1. Statistics for Business and Economics
Chapter 6
Inferences Based on a Single Sample: Tests of Hypothesis

2. Learning Objectives Distinguish Types of Hypotheses
Describe Hypothesis Testing Process
Explain p-Value Concept
Solve Hypothesis Testing Problems Based on a Single Sample
Explain Power of a Test
As a result of this class, you will be able to ...

3. Descriptive Statistics Inferential Statistics
Statistical Methods
Statistical Methods
Descriptive Statistics
Inferential Statistics
Hypothesis Testing
Estimation 5

4. Hypothesis Testing Concepts

5. Reject hypothesis! Not close.
Hypothesis Testing
I believe the population mean age is 50 (hypothesis).
Reject hypothesis! Not close.
Population 
Mean X = 20
Random sample 

6. I believe the mean GPA of this class is 3.5!
What’s a Hypothesis?
A belief about a population parameter
Parameter is population mean, proportion, variance
Must be stated before analysis
I believe the mean GPA of this class is 3.5!
© T/Maker Co.

7. Null Hypothesis What is tested
Has serious outcome if incorrect decision made
Always has equality sign: , , or 
Designated H0 (pronounced H-oh)
Specified as H0:   some numeric value
Specified with = sign even if  or 
Example, H0:   3

8. Alternative Hypothesis
Opposite of null hypothesis
Always has inequality sign: ,, or 
Designated Ha
Specified Ha:  ,, or  some value
Example, Ha:  < 3

9. Identifying Hypotheses Steps
Example problem: Test that the population mean is not 3
Steps:
State the question statistically (  3)
State the opposite statistically ( = 3)
Must be mutually exclusive & exhaustive
Select the alternative hypothesis (  3)
Has the , <, or > sign
State the null hypothesis ( = 3)

10. What Are the Hypotheses?
Is the population average amount of TV viewing 12 hours?
State the question statistically:  = 12
State the opposite statistically:   12
Select the alternative hypothesis: Ha:   12
State the null hypothesis: H0:  = 12

11. What Are the Hypotheses?
Is the population average amount of TV viewing different from 12 hours?
State the question statistically:   12
State the opposite statistically:  = 12
Select the alternative hypothesis: Ha:   12
State the null hypothesis: H0:  = 12

12. What Are the Hypotheses?
Is the average cost per hat less than or equal to $20?
State the question statistically:   20
State the opposite statistically:   20
Select the alternative hypothesis: Ha:   20
State the null hypothesis: H0:   20

13. What Are the Hypotheses?
Is the average amount spent in the bookstore greater than $25?
State the question statistically:   25
State the opposite statistically:   25
Select the alternative hypothesis: Ha:   25
State the null hypothesis: H0:   25

14. Sampling Distribution
Basic Idea
Sample Means m
= 50 H0
Sampling Distribution
It is unlikely that we would get a sample mean of this value ... 20
... therefore, we reject the hypothesis that  = 50.
... if in fact this were the population mean

15. Level of Significance Probability
Defines unlikely values of sample statistic if null hypothesis is true
Called rejection region of sampling distribution
Designated (alpha)
Typical values are .01, .05, .10
Selected by researcher at start

16. Rejection Region (One-Tail Test)Ho
Value
Critical a
Sample Statistic
Rejection
Region
Nonrejection
Sampling Distribution
1 – 
Level of Confidence
Observed sample statistic
Rejection region does NOT include critical value.

17. Rejection Region (One-Tail Test)
Sampling Distribution Ho
Value
Critical a
Sample Statistic
Rejection
Region
Nonrejection
Sampling Distribution
1 – 
Level of Confidence
Level of Confidence
Observed sample statistic
Rejection region does NOT include critical value.

18. Rejection Regions (Two-Tailed Test)Ho
Value
Critical
1/2 a
Sample Statistic
Rejection
Region
Nonrejection
Sampling Distribution
1 – 
Level of Confidence
Rejection region does NOT include critical value.
Observed sample statistic

19. Rejection Regions (Two-Tailed Test)Ho
Value
Critical
1/2 a
Sample Statistic
Rejection
Region
Nonrejection
Sampling Distribution
1 – 
Level of Confidence
Sampling Distribution
Level of Confidence
Observed sample statistic
Rejection region does NOT include critical value.

