1. Hypothesis Testing – Part I

2. Recall:
We learned how to describe data
Made no assumptions about where the data came from
Nor about method of sampling
We focused on methods of sampling
Probability samples
Learned how to calculate probabilities
Focus on specific probability distributions

3. We learned how to estimate unknown population parameters
Goal: to try to understand the characteristics (parameters) of the population that gave rise to our sample data
4. Now, we’ll learn how to evaluate alternative explanations for the data we have observed
The purpose is to test a research hypothesis

4. Examples of Research Hypotheses
1. Early treatment, compared to later treatment, of
patients with an evolving Ml will result in better
heart function (ejection fraction) at 24 hours.
2. The implementation of policy “A” will result in a
reduction in the inappropriate use of a particular
drug.
3. The delivery of an educational intervention to high school students will result in greater use of “safer” sex practices.
4. The average cost of a particular procedure is $X.

5. To answer such questions,
we collect data
then analyze the data
to see if they are compatible with the research hypothesis being true.
We reason by use of “proof by contradiction”
Proof by example won’t work.
The critic can always claim a counter example must exist.

6. The Logic of Statistical Hypothesis Testing
The investigator starts
by presuming the NULL explanation, eg:
The treatment as NO benefit
The new cost is the same as the old (there is NO difference between cost of new and old)
Data are then collected and evaluated for consistency with the NULL explanation

7. If the data are NOT consistent with the null explanation
then abandon the null explanation in favor of an alternative
Typically, it is the alternative explanation that the investigator would like to advance

8. If the null hypothesis is not true –
Then some alternative hypothesis must be true
This suggests some guidelines:
We’ll let “Ho” represent the null hypothesis “Ha” represent the alternative hypothesis
(called “H-naught” or “H-a”

9. The null is the one we hope to contradict
The null and alternative hypotheses are typically specified so that
The null is the one we hope to contradict
The null and alternative are
Mutually exclusive
Collectively exhaustive
Both should be specified in advance! – before the data is collected.

10. The Research Hypothesis  Ho and Ha
Examples
Early treatment post MI results in better function of the heart at 24 hours (better = higher)
Define study with:
Group 1 = “Early” m1 = true mean at 24 hours
Group 2 = “Late” m2 = true mean at 24 hours
Research hypothesis says m1 > m2
 This is the alternative hypothesis since it is the explanation the investigator seeks to advance.

11. Thus, with the alternative defined, the null is defined to be anything other than the alternative:
That is, m1  m2
Thus we have:
Note that we can rewrite the hypotheses to compare the difference between group means to zero

12. If our research hypothesis is that “early treatment leads to different heart function at 24 hours”
then our alternative is that the means are not equal: m1 ¹ m2

13. The first case is called a one-sided alternative
we are interested in a change in only one direction : m1 > m2
The second case is called a two-sided alternative
we consider change in either direction away from equality: m1  m2

14. Example 2:
The average cost of a particular procedure is $X
Suppose a medical insurance company wants to pay no more than $500 for a particular surgical procedure:
Let m = true average cost
The research hypothesis says m  500
 specify this as the alternative hypothesis
Thus, Ho: m ≥ 500
Ha: m < 500

15. Hypothesis test as “proof by contradiction”
Assume null hypothesis is true
Determine a “rejection region” corresponding to values unlikely to occur using this assumption (Ho true)
Either: “Reject” Ho if the observed data is in the rejection region. OR “Fail to Reject” Ho if data in the rejection region with assumption

16. For example we will reject Ho: m > 500
when X is low enough that we believe the true mean must be less than $500
Rejection Region X
500

17. Steps in Constructing a Statistical Hypothesis Test
1. Identify the research question
2. State the assumptions necessary for computing probabilities
3. Specify Ho and Ha and the α-level (usually α = 0.05)
Specify the test statistic
Specify a decision rule
6. Compute the test statistic and the achieved significance( or P - value) from sample data
7. Come to a “Statistical” Decision
8. Reach a Conclusion
9. Report a confidence interval

18. Example:
1. Identify the research question
Suppose the mean birth weight for 1998 of all US hospital births is known to be m = 3400 gm with s = 710 gm, based upon national birth certificate data.
How do births at Hospital A compare?
We are asking
Is the mean birth weight at Hospital A different from the national mean?

