Statistical tools in hypotheses testing design

Statistics

1. Statistics and Quantitative
Analysis U4320
Segment 7 :
Hypothesis Testing
Prof. Sharyn O’Halloran

2. Hypothesis Testing
I. Introduction
A. Review of Confidence Intervals
SE
-1.96*SE 1.96*SE
-1.96*SE 1.96*SEX

3. Introduction (cont.)
B. Hypothesis Testing: Basic Definitions
1. A Hypotheses is a statement about the
population
2. Null Hypothesis
The Null Hypothesis (Ho)-the statement about our
data that we want to test.
It is always stated as an equality. For instance;
Ho: = 82, where is the average test score
Or, H0: D= 0, where Dis the difference
between men's and women' salaries is zero.

4. Introduction (cont.)
3. Alternative Hypothesis
Every Null Hypothesis has an associated Alternative
Hypothesis, denoted Ha.
This is always stated as an inequality either , >, or
<.
For instances, the alternative hypothesis to the
test scores having a mean of 82 might be Ha: 
82.
The alternative hypothesis to men's and
women's' salaries being equal might be Ha: D>
0.

5. Introduction (cont.)
4. One Tail vs. Two Tail Tests
If the alternative hypothesis is in terms of a sign, it
is called a two-tailed test.
If the alternative hypothesis is in terms of a < or >
sign, it is called a one-tailed test.

6. Introduction (cont.)
C. Three Methods for Testing Hypothesis
1. Method I: Testing hypotheses using
confidence intervals.
2. Method II: Testing hypotheses using p-
values.
3. Method III: Testing hypotheses using critical
values.

7. Hypothesis Testing Using
Confidence Intervals
II. Method I: Hypothesis Testing Using
Confidence Intervals
Note: This method works only for two-tail testsH
0:  = 0 H
a:   0

8. Hypothesis Testing Using
Confidence Intervals (cont.)
A. Example: Differences in Means
In a large university, 10 male professors and 5
female professors were randomly sampled.
Their salaries were:Men (X
1) Women (X
2)
13 20 9
11 14 12
19 17 8
15 14 10
22 15 16
X
1 = 16 X
2 = 11

9. Hypothesis Testing Using
Confidence Intervals (cont.)
1. Step 1: Define Hypothesis
We are interested in the difference between the
means of men's and women's salaries. Call this
difference D= (1-2),
The males state that D= 0,
The females say that D= 7,
Do the data support both of these hypotheses,
one of them, or neither?
We will test these hypotheses at the 5 % a-level. H
0: D = 0, H
a: D  0. H
0: D = 7, H
a: D  7.

10. Hypothesis Testing Using
Confidence Intervals (cont.)
2. Step 2: Calculate a Confidence Interval
Form a 95% confidence interval:
Notice that our data are two samples, one of men
and other of women, from the same larger
population of university professors. So we can pool
our sample variances.D = (X
1-X
2)  t
.025 * sp *
11
1 2nn
 X
1 = 16
X
2 = 11
n
1 = 10
n
2 = 5 s
XX XX
n n
p
2 1 1
2
2 2
2
1 21 1

  

( ) ( )
( )( )
= 146 / [(10-1) + (5-1)]= 11.23

11. Hypothesis Testing Using
Confidence Intervals (cont.)
(cont.)
So the 95% confidence interval is from 1 to 9
thousand dollars. s
p = 146/13 = 3.35 SE = 3.35*
1
10
1
5
 = 3.35 * .548= 1.84 d.f. = 13 t
.025 = 2.16 (look in the t-tables) = 5  2.16 * 1.84 = 5  4.0 1 < (
1-
2) < 9

12. Hypothesis Testing Using
Confidence Intervals (cont.)
3. Step 3: Accept or Reject the Hypothesis
According to these data, is the claim that D= 0
plausible?
We must reject the hypothesis that D= 0 because it
falls outside the 95% confidence interval
What about the hypothesis that D= 7?
SE=1.84
1
D
9
-t
.025
*SE t
.025
*SE
5
reject
reject We cannot reject the null hypothesis H
0: D = 7 at the 5% level.

13. Hypothesis Testing Using
Confidence Intervals (cont.)
4.Summary: Step by Step Procedure
1.Step 1: Define Hypothesis


Pick a significance level; the usual one is 5%.
2.Step 2: Construct confidence interval
Formula depends on type of data, (matched or
pooled variance) and how confident you want to
be.
3.Step 3: Accept or Reject
If falls within this interval, then we fail to
reject the null, otherwise we reject it. Define the null hypothesis H
0:  = 
0. Define the alternative hypothesis H
a:   
0. 
0

