1. Chapter 8 Hypothesis Testing (假设检验)
The Null and Alternative Hypotheses and Errors in Hypothesis Testing.
z Tests about a Population Mean ( known):
t Tests about a Population Mean ( unknown)
z Tests about a Population Proportion

2. Hypothesis Testing        Population  
I believe the population mean age is 50 (hypothesis).
Reject hypothesis! Not close.
Population  
We develop The Procedure to test the validity of statement about the population parameter.
Based on sample evidence and probability theory
Used to determine whether the hypothesis is a reasonable statement and should not be rejected, or is unreasonable and should be rejected
Hypothesis testing is proof by contradiction.    
Random sample 
Mean X = 20  

3. What is a Hypothesis(假设)?
A hypothesis is a claim
(assumption) about a
population parameter:
population mean
population proportion
A statement about the value of a population parameter(母体参数) developed for the purpose of testing.
Example: The mean monthly cell phone bill of this city is μ = $42
Example: The proportion of adults in this city with cell phones is p = 0.68

4. The Null Hypothesis(零假设), H0
States the claim or assertion to be tested
Example: The average number of TV sets in U.S. Homes is equal to three ( )
Is always about a population parameter, not about a sample statistic
It is the statement or assertion about population parameter.
A Belief about a Population Parameter

5. The Null Hypothesis, H0 existing state
(continued)
Begin with the assumption that the null hypothesis is true
Similar to the notion of innocent until proven guilty
Refers to the
Always contains “=” , “≤” or “” sign
May or may not be rejected
The analogy to a criminal trail is apparent. An accused defendant is presumed innocent- the null hypothesis- unless sufficient strong evidence is produced to indicate guilt. The defendant may be found innocent either because he is innocent or because the evidence was not strong enough to convict.
Our approach starts with a hypothesis about the parameter-called null hypothesis –that will maintained unless there is strong evidence against this null hypothesis.
The equality part of the hypotheses always appears in the null hypothesis.
existing state

6. The Alternative Hypothesis(备择假设) Ha
Is the opposite of the null hypothesis
e.g., The average number of TV sets in U.S. homes is not equal to 3 ( Ha: μ ≠ 3 )
Challenges the existing state
Never contains the “=” , “≤” or “” sign
May or may not be proven
Is generally the hypothesis that the researcher is trying to prove
To conduct a statistics test regarding this statement, the first step is determine the null and the alternate hypothesis. It is important to remember that no matter how the problem is stated, the null hypothesis will always contain the equal sign. The equal sign will never appear in the alternate hypothesis. Why? Because Null hypothesis is the statement being tested, and we need a specified value include in our calculation
Important idea:
(If we seek strong evidence in favor of a particular outcome, we define the outcome as alternative hypothesis. When we reject H0, there is strong evidence in favor of H1 and we are confident that our decision is correct.) But failing to reject leads to great uncertainty.
But failure to reject null hypothesis leads to much greater uncertainty because we do not know the probability of type II error. Thus if we fail to reject, either the null hypothesis is true or our procedure for detecting a false null hypothesis does not have sufficient power..

7. Trash Bag Case
Example 8.1
Tests show that the current trash bag has a mean breaking strength m close to but not exceeding 50 lbs
The null hypothesis H0 is that the new bag has a mean breaking strength that is 50 lbs or less
The new bag’s mean breaking strength is not known and is in question, but it is hoped that the new bag is stronger than the current one
The alternative hypothesis Ha is that the new bag has a mean breaking strength that exceeds 50 lbs
H0:   vs. Ha:  > 50
In general, a hypothesis test about the value of a population mean  must take one of the following three forms (where 0 is the hypothesized value of the population mean).

8. Payment Time Case Example 8.2
With a new billing system, the mean bill paying time m is hoped to be less than 19.5 days
The alternative hypothesis Ha is that the new billing system has a mean payment time that is less than 19.5 days
With the old billing system, the mean bill paying time m was close to but not less than 39 days
The null hypothesis H0 is that the new billing system has a mean payment time close to but not less than 19.5 days
H0:   vs. Ha:  < 19.5

9. Types of Decisions
As a result of testing H0 vs. Ha, will have to decide either of the following decisions for the null hypothesis H0:
Do not reject H0 OR
Reject H0
After establishing the null hypothesis and alternative hypothesis (once we have specified null and alternative hypothese), the procedure of testing is based on Null… so the result is the decision about null …

