1. Chapter 9 Introduction to Hypothesis Testing
Business Statistics:
A Decision-Making Approach
8th Edition
Chapter 9 Introduction to Hypothesis Testing

2. Chapter Goals After completing this chapter, you should be able to:
Formulate null and alternative hypotheses involving a single population mean or proportion
Know what Type I and Type II errors are
Formulate a decision rule for testing a hypothesis
Know how to use the test statistic, critical value, and p-value approaches to test the null hypothesis
Compute the probability of a Type II error

3. What is Hypothesis Testing?
A statistical hypothesis is an assumption about a population parameter (assumption may or may not be true).
The best way to determine whether a statistical hypothesis is true would be to examine the entire population.
Often impractical. So, examine a random sample
If sample data are not consistent with the statistical hypothesis, the hypothesis is rejected.
We never 100% “prove” anything because of sampling error

4. Types of Hypotheses Null hypothesis. Alternative hypothesis.
Denoted by H0
Alternative hypothesis.
Denoted by H1 or Ha

5. The Null Hypothesis, H0 States the assumption to be tested
Example: The average number of TV sets in U.S. Homes is at least three ( )
Is always about a population parameter, not about a sample statistic (even though sample is used)

6. The Null Hypothesis, H0
(continued)
Begin with the assumption that the null hypothesis is true
Similar to the notion of “innocent until proven guilty”
Always contains “=” , “≤” or “” sign
May or may not be rejected
Based on the statistical evidence gathered

7. The Alternative Hypothesis, HA
Is the opposite of the null hypothesis
e.g.: The average number of TV sets in U.S. homes is less than 3 ( HA: µ < 3 )
Contains “≠”, “<” or “>” sign
Never contains the “=” , “≤” or “” sign
May or may not be accepted

8. Formulating Hypotheses
Example 1: The average annual income of buyers of Ford F150 pickup trucks is claimed to be $65,000 per year. An industry analyst would like to test this claim.
What is the appropriate test?
H0: µ = 65,000 (income is as claimed)
HA: µ ≠ 65,000 (income is different than claimed)
The analyst will believe the claim unless sufficient evidence is found to discredit it.

9. Formulating Hypotheses
Example 2: Ford Motor Company has worked to reduce road noise inside the cab of the redesigned F150 pickup truck. It would like to report in its advertising that the truck is quieter. The average of the prior design was 68 decibels at 60 mph.
What is the appropriate test?
H0: µ ≤ 67 (the truck is quieter: must be less than 68)
HA: µ > 68 (the truck is not quieter)

10. Research Hypothesis
More appropriate approach: standard and universal for new product or service test
H0: µ ≥ 68 (the truck is not quieter)
HA: µ < 68 (the truck is quieter) wants to support
Research Hypothesis: new product is superior than the original
If the null hypothesis is rejected, Ford has sufficient evidence to support that the truck is now quieter.

11. Steps for Formulating Hypothesis Tests
State the hypotheses.
The hypotheses must be mutually exclusive.
That is, if one is true, the other must be false. 
Formulate an analysis plan.
The analysis plan describes how to use sample data to evaluate the null hypothesis.
The evaluation often focuses around a single test statistic. 

12. Steps for Formulating Hypothesis Tests
Analyze sample data.
Find the value of the test statistic (mean score, proportion, t-score, z-score, etc.) described in the analysis plan and interpret result.
Apply the decision rule described in the analysis plan.
If the value of the test statistic is unlikely, based on the null hypothesis, reject the null hypothesis. 

13. Errors in Making Decisions
3 outcomes for a hypothesis test
No error (no need to discuss)
Type I error (Khan academy: type I error)
Type II error

14. Errors in Making Decisions
(continued)
Type I Error
Rejecting a true null hypothesis
Considered as a serious error
The probability of Type I Error is 
Called level of significance of the test
Set by researcher in advance

15. Errors in Making Decisions
(continued)
Type II Error
Failing to reject (i.e., accept) a false null hypothesis
The probability of Type II Error is β
β is a calculated value, the formula is discussed later in the chapter