14. Hypothesis Testing Using
Confidence Intervals (cont.)
B. Another Example: Matched Data
A firm producing plate glass has developed a less
expensive tempering process to allow glass for
fireplaces to rise to a higher temperature without
breaking. To test it, five different plates of glass
were drawn randomly from a production run, then
cut in half, with one half tempered by the new
process and one half tempered by the old. The two
halves were then heated until they broke. The
results of the experiment look like this: (next slide)

15. Hypothesis Testing Using
Confidence Intervals (cont.)
Matched Data (cont.)
We want to test the hypothesis that the two processes
are equal at the 95% confidence level or at the a= .05
significance level. New Old D
485 475 10
438 436 2
493 495 -2
486 483 3
433 426 7
D
D
n
i
i
n
 


1
4

16. Hypothesis Testing Using
Confidence Intervals (cont.)
1. Step 1: Define Hypothesis
H
0
: D= 0;
H
a
: D0;
Significance level a= 5%.
2. Step 2: Calculate a 95% Confidence
interval. (s
2
unknown) D = 4 s
2
D= (D-D)
2/(n-1) = (10-4)
2 + (2-4)
2 + (-2-4)
2 + (3-4)
2 +(7-4)
2 / (5-1) s
2
D= 21.5 s
D = 4.64 d.f. = 4 t
.025 = 2.78

17. Hypothesis Testing Using
Confidence Intervals (cont.)
Step 2 (cont.)SE =
S
n
D
 
464
5
208
.
. D = D  t
.025 * SE  4  2.78*2.08 D = 4  5.76  -1.76 to 9.77 -1.76 < (D
1-D
2) < 9.768

18. Hypothesis Testing Using
Confidence Intervals (cont.)
3. Step 3: Accept or reject null hypothesis?
So we do not rejectthe hypothesis that H
0
: D= 0
because 0 falls within that range. The two processes
are seen as indistinguishable.-1.76
9.77D=4

19. p-Values
III. Method II: p-Values
P-values are essentially the significance
level.
In essence, we are calculating the
probability that the hypothesis is true. It
summarizes the credibility of the null
hypothesis.

20. p-Values
A. sknown
1. Step 1: State the Hypothesis
A manufacturing process produces TV. tubes with an
average life
=1200 hours and s= 300 hours. A new process is
thought to give tubes a higher average life. And out
of a sample of 100 tubes we find that they have an
average life = 1265 hours. Is the new process
really any better than the old?X

21. p-Values
Step 1 (cont.)
H
0
: = 1200
H
a
: > 1200
a= .05 or 5% significance-level
This is a one-tailed test because we have put all the
area in one-tail of the distribution. We are interested
in those values that are greater than the mean.2
Reject Region
a

22. p-Values
2. Step 2: Calculate p-value
We know sand n is large so we can use the normal
distribution.

0= 1200, and s= 300 and n= 100
Standard error = s/n = 300/ 100 = 30.
The observed value = 1265.
a. Standardize
We then standardize (get the z-value ) X Z = (X- 
0) / (s/n) Z = 1265-1200 / 30 = 2.17

23. p-Values
b. Find z-score (probability of the event
occurring)
Pr (X 1265) = Pr(Z  2.17) = .015 (from the z-table) 2
Reject Region
X=1265
area=1.5%
area=5%

24. p-Values
3.Step 3: Accept or Reject the Hypothesis
This suggests that if the null hypothesis was true
that there would be only a 1.5% probability of
observing as larger as 1265.
Since 1.5% lies to the right of our initial 5%
significance level, we can reject the null hypothesis. X

25. p-Values
4. Two-Tailed Test
H
0
: = 1200
H
a
: 1200
a= .05 or 5% significance-level2
Reject Region
X=1265
area=1.5%
area=2.5%
Reject Region
area=2.5%

26. p-Values
Accept or Reject
Since the area to the right of 1265 is only 1.5%, we
can again reject H
0
.

27. p-Values
B. sunknown
Usually sis unknown and has to be estimated
with the sample standard deviation s. The test
statistic is then t instead of Z. t
X
sn
X
estimatedSE



  
/
.
t = estimate - null hypothesis
estimated SE

28. p-Values
1. Step 1: State Hypothesis (e.g., difference in
men's and women's salaries)
We know from the above example, ( -) = 5
Standard Error = 1.84
Is this a one or a two tailed test? H0: D = 0; Ha: D > 0 ; at a = 5%. X
1 X
2 D
Reject Region
area=5%

29. p-Values
2. Step 2: Calculate p-value
a.Standardizet = estimate-null = 5.0 - 0 = 2.72
SE 1.84

30. p-Values
b.Find probability of event from t-table
Degrees of freedom = (n-1) = 13
So the probability of observing a t-value of 2.72
lies beyond
This means that the tail probability is smaller
than .01. That is, p-value < .01.t
.01=2.65. D
Reject Region
area=5%
2.721.77

31. p-Values
3. Step 3: Accept or Reject Hypothesis
Since the p-value is a measure of the credibility of
H
0, such a low value (below a= 5%) leads us to
conclude that H
0is implausible.
Therefore, we reject the null hypothesis.