20. Rejection Regions (Two-Tailed Test)Ho
Value
Critical
1/2 a
Sample Statistic
Rejection
Region
Nonrejection
Sampling Distribution
1 – 
Level of Confidence
Sampling Distribution
Level of Confidence
Rejection region does NOT include critical value.
Observed sample statistic

21. Decision Making Risks

22. Errors in Making Decision
Type I Error
Reject true null hypothesis
Has serious consequences
Probability of Type I Error is (alpha)
Called level of significance
Type II Error
Do not reject false null hypothesis
Probability of Type II Error is (beta)

23. Decision Results H0: Innocent Jury Trial H0 Test Actual Situation
True
False
Accept
1 – a
Type II
Error
(b)
Reject
Type I
Error (a)
Power
(1 – b)
Actual Situation
Verdict
Innocent
Guilty
Innocent
Correct
Error
Guilty
Error
Correct

24.  &  Have an Inverse Relationship
You can’t reduce both errors simultaneously!  

25. Factors Affecting  True value of population parameter
Increases when difference with hypothesized parameter decreases
Significance level, 
Increases when decreases
Population standard deviation, 
Increases when  increases
Sample size, n
Increases when n decreases

26. Hypothesis Testing Steps

27. H0 Testing Steps State H0 State Ha Choose  Choose n Choose test
Set up critical values
Collect data
Compute test statistic
Make statistical decision
Express decision

28. One Population Tests One Population Z Test t Test Mean Proportion
(1 & 2
tail)
t Test
Mean
Proportion
Variance c 2
Test

29. Two-Tailed Z Test of Mean ( Known)

30. One Population Tests One Population Z Test t Test Mean Proportion
(1 & 2
tail)
t Test
Mean
Proportion
Variance c 2
Test

31. Two-Tailed Z Test for Mean ( Known)
Assumptions
Population is normally distributed
If not normal, can be approximated by normal distribution (n  30)
Alternative hypothesis has  sign
3. Z-Test Statistic

32. Two-Tailed Z Test for Mean Hypotheses
H0:=0 Ha: ≠ 0
Reject H
Reject H
a / 2
a / 2 Z

33. Two-Tailed Z Test Finding Critical Z
What is Z given  = .05? Z
.05
.07
1.6
.4505
.4515
.4525
1.7
.4599
.4608
.4616
1.8
.4678
.4686
.4693
.4744
.4756
.06
1.9
.4750
Standardized Normal Probability Table (Portion)   s
= 1
 / 2 = .025  
1.96
-1.96 Z

34. Two-Tailed Z Test Example
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 of significance.
368 gm.

35. Two-Tailed Z Test Solution
H0:
Ha:
 
n 
Critical Value(s):
 = 368
  368
Test Statistic:
Decision:
Conclusion:
.05 25
Do not reject at  = .05 Z
1.96
-1.96
.025
Reject H
No evidence average is not 368

36. Two-Tailed Z Test Thinking Challenge
You’re a Q/C inspector. You want to find out if a new machine is making electrical cords to customer specification: average breaking strength of 70 lb. with  = 3.5 lb. You take a sample of 36 cords & compute a sample mean of 69.7 lb. At the .05 level of significance, is there evidence that the machine is not meeting the average breaking strength?