10. Hypothesis Testing(假设检验) Process
Claim: the
population
mean age is 50.
(Null Hypothesis:
Population
H0: μ = 50 )
Now select a random sample X Is = 20
likely if μ = 50?
Suppose
the sample
If not likely,
REJECT
mean age
is 20: X = 20
Sample
Null Hypothesis

11. ... then we reject the null hypothesis that μ = 50.
Reason for Rejecting H0
Sampling Distribution of X X 20
μ = 50
If H0 is true
... then we reject the null hypothesis that μ = 50.
If it is unlikely that we would get a sample mean of this value ...
... if in fact this were the population mean…

12. Level of Significance(显著水平), 
Defines the unlikely values of the sample statistic if the null hypothesis is true
Defines rejection region(拒绝域) of the sampling distribution
Is designated by  , (level of significance)
Typical values are 0.01, 0.05, or 0.10
Is selected by the researcher at the beginning
Provides the critical value(s) (临界值) of the test
After establishing the null hypothesis and alternative hypothesis (once we have specified null and alternative hypothese), the next step is to select the level of significant.
Since hypothesis tests are based on sample data, we must allow for the possibility of errors.
The person conducting the hypothesis test specifies the maximum allowable probability of making a type I error, denoted by  and called the level of significance.
We know that sample mean is different population mean. With only one sample mean we cannot be certain of the value of the population mean. Thus, we know that the decision will have some chance of reaching an error conclusion. We define type I error as …….So our decision rule will be defined so that probability of rejecting a true null hypothesis, denoted as \alpha, is small.
Generally, we cannot control for the probability of making a Type II error, denoted by .
What is the unlikely value such that we reject null hypothesis
The value of the test statistic that established the boundary of the rejection region is called the critical value for the test.

13. Level of Significance and the Rejection Region
Represents
critical value a a
H0: μ = 3 Ha: μ ≠ 3 /2 /2
Rejection region is shaded
Two-tail test 3
Once we sepecified \alpha, we can determine the rejection region.
A decision rule is rule of the specific condition under which null hypothesis is rejected and the conditions under which it is not rejected
The dividing point between the region where the null hypothesis is rejected and the region where it is not rejected.
Two-Tailed Tests, naturally, reject region must be on the two tail
H0: μ ≤ 50 Ha: μ > 50 a 50
Upper-tail test
H0: μ ≥ Ha: μ < 19.5 a
Lower-tail test
19.5

14. Errors in Making Decisions
Type I Error(第一类误差)
Rejecting null hypothesis when it is true
Considered a serious type of error
The probability of Type I Error is 
Called level of significance of the test
Set by the researcher in advance
The meaning of \alpha.
It is also called the level of risk. This may be a more appropriate term because it is the risk you take of rejecting the null hypothesis when it is really true.

15. Errors in Making Decisions
(continued)
Type II Error(第二类误差)
Failing to reject the null hypothesis when it is false
The probability of Type II Error is β
So far we donot control \beta

16. Outcomes and Probabilities
Possible Hypothesis Test Outcomes
Actual Situation
Decision
H0 True
H0 False
Do Not
No error
( )
Type II Error
( β )
Reject
Key:
Outcome
(Probability) a H
Reject
Type I Error
( )
No Error
( 1 - β ) H a

17. Type I & II Error Relationship
Type I and Type II errors cannot happen at
the same time
Type I error can only occur if H0 is true
Type II error can only occur if H0 is false
If Type I error probability (  ) , then
Type II error probability ( β )

18. Hypothesis Tests for the Mean
 Known
 Unknown
(Z test)
(t test)

19. Z Test of Hypothesis for the Mean (σ Known)
Convert sample statistic ( ) to a Z test statistic(Z检验统计量) X
Hypothesis
Tests for 
 Known
σ Known
 Unknown
σ Unknown
(Z test)
(t test)
The test statistic is:

20. Critical Value Approach to Testing
Convert sample statistic ( ) to test statistic (Z statistic )
Determine the critical Z values for a specified level of significance  from a table or computer
Decision Rule: If the test statistic falls in the rejection region, reject H0 ; otherwise do not reject H0

21. Two-Tail Tests(双边检验) /2 /2 H0: μ = 3 H1: μ ¹ 3 X Z
There are two cutoff values (critical values), defining the regions of rejection
/2
/2 X 3
Reject H0
Do not reject H0
Reject H0 Z -Z +Z
Lower critical value
Upper critical value