32. p-Values
C. Getting t-values from Computers
(Review of Homework)
1. Calculate t-values
How does the computer calculate the t-value? The t-value =
(X-)
s
n
0

33. p-Values
2. Calculate p-value
The 2-tail probability gives the area to the right of
the t-value times two.
If this value is less than your significance level for a
2-tail test, then reject your null hypothesis.

34. p-Values
3. Example: Sample Homework
For example, the difference of means test between
men and women's incomes, produced a t-value =
6.60 and an associated p-value of .00.
Therefore, I can reject the hypothesis that 1-2 = 0
because .00 is less than .025. D
Reject Region
area=2.5%
6.601.96-1.96
area= .000

35. p-Values
D. Summary
1. Step 1: Define Hypothesis
Choose H
0, H
aand a significance level a(default is
5%).
2. Step 2: Calculate p-value
Calculate your p-value from the statistics
if sknown
if sis unknown Z
X
n
X
exactSE




s
 
/
.
t
X
sn
X
estimatedSE



  
/
.

36. p-Values
3. Step 3: Accept or Reject hypothesis
Reject H
0if p-value a
For a One-Tailed Test
Reject H
0if the p-value is less than the
significance level a.
Accept H
0otherwise.
For a Two-tailed Test
Reject H
0if the p-value is less than 1/2 the
significance level.(i.e., 1/2a= .025)
Accept H
0otherwise.

37. Critical Values
IV. Method III: Critical Values
Classical hypothesis testing is very similar
to the p-value approach.
A. Example: Manufacturing of TV tubes
1. State the Hypothesis:
H
0: = 1200 n=100
H
a: > 1200 
0=1200
a= 5%. s=300

38. Critical Values
2. Test Hypothesis: Find the Critical Values
A. In General
What z-value is associated with 5% of the area
under the curve?
From the z-tables we see that the area of 5% is
associated with a z-value of 1.64.
The question is what value on the x-axis
corresponds to a z-value of 1.64? 2
Reject Region
area=5%
z=1.64

39. Critical Values
B.Critical Value
The critical value is the X-value that corresponds to
a Z-value.
We obtain the critical value by arbitrarily setting a=
5% and calculating:
C. Calculating the Critical Value for Manufacturing TV
Tubes
We know that the 
0=1200, and SE=300/100=30.
The Critical Value then is: X
c = 
0 + Z.
05*SE X
c = 1200 + 1.64*30 = 1249.

40. Critical Values
3. Step 3: Reject or Accept the Hypothesis
To accept or reject our hypothesis we collect data and
see if our sample mean is greater then this critical
value.
From the above example we observed a sample mean
= 1265.
Therefore we reject H
0: =1200 because 1265>1249.
So we once again conclude that the new process is
better than the old. X 2
Reject Region
area=5%
z=1.64
X = 1249
c
X=1265 (observed)

41. Critical Values
B. Example of 2-tailed test
How do we construct a two-tailed test at the 5%
significance value?
1. Step 1: State Hypothesis
H0: = 1200
Ha: 1200
a= 5%.

42. Critical Values
2. Step 2: Calculate Critical Value
We use Z
.025instead of Z
.05.
In this case, we would get
c= 
0
Z
.025*SE.

c= 1200 1.96*30 = 1141 and 1259. X X

43. Critical Values
3 Step 3: Accept or reject null Hypothesis
We would reject H
0if the observed fell below 1141
or above 1259.
Again 1265 exceeds the critical value so we still
reject H
0. 2
Reject Region
area=2.5%
z=1.96
X = 1259
c
X=1265 (observed)
z=-1.96
X = 1141
c

44. Critical Values
C. Summary:
1. Step 1: Define Hypothesis
State H
0;
State H
a; and
Choose a significance level a.

45. Critical Values
2. Step 2: Calculate Critical Value
Draw a normal curve and find the critical values at
the level of significance you arbitrarily set. Usually
at the .05 significance-level.
For two-tailed test:
sknown: c = 
0±Z
.025*SE.
sunknown: c = 
0+ t
.025*SE(estimated)
For one-tailed test:
sknown: c = 
0+ Z
.05*SE.
sunknown: c = 
0+ t
.05*SE(estimated)

46. Critical Values
3. Step 3: Accept or Reject
Then collect sample data.
If the sample mean exceeds the critical value, then
reject H
0; otherwise accept H
0.

47. Notes About the Exam
V. Notes About the Exam
1. Hand in your homework at the beginning of
class
2. The exam will cover the material through
today's lecture.
3. Problems, no definitions.
4. You may bring a calculator and one 3 X 5 index
card with whatever you want written on it.
5. Z-tables and t-tables will be supplied.

48. Review Session
Review Session: Saturday March 8
11 to 1 PM
Room 411 IAB