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Chapter 23
Inferences About Means
1

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.

2
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,  .

3
Inference for 

Sampling distribution of y
Shape: Approximately normal
 Center: Mean,
 Spread: Standard Deviation,


SD y  

n
4
Problem
The population standard deviation,  is
unknown.

 Therefore, SD y  
is unknown as
n
well.

5
Solution

Use the sample standard deviation, s and
the standard error of y
s
SE y  
n
6
Problem

The distribution of the standardized
sample mean
y
SE y 
does not follow a normal model.
7
Solution

The distribution of the standardized
sample mean y  
SE y 
does follow a Student’s t-model with df =
n – 1.
8
Inference for


Do NOT use Table Z!
Table Z

Use Table T instead!
9
10
Conditions
Randomization condition.
 10% condition.
 Nearly normal condition.

11
Randomization Condition
Data arise from a random sample from
some population.
 Data arise from a randomized experiment.

12
10% Condition
The sample is no more than 10% of the
population.
 Not as critical for means as it is for
proportions.

13
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?

14
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
15
Table T
df
1
2
3
4
t
n–1
Confidence Levels 80%
90%
*
n 1
95%
98%
99%
16
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,  .

17
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
18
Summary of Data
 n = 25

y = 74.16 beats
 s = 5.375 beats
s
 SE y  
n
= 1.075 beats
19
Conditions
Randomization condition: met.
 10% condition: met.
 Nearly normal condition: met.

20
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
21
Nearly Normal Condition
Normal quantile plot – data follows
straight line for a normal model.
 Box plot – symmetric.
 Histogram – unimodal and symmetric.

22
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
23
Table T
df
1
2
3
4

2.064
24
Confidence Levels 80%
90%
95%
98%
99%
24
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
25
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
26
Interpretation
Plausible values for the population mean.
 95% of intervals produced using random
samples will contain the population mean.

27
JMP:Analyze – Distribution
Mean
74.16
Std Dev
5.375
Std Err Mean
1.075
Upper 95% Mean
76.38
Lower 95% Mean
71.94
N
25
28
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.

29
Test of Hypothesis for


Could the population mean heart rate of
young adults be 70 beats per minute or is
it something higher?
30
Test of Hypothesis for


Step 1: State your null and alternative
hypotheses.
H 0 :   70
H A :   70
31
Test of Hypothesis for


Step 2: Check conditions.
Randomization condition, met.
 10% condition, met.
 Nearly normal condition, met.

32
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
33
Summary of Data
 n = 25

y = 74.16 beats
 s = 5.375 beats

s
SE y  
n
= 1.075 beats
34
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.
35
Table T
One tail probability 0.10
df
1
2
3
4
0.05
0.025
0.01
0.005 P-value

24
2.064
2.492
2.797 3.87
The P-value is less than 0.005.
36
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.

37
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.

38
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 )
39
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
40
Another Example

Is the population mean octane rating 90
or is it something different?
41
Test of Hypothesis for


Step 1: State your null and alternative
hypotheses.
H 0 :   90
H A :   90
42
Test of Hypothesis for


Step 2: Check conditions.
Randomization condition, met.
 10% condition, met.
 Nearly normal condition, met.

43
Normal Quantile Plot
3
.99
2
.95
.90
.75
.50
1
0
.25
.10
.05
.01
-1
-2
-3
6
4
Count
8
2
87 88 89 90 91 92 93 94 95
Octane Rating
44
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
45
Summary of Data
 n = 40

y = 90.9475
 s = 1.530
s
 SE y  
n
= 0.2419
46
Value of Test Statistic
y  0 90.9475  90
t

SE y 
0.2419
t  3.92
Use Table T to find the P-value.
47
Table T
Two tail probability 0.20
0.10
0.05
0.02
0.01 P-value
df
1
2
3
4
39
40
3.92
2.021
2.423
2.704
The P-value is less than 0.01.
48
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.

49
Test of Hypothesis for

Step 5: State your conclusion within the
context of the problem.
 The population mean octane rating is not
90 but something different.

50
Test of Hypothesis for

y  t SE y  to y  t SE y 
*
n 1
*
n 1
90.9475  2.0210.2419 
90.95  0.49 to 90.95  0.49
90.46 to 91.44
51
Interpretation

We are 95% confident that the
population mean octane rating is
between
90.46 and 91.44
52
Interpretation
This confidence interval agrees with the
test of hypothesis.
 90 is not in the interval and so must be
rejected as a value for the population
mean.

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