22. One-Tail Tests(单边检验)
In many cases, the alternative hypothesis focuses on a particular direction
H0: μ ≥ 19.5
Ha: μ < 19.5
This is a lower-tail test since the alternative hypothesis is focused on the lower tail below the mean of 19.5
This is an upper-tail test since the alternative hypothesis is focused on the upper tail above the mean of 50
H0: μ ≤ 50
Ha: μ > 50

23. Lower-Tail Tests a H0: μ ≥ 19.5 Ha: μ < 19.5
There is only one critical value, since the rejection area is in only one tail a
Reject H0
Do not reject H0 Z -Z μ X
Critical value

24. Upper-Tail Tests a H0: μ ≤ 50
Ha: μ > 50
There is only one critical value, since the rejection area is in only one tail a
Do not reject H0
Reject H0 Z Zα _ μ X
Critical value

25. 6 Steps in Hypothesis Testing
State the null hypothesis, H0 and the alternative hypothesis, Ha
Choose the level of significance, , and the sample size, n
Determine the appropriate test statistic and sampling distribution
Determine the critical values that divide the rejection and non-rejection regions

26. 6 Steps in Hypothesis Testing
(continued)
Collect data and compute the value of the test statistic
Make the statistical decision and state the managerial conclusion. If the test statistic falls into the non-rejection region, do not reject the null hypothesis H0. If the test statistic falls into the rejection region, reject the null hypothesis. Express the managerial conclusion in the context of the problem

27. Hypothesis Testing Example
Test the claim that the true mean # of TV sets in US homes is equal to 3.
(Assume σ = 0.8)
1. State the appropriate null and alternative
hypotheses
H0: μ = Ha: μ ≠ 3 (This is a two-tail test)
2. Specify the desired level of significance and the sample size
Suppose that  = 0.05 and n = 100 are chosen for this test

28. Hypothesis Testing Example
(continued)
3. Determine the appropriate technique
σ is known so this is a Z test.
4. Determine the critical values
For  = 0.05 the critical Z values are ±1.96
5. Collect the data and compute the test statistic
Suppose the sample results are
n = 100, X = (σ = 0.8 is assumed known)
So the test statistic is:

29. Hypothesis Testing Example
(continued)
6. Is the test statistic in the rejection region?
 = 0.05/2
 = 0.05/2
Reject H0 if Z < or Z > 1.96; otherwise do not reject H0
Reject H0
Do not reject H0
Reject H0
-Z= -1.96
+Z= +1.96
Here, Z = -2.0 < -1.96, so the test statistic is in the rejection region

30. Hypothesis Testing Example
(continued)
6(continued). Reach a decision and interpret the result
 = 0.05/2
 = 0.05/2
Reject H0
Do not reject H0
Reject H0
-Z= -1.96
+Z= +1.96
-2.0
Since Z = -2.0 < -1.96, we reject the null hypothesis and conclude that there is sufficient evidence that the mean number of TVs in US homes is not equal to 3

31. Trash Bag Case ( is known)
Example 8.4
Tests show that the current trash bag has a mean breaking strength m close to but not exceeding 50 lbs, The new bag’s mean breaking strength is not known and is in question, but it is hoped that the new bag is stronger than the current one
Form hypothesis test:
H0: μ ≤ the average breaking strength is not over 50
lbs
Ha: μ > the average breaking strength for new trash
bag is greater than 50 lbs.

32. Case Study: Find Rejection Region
(continued)
Suppose that  = 0.05 is chosen for this test
Find the rejection region:
Reject H0
 = 0.05
Do not reject H0
Reject H0
1.645
Reject H0 if Z > 1.645

33. Review: One-Tail Critical Value
Standardized Normal Distribution Table (Portion)
What is Z given a = 0.05?
0.45
.04 Z
.03
.05
a = 0.05
1.5
.4370
.4382
.4394
1.6
.4484
.4495
.4505 z
1.645
1.7
.4582
.4591
.4599
Critical Value
= 1.645

34. Case Study: Test Statistic
(continued)
Obtain sample and compute the test statistic
Suppose a sample is taken with the following results: n = 40, X = (=1.65 was assumed known)
Then the test statistic is:

35. Case Study: Decision Reach a decision and interpret the result:
(continued)
Reach a decision and interpret the result:
Reject H0
 = 0.05
Do not reject H0
Reject H0
1.645
Z = 2.20
Reject H0 since Z = 2.20 ≥1.645
i.e.: there is strong evidence that the mean breaking
strength for new trash bag is over 50lbs

36. Discussion Question Using Critical Value Approach
In this payment time case, we assume that s is known and s = 4.2 days. A sample of n = 65,
= days
Have we strong evidence that the mean payment time for new billing system is less than 19.5 days at =0.01 significance level ?