19. Experiment: Collect birth weights of consecutive births at Hospital A and compute our sample mean of x = 3250 gm.
What Assumptions must we make about our data to compute probabilities? Assume:
a random sample of births from
a population with known s = 710 gm (known national standard deviation).
Thus, by the central limit theorem:

20. Specify null and alternative hypotheses:
Ho: The true mean birth weight at Hospital A is the same as the national mean.
Ha: The true mean birth weight at Hospital A is different from the national mean Or
Ho: m = 3400
Ha: m  3400

21. 4. Compute the Test Statistic
This is where the proof by contradiction thinking comes in.
We want to know:
If it is true that m = 3400 gm (Ho)
what are the chances of observing a sample mean as far away from m = 3400 as x = 3250?

22. (greater OR less than the value for Ho),
Since Ha is two-sided
(greater OR less than the value for Ho),
we want to know the probability represented by the following shaded area:
150
150
“What are the chances of observing x as far away from the pop. mean m=3400 as the one we have (3250) in either direction” ?

23. We want to compute: We know how to do this!
150
150
We know how to do this!
We can transform the probability calculation into an equivalent one for a standard normal:

24. When we use m = 3400, as presumed by Ho , the
resulting quantity is called a
a Test Statistic
More generally
if we let mo represent the value of m specified by Ho we have
Test Statistic:

25. 5. Specify Decision Rule
What is the probability of observing a sample mean as far away from mo as the mean we have observed?”
This probability calculation is known as:
the achieved significance
or the significance of the data
or the p-value

26. Our decision rule might be
Reject Ho if the achieved significance is less than 0.05
This is equivalent to saying,
If the probability of observing a sample mean, x, this far or farther from mo
is less than 5%,
then we will reject Ho in favor of Ha.

27. Compute the test statistic from the sample data:

28. The achieved significance (or P-value)is:

29. Statistical Decision:
Since is less than 0.05 we will reject Ho.
We are saying that x = 3250 is sufficiently different from µo = 3400
that it suggests that Ho is not true and should be abandoned.
That is, if Ho is true, the probability of a sample mean this far away is only or 3.5% – an unlikely outcome, so reject Ho.

30. 8. Conclusion:
Hospital A has babies of significantly different birth weight than the US average.
In fact, the mean birth weight at Hospital A appears to be lower than the US mean.

31. 9. Compute a Confidence Interval Estimate of the true mean birth weight for babies at Hospital A
We have all the ingredients to compute a confidence interval estimate:
x = 3250, s = 710, n=100
since the true standard deviation is known we use:
z.975 = 1.96 for a 95% confidence interval:
x  z (s/n) = 3250  (1.96)(710/10) = 3250  139.2
95% CI: ( , )

32. Interpretation:
The hypothesized mean mo = 3400 falls outside (above) the 95% confidence interval.
It therefore seems likely that the mean birth weight at Hospital A is less than the overall US mean.
Your confidence interval should give a consistent result with your hypothesis test.
If it doesn’t – check your work!

33. Comparing CI estimates and Hypothesis Testing
When conducting a hypothesis test, with an a=.05 decision rule, we are centering an interval around the hypothesized mean (m0):
When our observed sample mean (x) falls outside this interval, we interpret this as indicating, with 0.05 likelihood of error, that our sample comes from a distribution with a different mean
x ??
m0-z.975s/n m m0+z.975s/n

34. COMMENTS:
The “.05 rule” alone is very uninformative
it leads to a “reject” or “do not reject” with no information about the data.
A better approach is to report both
the achieved significance
confidence interval estimate
You can then interpret these, while also leaving room for your reader to interpret.

