1. One sample statistical tests, continued…

2. Recall: Single population mean (large n)
Hypothesis test:
Confidence Interval

3. Single population mean (small n, normally distributed trait)
Hypothesis test:
Confidence Interval

4. What is a T-distribution?
A t-distribution is like a Z distribution, except has slightly fatter tails to reflect the uncertainty added by estimating .
The bigger the sample size (i.e., the bigger the sample size used to estimate ), then the closer t becomes to Z.
If n>100, t approaches Z.

5. T-distribution with only 1 degree of freedom.

6. T-distribution with 4 degrees of freedom.

7. T-distribution with 9 degrees of freedom.

8. T-distribution with 29 degrees of freedom.

9. T-distribution with 99 degrees of freedom. Looks a lot like Z!!

10. Student’s t Distribution
Note: t Z as n increases
Standard Normal
(t with df = )
t (df = 13)
t-distributions are bell-shaped and symmetric, but have ‘fatter’ tails than the normal
t (df = 5) t
from “Statistics for Managers” Using Microsoft® Excel 4th Edition, Prentice-Hall 2004

11. Student’s t Table .05 2 t /2 = .05 2.920 Upper Tail Area df .25 .10 1
Let: n = df = n - 1 =  = /2 =.05 df
.25
.10
.05 1
1.000
3.078
6.314 2
0.817
1.886
2.920
/2 = .05 3
0.765
1.638
2.353
The body of the table contains t values, not probabilities t
2.920
from “Statistics for Managers” Using Microsoft® Excel 4th Edition, Prentice-Hall 2004

12. With comparison to the Z value
t distribution values
With comparison to the Z value
Confidence t t t Z
Level (10 d.f.) (20 d.f.) (30 d.f.) ____
Note: t Z as n increases
from “Statistics for Managers” Using Microsoft® Excel 4th Edition, Prentice-Hall 2004

13. The T probability density function
What does t look like mathematically? (You may at least recognize some resemblance to the normal distribution function…)
Where:      v is the degrees of freedom     (gamma) is the Gamma function     is the constant Pi ( )

14. The t-distribution in SAS
Yikes! The t-distribution looks like a mess! Don’t want to integrate!
Luckily, there are charts and SAS! MUST SPECIFY DEGREES OF FREEDOM!
The t-function in SAS is:
probt(t-statistic, df)

15. The normality assumption…
Ttests (and all linear models, in fact) have a “normality assumption”:
If the outcome variable is not normally distributed and the sample size is small, a ttest is inappropriate
it takes longer for the CLT to kick in and the sample means do not immediately follow a t-distribution…
This is the source of the “normality assumption” of the ttest…

16. Computer simulation of the distribution of the sample mean (non-normal, small n):
1. Pick any probability distribution and specify a mean and standard deviation.
2. Tell the computer to randomly generate 1000 observations from that probability distributions
E.g., the computer is more likely to spit out values with high probabilities
3. Calculate 1000 T-statistics:
4. Plot the T-statistics in histograms.
5. Repeat for different sample sizes (n’s).
IF the true mean was 128 with an average variability of 15 lbs….

17. n=2, underlying distribution is exponential (mean=1, SD=1)
This is NOT a t-distribution!

18. n=5, underlying distribution is exponential (mean=1, SD=1)
This is NOT a t-distribution!

19. n=10, underlying distribution is exponential (mean=1, SD=1)
This doesn’t yet follow a t-distribution!

20. n=30, underlying distribution is exponential (mean=1, SD=1)
Still not quite a t-distribution! Note the left skew.

21. n=100, underlying distribution is exponential (mean=1, SD=1)
Now, pretty close to a T-distribution!

22. Conclusions
If the underlying data are not normally distributed AND n is small**, the means do not follow a t-distribution (so using a ttest will result in erroneous inferences).
Data transformation or non-parametric tests should be used instead.
**How small is too small? No hard and fast rule—depends on the true shape of the underlying distribution. Here N>30 (closer to 100) is needed.

