Download 11: Inference About a Mean

Chapter 11:
Inference About a Mean
May 17
In Chapter 11:
11.1 Estimated Standard Error of the Mean
11.2 Student’s t Distribution
11.3 One-Sample t Test
11.4 Confidence Interval for μ
11.5 Paired Samples
11.6 Conditions for Inference
11.7 Sample Size and Power
§11.1 Estimated Standard Error of
the Mean
• We rarely know population standard deviation
σ  instead, we calculate sample standard
deviations s and use this as an estimate of σ
• We then use s to calculate this estimated
standard error of the mean:
SE x 
s
n
• Using s instead of σ adds a source of
uncertainty  z procedures no longer apply
 use t procedures instead
§11.2 Student’s t distributions
• A family of distributions identified by “Student”
(William Sealy Gosset) in 1908
• t family members are identified by their degrees
of freedom, df.
• t distributions are similar to z distributions but
with broader tails
• As df increases → t tails get skinnier → t
become more like z
t table (Table C)
Use Table C to look up t values and probabilities
Table C:
Entries  t values
Rows  df
Columns  probabilities
Understanding Table C
Let tdf,p ≡ a t value with df degrees of freedom and
cumulative probability p. For example, t9,.90 = 1.383
Table C. Traditional t table
Cumulative p
0.75
0.80
0.85
0.90
0.95
0.975
Upper-tail p
0.25
0.20
0.15
0.10
0.05
0.025
df = 9
0.703
0.883
1.100
1.383
1.833
2.262
Left tail:
Pr(T9 < -1.383) = 0.10
Right tail:
Pr(T9 > 1.383) = 0.10
§11.3 One-Sample t Test
A. Hypotheses. H0: µ = µ0 vs. Ha: µ ≠ µ0 (two-sided)
[ Ha: µ < µ0 (left-sided) or Ha: µ > µ0 (right-sided)]
B. Test statistic.
tstat 
x  0
s
with df  n  1
n
C. P-value. Convert tstat to P-value [table C or
software]. Small P  strong evidence against H0
D. Significance level (optional). See Ch 9 for
guidelines.
One-Sample t Test: Example
Statement of the problem:
• Do SIDS babies have lower than average birth
weights?
• We know from prior research that the mean birth
weight of the non-SIDs babies in this population
is 3300 grams
• We study n = 10 SIDS babies, determine their
birth weights, and calculate x-bar = 2890.5 and s
= 720.
• Do these data provide significant evidence that
SIDs babies have different birth weights than the
rest of the population?
One-Sample t Test: Example
A. H0: µ = 3300 versus Ha: µ ≠ 3300 (two-sided)
B. Test statistic
tstat
x   0 2890 .5  3300


 1.80
SE x
720 10
df  n  1  10  1  9
C. P = 0.1054 [next slide]
Weak evidence against H0
D. (optional) Data are not significant at α = .10
Converting the tstat to a P-value
tstat  P-value via Table C. Wedge |tstat| between
critical value landmarks on Table C. One-tailed
0.05 < P < 0.10 and two-tailed 0.10 < P < 0.20.
|tstat| = 1.80
Table C. Traditional t table
Cumulative p
0.75
0.80
0.85
0.90
0.95
0.975
Upper-tail p
0.25
0.20
0.15
0.10
0.05
0.025
df = 9
0.703
0.883
1.100
1.383
1.833
2.262
tstat  P-value via software. Use a software utility
to determine that a t of −1.80 with 9 df has twotails of 0.1054.
Two-sided P-value associated with a t
statistic of -1.80 and 9 df
§11.4 Confidence Interval for µ
(1   )100 % CI for   x  t n 1,1  
2
•
•
•
•
•
s
n
Typical point “estimate ± margin of error” formula
tn-1,1-α/2 is from t table (see bottom row for conf. level)
Similar to z procedure except uses s instead of σ
Similar to z procedure except uses t instead of z
Alternative formula:
x  t n 1,1   SEx where SEx 
2
s
n
Confidence Interval: Example 1
Let us calculate a 95% confidence interval for μ for
the birth weight of SIDS babies.
x  2890 .5
s  720 .0
n  10
s
95% CI for   x  t101,1 .0 5 
2
n
720
 2890 .5  2.262 
10
 2890 .5 ± 515.1
= (2375.4 to 3405.6) grams
Confidence Interval: Example 2
Data are “% of ideal body weight” in 18
diabetics: {107, 119, 99, 114, 120, 104, 88, 114,
124, 116, 101, 121, 152, 100, 125, 114, 95, 117}.
Based on these data we calculate a 95% CI for μ.
x  112 .778
s  14 .424
n  18
s
14 .242
SE x 