37. Two-Tailed Z Test Solution*
H0:
Ha:
 =
n =
Critical Value(s):
 = 70
  70
Test Statistic:
Decision:
Conclusion:
.05 36
Do not reject at  = .05 Z
1.96
-1.96
.025
Reject H
No evidence average is not 70

38. One-Tailed Z Test of Mean ( Known)

39. One-Tailed Z Test for Mean ( Known)
Assumptions
Population is normally distributed
If not normal, can be approximated by normal distribution (n  30)
Alternative hypothesis has < or > sign
3. Z-test Statistic

40. One-Tailed Z Test for Mean Hypotheses
H0:=0 Ha: < 0 Z a
Reject H
H0:=0 Ha: > 0
Small values satisfy H0 . Don’t reject! Z
Reject H a
Must be significantly below 

41. One-Tailed Z Test Finding Critical Z
What Is Z given  = .025? Z
.05
.07
1.6
.4505
.4515
.4525
1.7
.4599
.4608
.4616
1.8
.4678
.4686
.4693
.4744
.4756
.06
1.9
.4750
Standardized Normal Probability Table (Portion)   s
= 1
 = .025  
1.96 Z

42. One-Tailed Z Test Example
Does an average box of cereal contain more than 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 of significance.
368 gm.

43. One-Tailed Z Test Solution
H0:
Ha:
 =
n =
Critical Value(s):
 = 368
 > 368
Test Statistic:
Decision:
Conclusion:
.05 25
Do not reject at  = .05 Z
1.645
.05
Reject
No evidence average is more than 368

44. One-Tailed Z Test Thinking Challenge
You’re an analyst for Ford. You want to find out if the average miles per gallon of Escorts is at least 32 mpg. Similar models have a standard deviation of 3.8 mpg. You take a sample of 60 Escorts & compute a sample mean of 30.7 mpg. At the .01 level of significance, is there evidence that the miles per gallon is at least 32?

45. One-Tailed Z Test Solution*
H0:
Ha:
 =
n =
Critical Value(s):
 = 32
 < 32
Test Statistic:
Decision:
Conclusion:
.01 60
Reject at  = .01 Z
-2.33
.01
Reject
There is evidence average is less than 32

46. Observed Significance Levels: p-Values

47. p-Value
Probability of obtaining a test statistic more extreme (or than actual sample value, given H0 is true
Called observed level of significance
Smallest value of  for which H0 can be rejected
Used to make rejection decision
If p-value  , do not reject H0
If p-value < , reject H0

48. Two-Tailed Z Test p-Value Example
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. Find the p-Value.
368 gm.

49. Two-Tailed Z Test p-Value Solution
1.50 
Z value of sample statistic (observed)

50. Two-Tailed Z Test p-Value Solution
p-value is P(Z  or Z  1.50)  Z
1.50
-1.50
1/2 p-Value
From Z table: lookup 1.50
.4332  
Z value of sample statistic (observed)

51. Two-Tailed Z Test p-Value Solution
p-value is P(Z  or Z  1.50) = .1336 
1/2 p-Value
1/2 p-Value
.0668
.0668 Z
-1.50
1.50 
From Z table: lookup 1.50 
Z value of sample statistic

52. Two-Tailed Z Test p-Value Solution
(p-Value = .1336)  ( = .05). Do not reject H0.
Test statistic is in ‘Do not reject’ region
1/2 p-Value = .0668
1/2 p-Value = .0668
Reject H0
Reject H0
1/2  = .025
1/2  = .025 Z
-1.50
1.50

53. One-Tailed Z Test p-Value Example
Does an average box of cereal contain more than 368 grams of cereal? A random sample of 25 boxes showed x = The company has specified  to be 25 grams. Find the p-Value.
368 gm.

54. One-Tailed Z Test p-Value Solution
1.50 
Z value of sample statistic

55. One-Tailed Z Test p-Value Solution
p-Value is P(Z  1.50) 
p-Value Z
1.50 
Use alternative hypothesis to find direction
From Z table: lookup 1.50
.4332  
Z value of sample statistic

56. One-Tailed Z Test p-Value Solution
p-Value is P(Z  1.50) = .0668  Z
1.50
p-Value 
Use alternative hypothesis to find direction
.0668
.4332 
From Z table: lookup 1.50 
Z value of sample statistic

57. One-Tailed Z Test p-Value Solution
(p-Value = .0668)  ( = .05). Do not reject H0.
Test statistic is in ‘Do not reject’ region
p-Value = .0668
Reject H0
 = .05
1.50 Z

58. p-Value Thinking Challenge
You’re an analyst for Ford. You want to find out if the average miles per gallon of Escorts is at least 32 mpg. Similar models have a standard deviation of 3.8 mpg. You take a sample of 60 Escorts & compute a sample mean of 30.7 mpg. What is the value of the observed level of significance (p-Value)? .