23. Practice Problem:
A manufacturer of light bulbs claims that its light bulbs have a mean life of 1520 hours with an unknown standard deviation. A random sample of 40 such bulbs is selected for testing. If the sample produces a mean value of 1505 hours and a sample standard deviation of 86, is there sufficient evidence to claim that the mean life is significantly less than the manufacturer claimed?
Assume that light bulb lifetimes are roughly normally distributed.

24. Answer 1. What is your null hypothesis?
Null hypothesis: mean life = 1520 hours
Alternative hypothesis: mean life < 1520 hours
2. What is your null distribution?
Since we have to estimate the standard deviation, we need to make inferences from a T-curve with 39 degrees of freedom.
3. Empirical evidence: 1 random sample of 40 has a mean of hours
 5. Probably not sufficient evidence to reject the null. We cannot sue the light bulb manufacturer for false advertising! Notice that using t-distribution to calculate the p-value didn’t change much! With n>30, might as well use Z table.

25. Practice problem
You want to estimate the average ages of kids that ride a particular kid’s ride at Disneyland. You take a random sample of 8 kids exiting the ride, and find that their ages are: 2,3,4,5,6,6,7,7. Assume that ages are roughly normally distributed.
a. Calculate the sample mean.
b. Calculate the sample standard deviation.
c. Calculate the standard error of the mean.
d. Calculate the 99% confidence interval.

26. Answer (a,b) a. Calculate the sample mean.
b. Calculate the sample standard deviation.

27. Answer (c)
c. Calculate the standard error of the mean.

28. Answer (d)
d. Calculate the 99% confidence interval.
t7,.005=3.5

29. Example problem, class data:
A two-tailed hypothesis test:
A researcher claims that Stanford affiliates eat fewer than the recommended intake of 5 fruits and vegetables per week.
We have data to address this claim: 24 people in the class provided data on their daily fruit and vegetable intake.
Do we have evidence to dispute her claim?

30. Histogram fruit and veggie intake (n=24)…
Mean=3.7 servings
Median=3 servings
Mode=3 servings
Std Dev=1.7 servings

31. Answer 1. Define your hypotheses (null, alternative)
H0: P(average servings)=5.0
Ha: P(average servings)≠5.0 servings (two-sided)
2. Specify your null distribution

32. Answer, continued 3. Do an experiment
observed mean in our experiment = 3.7 servings
T23 critical value for p<.05, two tailed = 2.07
4.         Calculate the p-value of what you observed
  p-value < .05;
  5.  Reject or fail to reject (~accept) the null hypothesis
Reject! Stanford affiliates eat significantly fewer than the recommended servings of fruits and veggies.

33. 95% Confidence Interval H0: P(average servings)=5.0
The 95% CI excludes 5, so p-value <.05

34. Paired data (repeated measures)
Patient
BP Before (diastolic)
BP After 1
100 92 2 89 84 3 83 80 4 98 93 5
108 6 95 90
What about these data? How do you analyze these?

35. Example problem: paired ttest
Patient
Diastolic BP Before
D. BP After
Change 1
100 92 -8 2 89 84 -5 3 83 80 -3 4 98 93 5
108
-10 6 95 90
Null Hypothesis: Average Change = 0

36. Example problem: paired ttest
Change -8 -5 -3
-10
Null Hypothesis: Average Change = 0
With 5 df, T>2.571 corresponds to p<.05 (two-sided test)

37. Example problem: paired ttest
Change -8 -5 -3
-10
Note: does not include 0.

38. Summary: Single population mean (small n, normality)
Hypothesis test:
Confidence Interval

39. Summary: paired ttest Hypothesis test: Confidence Interval
Where d=change over time or difference within a pair.

40. Summary: Single population mean (large n)
Hypothesis test:
Confidence Interval

41. Examples of Sample Statistics:
Single population mean (known )
Single population mean (unknown )
Single population proportion
Difference in means (ttest)
Difference in proportions (Z-test)
Odds ratio/risk ratio
Correlation coefficient
Regression coefficient …
It turns out that if you were to go out and sample many, many times, most sample statistics that you could calculate would follow a normal distribution.
What are the 2 parameters (from last time) that define any normal distribution?
Remember that a normal curve is characterized by two parameters, a mean and a variability (SD)
What do you think the mean value of a sample statistic would be? The standard deviation?
Remember standard deviation is natural variability of the population
Standard error can be standard error of the mean or standard error of the odds ratio or standard error of the difference of 2 means, etc. The standard error of any sample statistic.