 3.400
n
18
t n 1,1   t181,1 .0 5  t17,.975  2.110 (from t table)
2
2
x  (t n 1,1  )( SE x )  112 .778  (2.110 )(3.44 )
2
 112.778 ± 7.17 = (105.6, 120.0)
§11.5 Paired Samples
• Paired samples: Each point in one
sample is matched to a unique point in the
other sample
• Pairs be achieved via sequential samples
within individuals (e.g., pre-test/post-test),
cross-over trials, and match procedures
• Also called “matched-pairs” and
“dependent samples”
Example: Paired Samples
• A study addresses whether oat bran reduce LDL
cholesterol with a cross-over design.
• Subjects “cross-over” from a cornflake diet to an
oat bran diet.
–
–
–
–
–
–
–
Half subjects start on CORNFLK, half on OATBRAN
Two weeks on diet 1
Measures LDL cholesterol
Washout period
Switch diet
Two weeks on diet 2
Measures LDL cholesterol
Example, Data
Subject CORNFLK OATBRAN
---------- ------1
4.61
3.84
2
6.42
5.57
3
5.40
5.85
4
4.54
4.80
5
3.98
3.68
6
3.82
2.96
7
5.01
4.41
8
4.34
3.72
9
3.80
3.49
10
4.56
3.84
11
5.35
5.26
12
3.89
3.73
Calculate Difference Variable “DELTA”
• Step 1 is to create difference variable “DELTA”
• Let DELTA = CORNFLK - OATBRAN
• Order of subtraction does not materially effect
results (but but does change sign of differences)
• Here are the first three observations:
ID
CORNFLK OATBRAN
---- ------- ------1
4.61
3.84
2
6.42
5.57
3
5.40
5.85
↓
↓
↓
DELTA
----0.77
0.85
-0.45
↓
Positive
values
represent
lower LDL
on oatbran
Explore DELTA Values
Here are all the twelve paired differences (DELTAs):
0.77, 0.85, −0.45, −0.26, 0.30, 0.86, 0.60, 0.62, 0.31, 0.72, 0.09, 0.16
Stemplot
|-0|42
|+0|0133
|+0|667788
×1
EDA shows a slight
negative skew, a median
of about 0.45, with results
varying from −0.4 to 0.8.
Descriptive stats for DELTA
Data (DELTAs): 0.77, 0.85, −0.45, −0.26, 0.30,
0.86, 0.60, 0.62, 0.31, 0.72, 0.09, 0.16
The subscript d will be used to denote statistics
for difference variable DELTA
n  12
xd  0.3808
sd  0.4335
95% Confidence Interval for µd
A t procedure directed toward the DELTA variable
calculates the confidence interval for the mean
difference.
sd
(1   )100 % CI for  d  xd  t n 1,1  
2
n
“Oat bran” data:
For 95% confidence use t121,1 .05  t11,.975  2.201 (from Table C)
2
95 % CI for  d  0.3808  2.201 
 0.3808  0.2754
 (0.105 to 0.656 )
.4335
12
Paired t Test
•
•
•
Similar to one-sample t test
μ0 is usually set to 0, representing “no
mean difference”, i.e., H0: μ = 0
Test statistic:
tstat 
xd   0
sd
df  n  1
n
Paired t Test: Example
“Oat bran” data
A. Hypotheses. H0: µd = 0 vs. Ha: µd  0
B. Test statistic.
xd   0 0.38083  0
tstat 

 3.043
s n
.4335 / 12
df  n  1  12  1  11
C. P-value. P = 0.011 (via computer). The
evidence against H0 is statistically significant.
D. Significance level (optional). The evidence
against H0 is significant at α = .05 but is not
significant at α = .01
SPSS Output: Oat Bran data
§11.6 Conditions for Inference
t procedures require these conditions:
• SRS (individual observations or DELTAs)
• Valid information (no information bias)
• Normal population or large sample (central
limit theorem)
The Normality Condition
The Normality condition applies to the sampling
distribution of the mean, not the population.
Therefore, it is OK to use t procedures when:
• The population is Normal
• Population is not Normal but is symmetrical and
n is at least 5 to 10
• The population is skewed and the n is at least 30
to 100 (depending on the extent of the skew)
Can a t procedures
be used?
• Dataset A is skewed
and small: avoid t
procedures
• Dataset B has a mild
skew and is moderate
in size: use t
procedures
• Data set C is highly
skewed and is small:
avoid t procedure
§11.7 Sample Size and Power
• Questions:
– How big a sample is needed to limit the
margin of error to m?
– How big a sample is needed to test H0 with
1−β power at significance level α?
– What is the power of a test given certain
conditions?
• In this presentation, we cover only the last
question
Power


|

|
n

1      z1  


2



where:
• α ≡ (two-sided) alpha level of the test
• Δ ≡ “the mean difference worth detecting” (i.e.,
the mean under the alternative hypothesis minus
the mean under the null hypothesis)
• n ≡ sample size
• σ ≡ standard deviation in the population
• Φ(z) ≡ the cumulative probability of z on a
Standard Normal distribution [Table B]
Power: Illustrative Example
SIDS birth weight example. Consider the SIDS
illustration in which n = 10 and σ is assumed to be
720 gms. Let α = 0.05 (two-sided). What is the
power of a test under these conditions to detect a
mean difference of 300 gms?


|

|
n

1      z1  


2





|
300
|
10

   1.96 


720


  0.64   0.2611
The power is about 26%