35. 2. Don’t forget the conclusion step!
Too often, only a p-value is reported or, worse
still, only a “reject” or “do not reject” is reported.
3. Statistical significance alone gives no clues
about biology.
Keep in mind that a standard error is a function
of sample size n. This means that by increasing n,
the SE can be made smaller and smaller.

36. Eventually, any observation can achieve statistical significance regardless of its biological relevance.
For example is a statistically significant change in blood pressure of 1 mm Hg very useful?
If we have a very large n, say n=1000 we might find such a difference of 1mmhg statistically significant, but it may not be a biologically meaningful distinction.

37. 4. A statistical hypothesis test uses probabilities based only on the null hypothesis (Ho) model!
The proof by contradiction thinking asks us to:
presume that Ho is true
then examine the plausibility of our data in light of this assumption.
We either reject it, or we fail to do so. We do not prove that Ho is correct.

38. Actually True Actually False
5. We can summarize the results of statistical hypothesis testing as follows:
NULL HYPOTHESIS
Actually True Actually False 
Type II error b
Type I error a
Fail to
Reject
DECISION

39. IF Ho is true and we (incorrectly) reject Ho
we have type I error
we can calculate Pr[type I error] = a
IF Ha is true and we (incorrectly) fail to reject Ho
we have type II error
we must have a specific Ha model before
we can calculate Pr[type II error] = b
IF Ha is true and we (correctly) reject Ho
This occurs with probability = (1-b)
which we call the “POWER” of a test

40. Example 2
Does a new treatment for cancer increase the survival time from diagnosis significantly beyond 38.3 months?
A sample of 100 subjects given the new treatment had a mean survival time of 46.9 months.
Assume the data are a random sample of survival times from a N(m, s2) with s = 43.3 months.
(e.g., we may know the distribution of survival times from prior studies)

41. SOLUTION.
2. Assumptions
We have a random sample of n=100 survival times from a population with s = 43.3.
Thus,
3. Specify Ho and Ha
Research hypothesis suggests an increase in survival
Ho: m £ 38.3
Ha: m > (one sided!)

42. 4. Specify Test Statistic:
Since s = 43.3 is known, we’ll use
5. Decision Rule
We’ll calculate z using observed data
compute the achieved significance (p-value)
and compare this to 0.05
If it is less than 0.05 we will reject Ho
otherwise we will “fail to reject” Ho

43. Calculations – Achieved significance
Be careful! For a one-sided test, we are concerned with a probability in only 1 direction from mo!
.0233
38.3
46.9 z
1.986

44. 7. Statistical Decision
.023 <  Reject Ho
8. Conclusion
It is unlikely that the improvements in survival time are due to chance. The new treatment appears to significantly improves survival.
Confidence Interval on True Mean survival using new treatment:
z.975 = 1.96 for a 95% confidence interval, known s:
x  z (s/n) = 46.9  (1.96)(43.3/10) = 46.9  8.49
95% CI: (38.41, 55.39)

45. A note on One-sided hypothesis tests:
Quite often, we are interested in a change in only one direction:
Does a new drug increase the proportion of patients cured?
Does a new policy decrease the hospital length of stay (LOS)?
A test that looks at a change in only one direction seems to make sense.
However in practice this is rarely done.

46. If it is possible for the change to occur in either direction
then a test should look for the change in either direction.
For example,
the new drug could actually decrease the proportion of patients cured,
or the new policy could potentially result in increased LOS due to unexpected side effects.
Standard practice is to use a two-sided test!