59. p-Value Solution*     p-Value .004 Z -2.65
p-Value is P(Z  -2.65) = p-Value < ( = .01). Reject H0.
p-Value
.004 
Use alternative hypothesis to find direction 
From Z table: lookup 2.65
.4960  Z
-2.65 
Z value of sample statistic

60. Two-Tailed t Test of Mean ( Unknown)

61. One Population Tests One Population Z Test t Test Mean Proportion
(1 & 2
tail)
t Test
Mean
Proportion
Variance c 2
Test

62. t Test for Mean ( Unknown)
Assumptions
Population is normally distributed
If not normal, only slightly skewed & large sample (n  30) taken
Parametric test procedure
t test statistic

63. Two-Tailed t Test Finding Critical t Values
Given: n = 3;  = .10 v t
.10
.05
.025 1
3.078
6.314
12.706 2
1.886
2.920
4.303 3
1.638
2.353
3.182
Critical Values of t Table (Portion)  
df = n - 1 = 2
 /2 = .05  t
2.920
-2.920 

64. Two-Tailed t Test Example
Does an average box of cereal contain 368 grams of cereal? A random sample of 36 boxes had a mean of and a standard deviation of 12 grams. Test at the .05 level of significance.
368 gm.

65. Two-Tailed t Test Solution
H0:
Ha:
 =
df =
Critical Value(s):
 = 368
  368
Test Statistic:
Decision:
Conclusion:
.05
= 35
Reject at  = .05 t
2.030
-2.030
.025
Reject H
There is evidence population average is not 368

66. Two-Tailed t Test Thinking Challenge
You work for the FTC. A manufacturer of detergent claims that the mean weight of detergent is 3.25 lb. You take a random sample of 64 containers. You calculate the sample average to be lb. with a standard deviation of .117 lb. At the .01 level of significance, is the manufacturer correct?
3.25 lb.
Allow students about 10 minutes to finish this.

67. Two-Tailed t Test Solution*
H0:
Ha:
 
df 
Critical Value(s):
 = 3.25
  3.25
Test Statistic:
Decision:
Conclusion:
.01
= 63
Do not reject at  = .01 t
2.656
-2.656
.005
Reject H
There is no evidence average is not 3.25

68. One-Tailed t Test of Mean ( Unknown)

69. One-Tailed t Test Example
Is the average capacity of batteries at least 140 ampere-hours? A random sample of 20 batteries had a mean of and a standard deviation of Assume a normal distribution. Test at the .05 level of significance.

70. One-Tailed t Test Solution
H0:
Ha:
 =
df =
Critical Value(s):
 = 140
 < 140
Test Statistic:
Decision:
Conclusion:
.05
= 19
Reject at  = .05 t
-1.729
.05
Reject H0
There is evidence population average is less than 140

71. One-Tailed t Test Thinking Challenge
You’re a marketing analyst for Wal-Mart. Wal-Mart had teddy bears on sale last week. The weekly sales ($ 00) of bears sold in 10 stores was: At the .05 level of significance, is there evidence that the average bear sales per store is more than 5 ($ 00)?
Assume that the population is normally distributed.
Allow students about 10 minutes to solve this.

72. One-Tailed t Test Solution*
H0:
Ha:
 =
df =
Critical Value(s):
 = 5
 > 5
Test Statistic:
Decision:
Conclusion:
.05
= 9
Note: More than 5 have been sold (6.4), but not enough to be significant. t
1.833
.05
Reject H0
Do not reject at  = .05
There is no evidence average is more than 5

73. Z Test of Proportion

74. Data Types
Data
Quantitative
Qualitative
Continuous
Discrete

75. Qualitative Data
Qualitative random variables yield responses that classify
e.g., Gender (male, female)
Measurement reflects number in category
Nominal or ordinal scale
Examples
Do you own savings bonds?
Do you live on-campus or off-campus?

