1. This Week Review of estimation and hypothesis testing
Reading Le (review)
Chapter 4: Sections 4.1 – 4.3
Chapter 5: Sections 5:1 and 5:4
Chapter 7: Sections 7:1 – 7.3
Reading C &S
Chapter 2:A-E
Chapter 6: A,B,F

2. Point Estimate Population Parameter Point Estimate m Sample mean p
Sample proportion r
Sample correlation
m1 - m2
Difference between 2 sample means
p1 - p2
Difference between 2 sample proportions s
Sample standard deviation
Sampling error: True value – estimate (unknown)

3. Statistical Inference
Population
with mean
m = ?
A simple random sample
of n elements is selected
from the population.
The value of is used to
make inferences about
the value of m.
The sample data
provide a value for
the sample mean .

4. Interval Estimation In general, confidence intervals are of the form:
Estimate = mean, proportion, regression coefficient, odds ratio...
SE = standard error of your estimate
1.96 = for 95% CI based on normal distribution

5. Standard normal distribution
2.5% probability
2.5% probability

6. Estimation for Population Mean m
Point estimate:
Estimate of variability
in population
A slightly larger number based on the t-distribution is used for smaller n
Estimate of variability
in point estimate
(SE)
95% Confidence Interval

7. Assumptions Data in population follows a normal distribution or
Sample size is large enough to apply central limit theorem (CLT)
CLT – no matter the shape of the population distribution of the sample mean approaches a normal distribution as the sample size gets large

8. Meaning of Confidence Interval
There is a 95% chance that your interval contains m. (That you “captured” the true value m with your interval)

9. Example
Suppose sample of n=100 persons mean = 215 mg/dL, standard deviation = 20 95% CI =
Lower Limit: 215 – 1.96*20/10
Upper Limit: *20/10
= (211, 219)
“We are about 95% confident that the interval contains m”
We can pretty much rule out that m > 220

10. Properties of Confidence Intervals
As sample size increases, CI gets smaller
Because SE gets smaller;
Can use different levels of confidence
90, 95, 99% common
More confidence means larger interval; so a 90% CI is smaller than a 99% CI
What would a 100% CI look like?
Changes with population standard deviation
More variable population means larger interval

11. Effect of sample size Suppose we had only 10 observations
What happens to the confidence interval?
For n = 100,
For n = 10,
Larger sample size = smaller interval

12. Effect of confidence level
Suppose we use a 90% interval
What happens to the confidence interval?
90%:
Lower confidence level = smaller interval
(A 99% interval would use 2.58 as multiplier and
the interval would be larger)

13. Effect of standard deviation
Suppose we had a SD of 40 (instead of 20)
What happens to the confidence interval?
More variation = larger interval

14. Effect of different sample
Suppose new sample with mean of 212 (but same standard deviation)
What happens to the confidence interval?
Same size, moves a little

15. How Big A Sample To Take? Depends on the variability in the population
Depends on how precise an estimate you want
Cost - if it doesn’t cost much to sample an element then sample many

16. 95% Confidence Intervals for m Using SAS
PROC MEANS DATA = datasetname CLM ;
VAR list of variables
This will display the following statistics N
Mean
Standard Deviation
Standard Error of Mean
Lower 95% Confidence Limit
Upper 95% Confidence Limit
Confidence Limits

17. Assessing Normality with Graphs
Boxplots and stem-and-leaf plots, histograms
Look for skewness (non-symmetry)
Hard to get normal looking graphs with small sample sizes
Can check effect of transformations
Normal probability plots
x-axis: related to inverse of standard normal distribution
y-axis: actual data
* actual data
+ what we would expect if data were really normal

18. Assessing normality - PROC UNIVARIATE
PROC UNIVARIATE DATA = demo NORMAL PLOT;
VAR ursod;
* Ursod is urinary sodium excretion in 8-hours
RUN;
NORMAL and PLOT are two options that test for normality and display simple graphs
Plots are best - with enough data, tests for normality almost always reject normality assumption

19. STEM AND LEAF PLOT
Stem Leaf # Boxplot | | | | |
| + |
*-----* | |
Multiply Stem.Leaf by 10**+1

20. The UNIVARIATE Procedure
Variable: ursod
Normal Probability Plot *
| *
| * *
| *** +++
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* +++
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***
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******
| *****
| ********
15+* * ** ** +++

21. Log transformed value shows a better linear pattern
Variable: lursod
Normal Probability Plot *
| *++
| **++
| **++
| ** +
* ++
| *++
| *+
| ***
| ** **
| *
| **
| ***
| *** **
| **
| *
| ****
| **
**+
| *+ |
| **+**
| * +
2.65+* ++
Log transformed value shows a better linear pattern

22. Hypothesis Testing
Hypothesis: A statement about parameters of population or of a model (m=200 ?)
Test: Does the data agree with the hypothesis? (sample mean 220)
Measure the agreement with probability

23. Steps in hypothesis testing
State null and alternative hypothesis
(Ho and Ha)
Ho usually a statement of no effect or no difference between groups
Choose α level
Probability of falsely rejecting Ho (Type I error)

24. Steps in hypothesis testing
Calculate test statistic, find p-value (p)
Measures how far data are from what you expect under null hypothesis
State conclusion:
p < α, reject Ho
p > α, insufficient evidence to reject Ho

25. Possible results of tests
What we decide
Reality

26. Details α related to confidence level Commonly set at 0.05 or 0.01
β usually predetermined by sample size

27. One sample t-test; test for population mean
Simple random sample from a normal population (or n large enough for CLT)
Ho: μ = μo
Ha : μ  μo , pick α
test statistic:

28. Matched pairs data d = X2 - X1 Ho: d = 0, Ha: d  0
Recall independence requirement for CIs
Similar issue for t-tests
Observations not independent
Examples; pre and post test, left and right eyes, brother-sister pairs
Solution: look at paired differences, do one sample test on differences
d = X2 - X1
Ho: d = 0, Ha: d  0

29. PROC TTEST, one sample test
PROC TTEST DATA = DEMO;
VAR age;
RUN;
Tests if mean age is different than zero. Not very useful
Need to be tricky...

30. Use a Data step to calculate a new variable
Subtract value of mean under null hypothesis
Test new variable for difference from zero
DATA DEMO;
SET DEMO;
dage = age - 25;
RUN;
PROC TTEST DATA=DEMO ;
VAR dage;
This tests whether the mean age is different from 25

31. PROC TTEST one sample output
T-Tests
Variable DF t Value Pr > |t|
dage
Conclusion: We have insufficient evidence to claim that
the mean age is different than 25 (p=0.69)