37. Exercise
Does an average box of cereal contain 368 grams of cereal? A random sample of 25 boxes showedX = The company has specified  to be 25 grams. Test at the .05 level.
368 gm.

38. Two-Tailed Z Test Solution
H0:  = 368
Ha:   368
  .05
n  25
Critical Value(s):
Test Statistic:
Decision:
Conclusion:
Do not reject at  = .05
No evidence average is not 368

39. Trash Bag Case ( is known)
Example 8.4
Tests show that the current trash bag has a mean breaking strength m close to but not exceeding 50 lbs, The new bag’s mean breaking strength is not known and is in question, but it is hoped that the new bag is stronger than the current one
Form hypothesis test:
H0: μ ≤ the average breaking strength is not over 50
lbs
Ha: μ > the average breaking strength for new trash
bag is greater than 50 lbs.

40. Case Study: Test Statistic
(continued)
Obtain sample and compute the test statistic
Suppose a sample is taken with the following results: n = 40, X = (=1.65 was assumed known)
Then the test statistic is:

41. p -Value Solution for trash bag case
Example 8.6
Calculate the p-value and compare to 
(assuming that μ = 50)
p-value =
Reject H0
There is a another procedure for considering test.
Calculate the area under the standard normal curve to the right of z
P-value is the probability , assuming H0 is true of observing a value of the test statistics that is greater than or equal to the value….. That have actually computer from the sample data.
 = 0.05
Do not reject H0
Reject H0
1.645
Z = 2.20
Reject H0 since p-value = ＜  = 0.05

42. p-Value (p值) Approach to Testing
p-value: Probability of obtaining a test statistic more extreme ( ≤ or  ) than the observed sample value given H0 is true
Also called observed level of significance
Smallest value of  for which H0 can be rejected

43. p-Value Approach to Testing
(continued)
Convert Sample Statistic (e.g., ) to Test Statistic (e.g., Z statistic )
Obtain the p-value from a table or computer based on test statistic.
Compare the p-value with 
If p-value <  , reject H0
If p-value   , do not reject H0 X

44. Weight of Evidence Against the Null
Calculate the test statistic and the corresponding p-value
Rate the strength of the conclusion about the null hypothesis H0 according to these rules:
If p < 0.10, then there is some evidence to reject H0
If p < 0.05, then there is strong evidence to reject H0
If p < 0.01, then there is very strong evidence to reject H0
If p < 0.001, then there is extremely strong evidence to reject H0
The value of p indicates how strong the evidence is. We can see the samller the p –value, the strong the evidence. Because p-value is samller that means the probobility of sample result is smaller if H0 is true.
P value measures the likelihood of the sample results if the null hy.. Is true.
P-value is the probability , assuming H0 is true of observing a value of the test statistics that is greater than or equal to the value….. That have actually computer from the sample data.
If the H0- true , then only 1 in 1000 of all possible test statistic are at least as large as the value z=…

45. p-Value Example Example 8.5
How likely is it to see a sample mean of 2.84 (or something further from the mean, in either direction) if the true mean is  = 3.0? (σ = 0.8 and n = 100)
X = 2.84 is translated to a Z score of Z = -2.0
/2 = 0.025
/2 = 0.025
0.0228
0.0228
p-value
= =
-1.96
1.96 Z
-2.0
2.0

46. p-Value Example Compare the p-value with 
(continued)
Compare the p-value with 
If p-value <  , reject H0
If p-value   , do not reject H0
Here: p-value =
 = 0.05
Since < 0.05, we reject the null hypothesis
/2 = 0.025
/2 = 0.025
0.0228
0.0228
-1.96
1.96 Z
-2.0
2.0

47. Connection to Confidence Intervals
For X = 2.84, σ = 0.8 and n = 100, the 95% confidence interval is:
≤ μ ≤
Since this interval does not contain the hypothesized mean (3.0), we reject the null hypothesis at  = 0.05

48. Question Using p-Value Approach
In this payment time case, we assume that s is known and s = 4.2 days. A sample of n = 65,
= days
Are there strong evidence that the mean payment time for new billing system is less than 19.5 days at =0.01 significance level ?

