than 0.005.
37
Test of Hypothesis for 
 Step
4: Use the P-value to
reach a decision.
 The P-value is very small,
therefore we should reject the
null hypothesis.
38
Test of Hypothesis for 
 Step
5: State your conclusion
within the context of the
problem.
 The mean heart rate of all
young adults is more than 70
beats per minute.
39
Alternatives
H 0 :   0
H A :   0 , One tail prob (Pr  t )
H A :   0 , One tail prob (Pr  t )
H A :   0 , Two tail prob (Pr  t )
40
JMP:Analyze – Distribution
 Test
Mean
t-test
Hypothesized
value
Actual Estimate
df
Std Dev
70 Test
statistic
74.16 Prob > |t|
3.87
0.0007
24 Prob > t
0.0004
5.375 Prob < t
0.9996
41
Example
 What
is the mean alcohol
content of beer?
 A random sample of 10 beers is
taken and the alcohol content
(%) is measured.
42
Sample Data – Alcohol (%)
Molson
Canadian
5.19
Heineken
Dark
5.17
Michelob
Dark
4.76
O’Keefe
Canadian
4.96
Big Barrel
Lager
4.32
Olympia
Lager
4.78
Hamm’s
4.53
Miller Draft
4.85
Tsingtao
4.79
Guinness
Stout
4.27
43
Test of Hypothesis for 
 Step
1: State your null and
alternative hypotheses.
H0 :   5
HA :   5
44
Test of Hypothesis for 
 Step
2: Check conditions.
–Randomization condition, met.
–10% condition, met.
–Nearly normal condition, met.
45
Test of Hypothesis for 
 Step
3: Calculate the test
statistic and convert to a Pvalue.
y  0
t
SE y 
s
SE y  
n
46
Test of Hypothesis for 
 Test
statistic,
y  0 4.762  5  0.238
t


 2.397
 s   0.314  0.0993

 

 n   10 
47
Table T
Two tail probability 0.20
0.10
0.05 P-value 0.02
df
1
2
3
4

9
2.262 2.397 2.821
The P-value is between 0.02 and 0.05.
48
Test of Hypothesis for 
 Step
4: Use the P-value to
reach a decision.
 The P-value is smaller than
0.05, therefore we should reject
the null hypothesis.
49
Test of Hypothesis for 
 Step
5: State your conclusion
within the context of the
problem.
 The population mean alcohol
content of beer is not 5%.
50
JMP Output
Moments
Mean
Std Dev
Std Err Mean
upper 95% Mean
lower 95% Mean
N
4.762
0.3142823
0.0993848
4.986824
4.537176
10
4.7 4.8 4.9 5.0 5.1 5.2 5.3
Test Mean=value
Hypothesized Value
Actual Estimate
df
Std Dev
5
4.762
9
0.31428
Test Statistic
Prob > |t|
Prob > t
Prob < t
t Test
-2.3947
0.0402
0.9799
0.0201
51
Confidence Interval for
 s 
 s 
y  t* 
 to y  t * 

 n
 n
 0.314 
 0.314 
4.762  2.262
 to 4.762  2.262

 10 
 10 
4.762  0.225 to 4.762  0.225
4.537 to 4.987
52
Interpretation
 We
are 95% confident that the
population mean alcohol
content of beer is between
4.537% and 4.987%.
53
Interpretation
 The
population mean alcohol
content of beer could be any value
between 4.537% and 4.987%.
 If we repeat the procedure that
produces a confidence interval,
95% of intervals produced will
capture the population mean.
54