16. Outcomes and Probabilities
Possible Hypothesis Test Outcomes
State of Nature
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 (mutually exclusive)
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. Level of Significance, 
The maximum allowable probability of committing a Type I error (is selected by researcher at the beginning).
Defines rejection region (Cutoff) of the sampling distribution
Is designated by  , (level of significance)
Typical values are 0.01, 0.05, or 0.10
Based on COST!
Want small probability of Type I error  high cost
Provides the critical value(s) of the test
The value corresponding to a significance level

19. Hypothesis Tests for the Mean
σ Known
σ Unknown
Assume first that the population standard deviation σ is known

20. Level of Significance and the Rejection Region
*comparison operator of each test*
Lower tail test
Upper tail test
Two tailed test
Example:
Example:
Example:
H0: μ ≥ 3 HA: μ < 3
H0: μ ≤ 3 HA: μ > 3
H0: μ = 3 HA: μ ≠ 3 a a a a /2 /2
-zα zα
-zα/2
zα/2
Do not reject H0
Do not reject H0
Do not reject H0
Reject H0
Reject H0
Reject H0
Reject H0

21. Critical Value for Lower Tail Test
H0: μ ≥ 3
HA: μ < 3
The cutoff value, or ,
is called a critical value
-zα xα a
based on a
Reject H0
Do not reject H0
-zα xα
µ=3

22. Critical Value for Upper Tail Test
H0: μ ≤ 3
HA: μ > 3
The cutoff value, or ,
is called a critical value zα xα a
Do not reject H0
Reject H0 zα
µ=3 xα

23. Critical Values for Two Tailed Tests
H0: μ = 3 HA: μ ¹ 3
There are two cutoff values (critical values): or
± zα/2
/2
/2
xα/2
Lower
Upper
Reject H0
Do not reject H0
Reject H0
xα/2
-zα/2
zα/2
µ=3
xα/2
xα/2
Lower
Upper

24. Two Equivalent Approaches to Hypothesis Testing
z-units (Not as universal as using p-value):
For given , find the critical z value(s):
-zα , zα ,or ±zα/2
Convert the sample mean x to a z test statistic:
Reject H0 if z is in the rejection region,
otherwise do not reject H0

25. Two Equivalent Approaches to Hypothesis Testing
x units (Not Recommended and Not Necessary):
Given , calculate the critical value(s)
xα , or xα/2(L) and xα/2(U)
The sample mean is the test statistic. Reject H0 if x is in the rejection region, otherwise do not reject H0

26. Detail Process of Hypothesis Testing
Specify population parameter of interest
Formulate the null and alternative hypotheses
Specify the desired significance level, α
Define the rejection region
Take a random sample and determine whether or not the sample result is in the rejection region
Reach a decision and draw a conclusion
Example  next slide

27. Hypothesis Testing Example
Test the claim that the true mean # of TV sets in US homes is at least 3. (Assume σ = 0.8)
1. Specify the population value of interest
Testing mean number of TVs in US homes
2. Formulate the appropriate null and alternative hypotheses
H0: μ  HA: μ < 3 (This is a lower (left) tail test)
3. Specify the desired level of significance
Suppose that  = 0.05 is chosen for this test: probability is given, need to find z value

28. Formula Table We have X Z p We want X=μ+zσ Norminv(p,μ,σ)
Normsinv(p as )
Normdist(x,μ,σ,1)
Normsdist(z)

29. Hypothesis Testing Example
(continued)
4. Determine the rejection region
 = .05
Reject H0
Do not reject H0
Normsinv(.05) = 1.645
-zα=
This is a one-tailed test with  = 0.05
Since σ is known, the cutoff value is a z value:
Reject H0 if z < z = ; otherwise do not reject H0

30. Hypothesis Testing Example
5. Obtain sample evidence and compute the test statistic
Suppose a sample is taken with the following results: n = 100, x = ( = 0.8 is assumed known)
Then the test statistic is:

31. Hypothesis Testing Example
(continued)
6. Reach a decision and interpret the result
 = .05 z
Reject H0
Do not reject H0
-1.645
-2.0
Since z = -2.0 < , we reject the null hypothesis that the mean number of TVs in US homes is at least 3. There is sufficient evidence that the mean is less than 3.