47. Recap of Significance Testing So Far
The Basic Idea
Compute the “probability of the data” (achieved significance) presuming Ho to be true.
Large Probabilities are consistent with Ho -- do not reject
Small Probabilities are NOT consistent with Ho -- reject

48. “Probability of the Data”
We want to know the probability of a sample statistic as extreme or more extreme than the one observed.
One Sided Alternative
Two Sided Alternative
Distribution
determined
by Ho
t or z = observed
sample statistic

49. Next we will consider a couple of examples that parallel the situations we have discussed so far for confidence interval estimation.
We will also focus on computer analysis for conducting hypothesis tests

50. Application 1: One Population, s2 Known, Test of hypothesis on mean, m
1. Research Question: Serum enzyme A levels are measured in 10 patients with a sample mean of 22. If it is known that the population variance is 45 and if normality is assumed, are the data consistent with a population mean of 25?
That is, we have
n=10
s2=45
x=22
mo=25

51. Random sample of serum enzyme A levels
2. Assumptions
Random sample of serum enzyme A levels
from a Normal distribution with s2 = 45.
Thus,
Why must we assume normality of the data for this example?
Is n particularly large for Central Limit Theorem to hold?

52. 3. Specify Ho and Ha :
The wording “are the data consistent with” suggests a two sided alternative:
Ho: m = 25
Ha: m ¹ 25
4. Test Statistic
s2 known suggests use of Normal or z-transformation:

53. Decision Rule
We’ll calculate the achieved significance (p-value) and compare to a = .05
Reject Ho for p<.05, else fail to reject.
Calculations
Test Statistic:

54. Achieved Significance:
For 2-sided test:
Total area is the
achieved level
of significance or the p-value
Minitab examples

55. Statistical Decision
represents the probability of a sample mean at least as far away from 25 as 22, if in fact the true mean is 25.
0.1586, or 15.86% is reasonably high.
This suggests the data are consistent with the null hypothesis
Þ Do NOT reject Ho
.1586 > (or > a)

56. Conclusion
The data are consistent with an hypothesized mean serum A level of m = 25.
Note
we have not proven Ho is true, merely that
with the evidence of our sample, m = 25 is a reasonable possibility
and we cannot reject it.

57. 95% Confidence Interval
x  z.975(s/n) = 22  1.96(2.12) = (17.5, 26.5)
Note that:
mo = 25 is within the interval

58. Using Minitab: Stat  Basic Stats  1-Sample Z
Select variable to test
Test mean: Enter mo
Sigma: Enter s

59. Z-Test
Test of mu = vs mu not = 25.00
The assumed sigma = 6.71
Variable N Mean StDev SE Mean Z P
SerA

60. Application 2: One Normal Population, s2 UNknown
Test of m
1. Research Question
A drug company claims a certain capsule contains 2.5 milligrams of a drug X. An independent laboratory obtained a random sample of 20 capsules and measured the amount of the drug in each. The measurements were as follows:
Is the drug company claim correct?

61. s2 UNknown suggests use of the t-transformation:
Assumptions
We have data from a random sample from a normal distribution, s2 unknown.
Specify Ho and Ha
Ho: mo = 2.5 mg
Ha: mo  2.5 mg (Two-sided)
Test Statistic
s2 UNknown suggests use of the t-transformation:
Why must we assume normality of the data?
So the t-distribution is applicable!

62. Decision Rule
We’ll calculate the achieved significance level (two-sided) and compare this to a type I error of a = 0.05
Calculations (x = 2.00, s = .787)
Test Statistic:
Achieved significance:

63. Statistical Decision
< p-value < type I error (a)
Therefore, REJECT Ho
Conclusion
The mean amount of drug in the capsules appears to be significantly different from 2.5 mg.
In fact, the mean is significantly less than 2.5 mg.

64. Confidence Interval Estimate
x  t19; .975 (se) =  2.093(.176) = (1.66, 2.35).
Note that The confidence interval does not include the hypothesized mean amount of 2.50

65. The particular test just conducted is known as a
ONE-SAMPLE t-TEST.
We have a sample from a single population
We are comparing our observed sample mean to some hypothesized value for the mean.

66. Using Minitab: Stats  Basic Stats  1-Sample t
Select variable to test
Enter mo

67. T-Test of the Mean
Test of mu = vs mu not = 2.500
Variable N Mean StDev SE Mean T P
drugx
Note that the one sample t-test provides you with estimates of the mean, standard deviation and standard error.
By checking the Confidence Interval option, you can get a confidence interval rather than a hypothesis test.