49. t Test of Hypothesis for the Mean (σ Unknown)
Convert sample statistic ( ) to a t test statistic X
Hypothesis
Tests for 
 Known
σ Known
 Unknown
σ Unknown
(Z test)
(t test)
The test statistic is:

50. Two-Tail Test ( Unknown)
Example 8.7
The average cost of a hotel room in New York is said to be $168 per night. A random sample of 25 hotels resulted in X = $ and
S = $ Test at the
 = level.
(Assume the population distribution is normal)
H0: μ = Ha: μ ¹ 168

51. Example Solution: Two-Tail Test
Reject H0
a/2=.025
-t n-1,α/2
Do not reject H0
t n-1,α/2
H0: μ = Ha: μ ¹ 168
a = 0.05
n = 25
 is unknown, so
use a t statistic
Critical Value:
t24 = ±
2.0639
1.46
Do not reject H0: not sufficient evidence that true mean cost is different than $168

52. Connection to Confidence Intervals
For X = 172.5, S = and n = 25, the 95% confidence interval is:
(2.0639) 15.4/ to (2.0639) 15.4/ 25
≤ μ ≤
Since this interval contains the Hypothesized mean (168), we do not reject the null hypothesis at  = 0.05

53. t Test of Hypothesis for the Mean (σ Unknown) Continued
Alternative
Reject H0 if:
p-value
Ha: m > m0
t > ta
Area under t distribution to right of t
Ha: m < m0
t < –ta
Area under t distribution to left of –t
Ha: m  m0
|t| > t a/2 *
Twice area under t distribution to right of |t|
ta, ta/2, and p-values are based on n – 1 degrees of freedom (for a sample of size n)
* either t > ta/2 or t < –ta/2

54. Hypothesis Tests for Proportions
Involves categorical variables
Two possible outcomes
“Success” (possesses a certain characteristic)
“Failure” (does not possesses that characteristic)
Fraction or proportion of the population in the “success” category is denoted by p

55. Proportions Sample proportion in the success category is denoted by
(continued)
Sample proportion in the success category is denoted by
When both np and n(1-p) are at least 5, can be approximated by a normal distribution with mean and standard deviation

56. Hypothesis Tests for Proportions
The sampling distribution of is approximately normal, so the test statistic is a Z value:
Hypothesis
Tests for p
np  5
and
n(1-p)  5
np < 5 or
n(1-p) < 5
Not discussed in this chapter

57. Z Test for Proportion in Terms of Number of Successes
An equivalent form to the last slide, but in terms of the number of successes, X:
Hypothesis
Tests for X
X  5
and
n-X  5
X < 5 or
n-X < 5
Not discussed in this chapter

58. Z Test for Proportion
Example 8.8
A marketing company claims that it receives 8% responses from its mailing. To test this claim, a random sample of 500 were surveyed with 25 responses. Test at the  = 0.05 significance level.
Check:
n p= (500)(.08) = 40
n(1-p) = (500)(.92) = 460 

59. Z Test for Proportion: Solution
Test Statistic:
H0: p = Ha: p ¹ 0.08
a = 0.05
n = 500, = 0.05
Decision:
Critical Values: ± 1.96
Reject H0 at  = 0.05
Reject
Reject
Conclusion:
.025
.025
There is sufficient evidence to reject the company’s claim of 8% response rate. z
-1.96
1.96
-2.47

60. p-Value Solution Calculate the p-value and compare to 
(continued)
Calculate the p-value and compare to 
(For a two-tail test the p-value is always two-tail)
Do not reject H0
Reject H0
Reject H0
p-value = :
/2 = .025
/2 = .025
0.0068
0.0068
-1.96
1.96
Z = -2.47
Z = 2.47
Reject H0 since p-value = <  = 0.05

61. Hypothesis Tests about a Population Proportion Continued
Alternative
Reject H0 if:
p-value
Area under standard normal to the right of z
Area under standard normal to the left of –z
Twice the area under standard normal to the right of |z| *
where the test statistic is:
* either z > za/2 or z < –za/2

62. Chapter Summary Addressed hypothesis testing methodology
Performed Z Test for the mean (σ known)
Discussed critical value and p–value approaches to hypothesis testing
Performed one-tail and two-tail tests
Performed t test for the mean (σ unknown)
Performed Z test for the proportion