32. Hypothesis Testing Example
(continued)
An alternate way (not recommend) of constructing rejection region:
Now expressed in x, not z units
 = .05 x
Reject H0
Do not reject H0
2.8684 3
2.84
Since x = 2.84 < , we reject the null hypothesis
Not enough statistical evidence to conclude that the number of TVs is at least 3

33. Using p-Value
p-value: The p-value is the probability that your null hypothesis is actually correct. 
Simple and easy to apply: always reject null hypothesis if p-value <  .
Best and Universal approach: because p-values are usually computed by statistical SW packages (i.e., Excel, Minitab)

34. p-Value Approach to Testing
(continued)
Convert Sample Statistic ( ) to Test Statistic (a z value, if σ is known)
Determine the p-value from a table or computer
Compare the p-value with 
If p-value <  , reject H0
If p-value   , do not reject H0 x

35. p-Value Approach to Testing
(continued)
More than just a simple “reject”
Can now determine how strongly we “reject” or “accept”
The further the p-value is from a, the stronger the decision
Just compare “absolute value” whether it is one-tail or two-tail test.

36. p-value Example
Example: Based on previous example, how likely is it to see a sample mean of 2.84 (or something further below the mean) if the true mean is  = 3.0?
Normsinv(.05) = 1.645
 = 0.05
p-value =0.0228
Normsdist(-2) =
Reject H0
Do not reject H0 z
-1.645
-2.0: z value from previous example

37. 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
 = 0.05
Here: p-value =
 = 0.05
Since < 0.05, we reject the null hypothesis
p-value =0.0228
Reject H0
-1.645
-2.0

38. Example: Upper Tail z Test for Mean ( Known)
A phone industry manager thinks that customer monthly cell phone bill have increased, and now average over $52 per month. The company wishes to test this claim. (Assume  = 10 is known)
Form hypothesis (upper-tail) test:
H0: μ ≤ the average is not over $52 per month
HA: μ > the average is greater than $52 per month
(i.e., sufficient evidence exists to support the
manager’s claim)

39. Example: Find Rejection Region
(continued)
Suppose  = 0.10 is chosen for this test
Find the rejection region:
Reject H0
 = 0.10
Do not reject H0
Reject H0
zα=1.28
Normsinv(0.1) = 1.28
Reject H0 if z > 1.28

40. Example: Test Statistic
(continued)
Obtain sample evidence and compute the test statistic
Suppose a sample is taken with the following results: n = 64, x = (=10 was assumed known)
Then the test statistic is:

41. Example: Decision Reach a decision and interpret the result:  = 0.10
(continued)
Reach a decision and interpret the result:
Reject H0
 = 0.10
Do not reject H0
Reject H0
1.28
z = 0.88
Do not reject H0 since z = 0.88 ≤ 1.28
i.e.: there is not sufficient evidence that the
mean bill is over $52

42. p -Value Solution Calculate the p-value and compare to 
(continued)
Calculate the p-value and compare to 
p-value =
Reject H0
 = 0.10
Do not reject H0
Reject H0
1.28
z = 0.88
1-NORMSDIST(0.88) =
Do not reject H0 since p-value = >  = 0.10

43. Using t value
When Pop. Dist. Type is unknown and sample size is smaller, convert sample statistic ( ) to a t – statistic. x
Hypothesis
Tests for 
 Known
 Unknown
The test statistic is:

44. When to apply t-statistic
When σ is unknown, convert sample statistic (x) to a t – statistic. This is what the textbook said….based on assumption that the population is approximately normal.
In reality, we do not know whether the population is almost or even half normal or not. Sol, forget about what the textbook says.
Thus, if n is less than 30, apply t – statistic.
Of course, if n is greater than equal to 30, apply z – statistic.

45. Hypothesis Tests for μ, σ Unknown
Specify the population value of interest
Formulate the appropriate null and alternative hypotheses
Specify the desired level of significance
Determine the rejection region (critical values are from the t-distribution with n-1 d.f.)
Obtain sample evidence and compute the test statistic
Reach a decision and interpret the result

46. Example: Two-Tail Test
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

47. Example Solution: Two-Tail Test
H0: μ = HA: μ ¹ 168
a/2=0.025
a/2=0.025
a = 0.05
n = 25
Critical Values:
t24 = ±
 is unknown, so
use a t statistic
Reject H0
Do not reject H0
Reject H0
-tα/2
tα/2
2.0639
1.46
Do not reject H0: not sufficient evidence that true mean cost is different than $168

48. TINV
NORMSINV(p) gives the z-value that puts probability (area) p to the left of that value of z.
TINV(p,DF) gives the t-value that puts one-half the probability (area) to the right with DF degrees of freedom.
The reason for this is that TINV is used mainly to give the t-value used in confidence intervals.
Remember in confidence intervals we use zα/2 or tα/2. So to get tα/2, you put in α and TINV automatically splits it when giving the appropriate t-value.

