1. The p-value approach to Hypothesis Testing

2. In hypothesis testing we need
A test statistic
A Critical and Acceptance region for the test statistic
The Critical Region is set up under the sampling distribution of the test statistic.
Area = a (0.05 or 0.01) above the critical region. The critical region may be one tailed or two tailed

3. The Critical region:
a/2
a/2
Reject H0
Accept H0

4. In test is carried out by
Computing the value of the test statistic
Making the decision
Reject if the value is in the Critical region and
Accept if the value is in the Acceptance region.

5. The value of the test statistic may be in the Acceptance region but close to being in the Critical region, or
The it may be in the Critical region but close to being in the Acceptance region.
To measure this we compute the p-value.

6. Definition – Once the test statistic has been computed form the data the p-value is defined to be:
p-value = P[the test statistic is as or more extreme than the observed value of the test statistic]
more extreme means giving stronger evidence to rejecting H0

7. Example – Suppose we are using the z –test for the mean m of a normal population and a = 0.05.
Thus the critical region is to reject H0 if
Z < or Z >
Suppose the z = 2.3, then we reject H0
p-value = P[the test statistic is as or more extreme than the observed value of the test statistic]
= P [ z > 2.3] + P[z < -2.3]
= =

8. Graph
p - value
-2.3
2.3

9. If the value of z = 1.2, then we accept H0
p-value = P[the test statistic is as or more extreme than the observed value of the test statistic]
= P [ z > 1.2] + P[z < -1.2]
= =
23.02% chance that the test statistic is as or more extreme than 1.2. Fairly high, hence 1.2 is not very extreme

10. Graph
p - value
-1.2
1.2

11. Properties of the p -value
If the p-value is small (<0.05 or 0.01) H0 should be rejected.
The p-value measures the plausibility of H0.
If the test is two tailed the p-value should be two tailed.
If the test is one tailed the p-value should be one tailed.
It is customary to report p-values when reporting the results. This gives the reader some idea of the strength of the evidence for rejecting H0

12. Summary
A common way to report statistical tests is to compute the p-value.
If the p-value is small ( < 0.05 or < 0.01) then H0 is rejected.
If the p-value is extremely small this gives a strong indication that HA is true.
If the p-value is marginally above the threshold 0.05 then we cannot reject H0 but there would be a suspicion that H0 is false.

13. Testing and Estimation of Variances

14. Let x1, x2, x3, … xn, denote a sample from a Normal distribution with mean m and standard deviation s (variance s2)
The point estimator of the variance s2 is:
The point estimator of the standard deviation s is:

15. The sampling distribution of s2
The c2 distribution

16. The c2 distribution
Let z1, z2, z3, … zn denote a sample from the Standard Normal distribution
Let
Then the distribution of U is called the Chi-square (c2) distribution with n degrees of freedom

17. c 2 distribution
n =1 df
n =2 df
n =4 df

18. comments
Usually statistics that are “sum of squares” of observations have a distribution that is related to the c2 distribution.
The degrees of freedom are the number of “independent” terms in the sum of squares

19. Let x1, x2, x3, … xn, denote a sample from a Normal distribution with mean m and standard deviation s (variance s2)
Let
Then
has a c2 distribution with n = n – 1 degrees of freedom

20. Critical Points of the c2 distribution

21. Confidence intervals for s2 and s.

22. Confidence intervals for s2 and s.
It is true that
from which we can show
and

23. Hence (1 – a)100% confidence limits for s2 are:
and (1 – a)100% confidence limits for s are:

24. Example
A study was interested in determining if administration of a drug reduces cancerous tumor size.
For this purpose n +m = 9 test animals are implanted with a cancerous tumor.
n = 3 are selected at random and administered the drug.
The remaining m = 6 are left untreated.
Final tumour sizes are measured at the end of the test period

25. Suppose the data has been collected and:

26. (1 – a)100% confidence limits for s2 are:
Now:
(1 – a)100% confidence limits for s2 are:
and (1 – a)100% confidence limits for s are:

27. The drug treated group 95 % confidence limits for s2 are:

28. The control group 95 % confidence limits for s2 are:

29. Testing for the equality of variances
The F test

30. Situation:
Let x1, x2, x3, … xn, denote a sample from a Normal distribution with mean mx and standard deviation sx
Let y1, y2, y3, … ym, denote a second independent sample from a Normal distribution with mean my and standard deviation sy
We want to test for the equality of the two variances

31. i.e.:
Test
(Two sided alternative) or
Test
(one sided alternative) or
Test
(one sided alternative)

32. The sampling distribution of the test statistic
The test statistic (F)
The sampling distribution of the test statistic
If the Null Hypothesis (H0) is true then the sampling distribution of F is called the F-distribution with
n1 = n - 1 degrees in the numerator
and n2 = m - 1 degrees in the denominator

