1. Chapter 11: Inference About a Mean
12/2/2018

2. In Chapter 11: 11.1 Estimated Standard Error of the Mean
12/2/2018
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
12/2/2018
Basic Biostat

3. σ not known
Prior chapter: σ was known before collecting data  z procedures used to help infer µ
When σ NOT known, calculate sample standard deviations s and use it to calculate this standard error:
12/2/2018

4. Additional Uncertainty
The Normal distribution doesn’t fit well
Using s instead of σ adds uncertainty to inferences  can NOT use z procedures
Instead, rely on Student’s t procedures
William Sealy Gosset (1876–1937)
12/2/2018

5. Student’s t distributions
Family of probability distributions
Family members identified by degrees of freedom (df)
Similar to “Z”, but with broader tails
As df increases → tails get skinnier → t become like z
A t distribution with infinite degrees of freedom is a Standard Normal Z distribution
12/2/2018

6. Table C (t table) Rows  df Columns  probabilities Entries  t values
Notation: tcum_prob,df = t value
Example: t.975, 9 = 2.262
12/2/2018

7. One-Sample t Test Objective: test a claim about population mean µ
Conditions :
Simple Random Sample
Normal population or “large sample”
12/2/2018

8. Hypothesis Statements
Null hypothesis H0: µ = µ0
where µ0 represents the pop. mean expected by the null hypothesis
Alternative hypotheses
Ha: µ < µ0 (one-sided, left)
Ha: µ > µ0 (one-sided, right)
Ha: µ ≠ µ0 (two-sided)
12/2/2018

9. Example
Do SIDS babies have lower average birth weights than a general population mean µ of 3300 gms?
H0: µ = 3300
Ha: µ < 3300 (one-sided) or Ha: µ ≠ 3300 (two-sided)
12/2/2018

10. One-Sample t Test Statistic
where
This t statistic has n – 1 degrees of freedom
12/2/2018

11. Example (Data)
2998
3740
2031
2804
2454
2780
2203
3803
3948
2144
SRS n = 10 birth weights (grams) of SIDS cases
12/2/2018

12. Example
Testing H0: µ = 3300
12/2/2018

13. P-value via Table C Bracket |tstat| between t critical values Table C.
For |tstat| = 1.80 with 9 df
Table C.
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| = 1.80
Thus  One-tailed: 0.05 < P < 0.10
Two-tailed: 0.10 < P < 0.20
12/2/2018

14. For a more precise P-value use a computer utility
Here’s output from the free utility StaTable
Graphically:
12/2/2018

15. Interpretation Testing H0: µ = 3300 gms Two-tailed P > .10
Conclude: weak evidence against H0
The sample mean (2890.5) is NOT significantly different from 3300
12/2/2018

16. (1− α)100% CI for µ
where
12/2/2018

17. Same Data
Interpretation: Population mean µ is between 2375 and 3406 grams with 95% confidence
12/2/2018

18. §11.5 Paired Samples Two samples
Each data point in one sample uniquely matched to a data point in the other sample
Examples of paired samples
“Pre-test/post-test”
Cross-over trials
Pair-matching
12/2/2018

19. Example Does oat bran reduce LDL cholesterol?
Start half of subjects on CORNFLK diet
Start other half on OATBRAN
Two weeks  LDL cholesterol
Washout period
Cross-over to other diet
12/2/2018

20. Oat bran data LDL cholesterol mmol
Subject CORNFLK OATBRAN
12/2/2018

21. Within-pair difference “DELTA”
Let DELTA = CORNFLK - OATBRAN
First three observations in OATBRAN data:
ID CORNFLK OATBRAN DELTA
etc.
All procedures are now directed toward difference variable DELTA
12/2/2018

22. Exploratory and descriptive stats
DELTA: 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
|-0f|4 |-0*|2
|+0*|01 |+0t|33
|+0f|
|+0s|6677
|+0.|88 ×1 LDL (mmol)
subscript d denotes “difference”
12/2/2018

23. 95% CI for µd
 95% confident population mean difference µd is between and mmol/L
12/2/2018

24. Hypothesis Test
Claim: oat bran diet is associated with a decline (one-sided) or change (two-sided) in LDL cholesterol.
Test H0: µd = µ0 where µ0 = 0 Ha: µd > µ0 (one-sided) Ha: µ ≠ µ0 (two-sided)
12/2/2018

25. Paired t statistic
12/2/2018

26. P-value via Table C Table C. Thus  One-tailed: .005 < P < .01
|tstat| =
Table C.
Upper-tail P
.01
.005
.0025
df = 11
2.718
3.106
3.497
Thus  One-tailed: .005 < P < .01
Two-tailed: .01 < P < .02
12/2/2018

27. P-value via Computer
12/2/2018

28. SPSS Output: “Oat Bran”
12/2/2018

29. My P value is smaller than yours!
Interpretation
My P value is smaller than yours!
Testing H0: µ = 0
Two-tailed P = 0.011
 Good reason to doubt H0
(Optional) The difference is “significant” at α = .05 but not at α = .01
12/2/2018

30. The Normality Condition
t Procedures require Normal population or large samples
How do we assess this condition?
Guidelines. Use t procedures when:
Population Normal
population symmetrical and n ≥ 10
population skewed and n ≥ ~45 (depends on severity of skew)
12/2/2018

31. Can a t procedures be used?
Skewed small sample  avoid t procedures
12/2/2018

32. Can a t procedures be used?
Mild skew in moderate sample  t OK
12/2/2018

33. Can a t procedures be used?
Skewed moderate sample  avoid t
12/2/2018

34. Sample Size and Power Methods:
(1) n required to achieve m when estimating µ
(2) n required to test H0 with 1−β power
(3) Power of a given test of H0
12/2/2018

35. Power α ≡ alpha (two-sided) Δ ≡ “difference worth detecting” = µa – µ0
n ≡ sample size
σ ≡ standard deviation
Φ(z) ≡ cumulative probability of Standard Normal z score
with .
12/2/2018

36. Power: SIDS Example Let α = .05 and z1-.05/2 = 1.96
Test: H0: μ = 3300 vs. Ha: μ = Thus: Δ ≡ µ1 – µ0 = 3300 – 3000 = 300
n = 10 and σ ≡ 720 (see prior SIDS example)
Use Table B to look up cum prob  Φ(-0.64) = .2611
12/2/2018

37. Power: Illustrative Example
12/2/2018