49. TDIST NORMSDIST(z) gives the probability (area) to the left of z.
TDIST gives the area to the right of t; and on top of that it only works for positive values of t
if you do want the area to the left of t?
1 – (area to the right of t).
The general form TDIST(t, degrees of freedom, 1 or 2).
The last argument (1 or 2)
1 for one-tail test or 2 for two-tail test.
This is because that, in general, TDIST is used to get the p-values for t-tests.

50. Using Excel….
Download and review following Word files from the class website
Implication of TDIST and TINV
Using TDIST and TINV for Hypothesis
Download Finding Critical Value and P-Value Excel file….
Last slide for today’s lecture
Review example from page 349 to 365

51. Hypothesis Tests in PHStat
Options

52. Sample PHStat Output
Input
Output

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

54. Proportions The sample proportion of successes is denoted by p :
When both nπ and n(1- π) are at least 5, p is approximately normally distributed with mean and standard deviation

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

56. Example: z Test for Proportion
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 π = (500)(0.08) = 40
n(1-π) = (500)(0.92) = 460 
Both > 5, so assume normal

57. Z Test for Proportion: Solution
Test Statistic:
H0: π = HA: π ¹ 0.08
a = 0.05
n = 500, p = 0.05
Decision:
Critical Values: ± 1.96
Reject H0 at  = 0.05
Reject
Reject
Conclusion:
0.025
0.025
There is sufficient evidence to reject the company’s claim of 8% response rate. z
-1.96
1.96
-2.47

58. p -Value Solution Calculate the p-value and compare to 
(For a two tailed test the p-value is always two tailed)
Do not reject H0
Reject H0
Reject H0
p-value = .0136:
/2 = 0.025
/2 = 0.025
0.0068
0.0068
-1.96
1.96
z = -2.47
z = 2.47
Reject H0 since p-value = <  = 0.05

59. Type II Error Suppose we fail to reject H0: μ  52
Type II error is the probability of
failing to reject a false H0
Suppose we fail to reject H0: μ  52
when in fact the true mean is μ = 50  50 52
Reject
H0: μ  52
Do not reject
H0 : μ  52

60. Type II Error
(continued)
Suppose we do not reject H0:   52 when in fact the true mean is  = 50
This is the range of x where H0 is not rejected
This is the true distribution of x if  = 50 50 52
Reject
H0:   52
Do not reject
H0 :   52

61. Type II Error
(continued)
Suppose we do not reject H0: μ  52 when in fact the true mean is μ = 50
Here, β = P( x  cutoff ) if μ = 50 β  50 52
Reject
H0: μ  52
Do not reject
H0 : μ  52

62. Steps for Calculating b
Specify the population parameter of interest
Formulate the hypotheses
Specify the significance level
Determine the critical value
Determine the critical value for an upper or lower tail test
Specify the “true” population mean value of interest
Compute z-value based on stipulated population mean
Use standard normal table to find b

63. Calculating β Suppose n = 64 , σ = 6 , and  = 0.05
(for H0 : μ  52)
So β = P( x  ) if μ = 50  50
50.766 52
Reject
H0: μ  52
Do not reject
H0 : μ  52

64. Calculating β Suppose n = 64 , σ = 6 , and  = 0.05 (continued)
Probability of type II error:
β =  50 52
Reject
H0: μ  52
Do not reject
H0 : μ  52

65. Power Of Hypothesis Testing
Would like b to be as small as possible
If the null hypothesis is FALSE, we want to reject it
i.e., we want the hypothesis test to have a high probability of rejecting a false null hypothesis
Referred to as the power of the test
Power = 1 - b

66. Chapter Summary Addressed hypothesis testing methodology
Performed z Test for the mean (σ known)
Discussed p–value approach to hypothesis testing
Performed one-tail and two-tail tests

67. Chapter Summary Performed t test for the mean (σ unknown)
(continued)
Performed t test for the mean (σ unknown)
Performed z test for the proportion
Discussed Type II error and computed its probability