42. Recall: normal approximation to the binomial…
Statistics for proportions are based on a normal distribution, because the binomial can be approximated as normal if np>5

43. Recall: stats for proportions
For binomial:
Differs by a factor of n.
Differs by a factor of n.
For proportion:
P-hat stands for “sample proportion.”

44. Sampling distribution of a sample proportion
p=true population proportion.
BUT… if you knew p you wouldn’t be doing the experiment!
Always a normal distribution!

45. Practice Problem
A fellow researcher claims that at least 15% of smokers fail to eat any fruits and vegetables at least 3 days a week. You find this hard to believe and decide to check the validity of this statistic by taking a random (representative) sample of smokers. Do you have sufficient evidence to reject your colleague’s claim if you discover that 17 of the 200 smokers in your sample eat no fruits and vegetables at least 3 days a week?

46. Answer 1. What is your null hypothesis?
Null hypothesis: p=proportion of smokers who skip fruits and veggies frequently >= .15
Alternative hypothesis: p < .15
2. What is your null distribution?
Var( ) = .15*.85/200 = SD( ) = .025
~ N (.15, .025)
3. Empirical evidence: 1 random sample: = 17/200 = .085
4. Z = ( )/.025 = -2.6
p-value = P(Z<-2.6) = .0047
5. Sufficient evidence to reject the claim.

47. OR, use computer simulation…
1. Have SAS randomly pick 200 observations from a binomial distribution with p=.15 (the null).
2. Divide the resulting count by 200 to get the observed sample proportion.
3. Repeat this 1000 times (or some arbitrarily large number of times).
4. Plot the resulting distribution of sample proportions in a histogram:

48. How often did we get observed values of 0. 085 or lower when true p=
Only 4/1000 times!
Emprical p-value=.004

49. Practice Problem
In Saturday’s newspaper, in a story about poll results from Ohio, the article said that 625 people in Ohio were sampled and claimed that the margin of error in the results was 4%. Can you explain where that 4% margin of error came from?

50. Answer

51. Paired data proportions test…
Analogous to paired ttest…
Also takes on a slightly different form known as McNemar’s test (we’ll see lots more on this next term…)

52. Paired data proportions test…
1000 subjects were treated with antidepressants for 6 months and with placebo for 6 months (order of tx was randomly assigned)
Question: do suicide attempts (yes/no) differ depending on whether a subject is on antidepressants or on placebo?

53. Paired data proportions test…
15 subjects attempted suicide in both conditions (non-informative)
10 subjects attempted suicide in the antidepressant condition but not the placebo condition
5 subjects attempted suicide in the placebo condition but not the antidepressant condition
970 did not attempt suicide in either condition (non-informative)
Data boils down to 15 observations…
In 10/15 cases (66.6%), antidepressant>placebo.

54. Paired proportions test…
Single proportions test:
Under the null hypothesis, antidepressants and placebo work equally well. So,
Ho: among discordant cases, p (antidepressant>placebo) = 0.5
Observed p = .666
Not enough evidence to reject the null!

55. Key one-sample Hypothesis Tests…
Test for Ho: μ = μ0
Test for Ho: p = po:
Tn-1 approaches Z for large n.
** If np (expected value)<5, use exact binomial rather than Z approximation…

56. Corresponding confidence intervals…
For a mean:
For a proportion:
Tn-1 approaches Z for large n.
** If np (expected value)<5, use exact binomial rather than Z approximation…

57. Symbol overload! n: Sample size Z: Z-statistic (standard normal)
tdf: T-statistic (t-distribution with df degrees of freedom)
p: (“p-hat”): sample proportion
X: (“X-bar”): sample mean
s: Sample standard deviation
p0: Null hypothesis proportion
0: Null hypothesis mean