33. The F distribution n1 = n - 1 degrees in the numerator
n2 = m - 1 degrees in the denominator a
Fa(n1, n2)

34. Note: If has F-distribution with n1 = n - 1 degrees in the numerator
and n2 = m - 1 degrees in the denominator
then
has F-distribution with
n1 = m - 1 degrees in the numerator
and n2 = n - 1 degrees in the denominator

35. Critical region for the test:
has F-distribution with
n1 = n - 1 degrees in the numerator
and n2 = m - 1 degrees in the denominator
then
has F-distribution with
n1 = m - 1 degrees in the numerator
and n2 = n - 1 degrees in the denominator

36. Critical region for the test:
(Two sided alternative)
Reject H0 if or

37. Critical region for the test (one tailed):
(one sided alternative)
Reject H0 if

38. Example
A study was interested in determining if administration of a drug reduces cancerous tumor size.
For this purpose n +m = 9 test animals are implanted with a cancerous tumor.
n = 3 are selected at random and administered the drug.
The remaining m = 6 are left untreated.
Final tumour sizes are measured at the end of the test period

39. Suppose the data has been collected and:

40. We want to test:
(H0 is assumed for the t-test for comparing the means )
Using a =0.05 we will reject H0 if or

41. Test statistic:
and
Therefore we accept

42. Comparing k Populations
Means – One way Analysis of Variance (ANOVA)

43. The F test – for comparing k means
Situation
We have k normal populations
Let mi and s denote the mean and standard deviation of population i.
i = 1, 2, 3, … k.
Note: we assume that the standard deviation for each population is the same.
s1 = s2 = … = sk = s

44. We want to test
against

45. The data Assume we have collected data from each of th k populations
Let xi1, xi2 , xi3 , … denote the ni observations from population i.
i = 1, 2, 3, … k.
Let

46. The pooled estimate of standard deviation and variance:

47. Consider the statistic comparing the sample means
where

48. To test
against
use the test statistic

49. Computing Formulae

51. Now
Thus

52. To Compute F:
Compute 1) 2) 3) 4) 5)

53. Then 1) 2) 3)

54. The sampling distribution of F
The sampling distribution of the statistic F when H0 is true is called the F distribution.
The F distribution arises when you form the ratio of two c2 random variables divided by there degrees of freedom.

55. i.e. if U1 and U2 are two independent c2 random variables with degrees of freedom n1 and n2 then the distribution of
is called the F-distribution with n1 degrees of freedom in the numerator and n2 degrees of freedom in the denominator

57. Recall: To test
against
use the test statistic

58. We reject if
Fa is the critical point under the F distribution with n1 degrees of freedom in the numerator and n2 degrees of freedom in the denominator

59. Example
In the following example we are comparing weight gains resulting from the following six diets
Diet 1 - High Protein , Beef
Diet 2 - High Protein , Cereal
Diet 3 - High Protein , Pork
Diet 4 - Low protein , Beef
Diet 5 - Low protein , Cereal
Diet 6 - Low protein , Pork

61. Hence

62. Thus
Thus since F > we reject H0

63. A convenient method for displaying the calculations for the F-test
The ANOVA Table
A convenient method for displaying the calculations for the F-test

64. Anova Table Mean Square F-ratio Between k - 1 SSBetween MSBetween
Source
d.f.
Sum of Squares
Mean Square
F-ratio
Between
k - 1
SSBetween
MSBetween
MSB /MSW
Within
N - k
SSWithin
MSWithin
Total
N - 1
SSTotal

65. Diet Example

66. Equivalence of the F-test and the t-test when k = 2

67. the F-test

69. Hence

70. The c2 test for independence

71. Situation We have two categorical variables R and C.
The number of categories of R is r.
The number of categories of C is c.
We observe n subjects from the population and count xij = the number of subjects for which R = I and C = j.
R = rows, C = columns

72. Example
Both Systolic Blood pressure (C) and Serum Chlosterol (R) were meansured for a sample of n = 1237 subjects.
The categories for Blood Pressure are:
<
The categories for Chlosterol are:
<

73. Table: two-way frequency

74. The c2 test for independence
Define
= Expected frequency in the (i,j) th cell in the case of independence.

75. Justification - for Eij = (RiCj)/n in the case of independence
Let pij = P[R = i, C = j] = P[R = i] P[C = j] = rigj in the case of independence
= Expected frequency in the (i,j) th cell in the case of independence.

76. H0: R and C are independent
Then to test
H0: R and C are independent
against
HA: R and C are not independent
Use test statistic
Eij= Expected frequency in the (i,j) th cell in the case of independence.
xij= observed frequency in the (i,j) th cell

77. Sampling distribution of test statistic when H0 is true
- c2 distribution with degrees of freedom n = (r - 1)(c - 1)
Critical and Acceptance Region
Reject H0 if :
Accept H0 if :

79. Standardized residuals
Test statistic
degrees of freedom n = (r - 1)(c - 1) = 9
Reject H0 using a = 0.05