76. Proportions Involve qualitative variables
Fraction or percentage of population in a category
If two qualitative outcomes, binomial distribution
Possess or don’t possess characteristic
4. Sample Proportion (p) ^

77. Sampling Distribution of Proportion
Approximated by Normal Distribution
Excludes 0 or n
Mean
Standard Error
Sampling Distribution ^
P(P ) .3 .2 .1 ^ .0 P .0 .2 .4 .6 .8
1.0
where p0 = Population Proportion

78. Standardizing Sampling Distribution of Proportion^       ( ) 1
Sampling Distribution
Standardized Normal Distribution   Z
= 0  z 
= 1 P ^ 

79. One Population Tests One Population Z Test t Test Mean Proportion
(1 & 2
tail)
t Test
Mean
Proportion
Variance c 2
Test

80. One-Sample Z Test for Proportion
1. Assumptions
Random sample selected from a binomial population
Normal approximation can be used if
2. Z-test statistic for proportion
Hypothesized population proportion

81. One-Proportion Z Test Example
The present packaging system produces 10% defective cereal boxes. Using a new system, a random sample of 200 boxes had11 defects. Does the new system produce fewer defects? Test at the .05 level of significance.

82. One-Proportion Z Test Solution
H0:
Ha:
 =
n =
Critical Value(s):
p = .10
p < .10
Test Statistic:
Decision:
Conclusion:
.05
200
Reject at  = .05 Z
-1.645
.05
Reject H0
There is evidence new system < 10% defective

83. One-Proportion Z Test Thinking Challenge
You’re an accounting manager. A year-end audit showed 4% of transactions had errors. You implement new procedures. A random sample of 500 transactions had 25 errors. Has the proportion of incorrect transactions changed at the .05 level of significance?

84. One-Proportion Z Test Solution*
H0:
Ha:
 =
n =
Critical Value(s):
p = .04
p  .04
Test Statistic:
Decision:
Conclusion:
.05
500 Z
1.96
-1.96
.025
Reject H
Do not reject at  = .05
There is evidence proportion is not 4%

85. Calculating Type II Error Probabilities

86. Power of Test Probability of rejecting false H0 Designated 1 - 
Correct decision
Designated 1 - 
Used in determining test adequacy
Affected by
True value of population parameter
Significance level 
Standard deviation & sample size n

87. Finding Power Step 1   X  = 368 Reject H0 Do Not
= 368
Reject H0
Do Not
Hypothesis: H0: 0  368 Ha: 0 < 368
 = .05 
Draw

88. Finding Power Steps 2 & 3     X  1-  = 360a
= 360
‘True’ Situation: a = 360 (Ha)  
Draw
Specify 
1-
= 368
Reject H0
Do Not
Hypothesis: H0: 0  368 Ha: 0 < 368
 = .05 

89. Finding Power Step 4      X 1- 363.065  = 360a
= 360
‘True’ Situation: a = 360 (Ha)  
Draw
Specify
1-
= 368
Reject H0
Do Not
Hypothesis: H0: 0  368 Ha: 0 < 368
 = .05 

90. Finding Power Step 5       X  X  = 368 Reject H0 Do Not
= 368
Reject H0
Do Not
Hypothesis: H0: 0  368 Ha: 0 < 368
 = .05 
Draw 
‘True’ Situation: a = 360 (Ha)
Draw
 = .154   
1- =.846
Specify
Z Table 
= 360  X a

91. Power Curves H0:  0 H0:  0 H0:  =0 Power Power
Possible True Values for a
Possible True Values for a
H0:  =0
Power
 = 368 in Example
Possible True Values for a

92. Chi-Square (2) Test of Variance

93. One Population Tests One Population Z Test t Test Mean Proportion
(1 & 2
tail)
t Test
Mean
Proportion
Variance c 2
Test

94. Chi-Square (2) Test for Variance
Tests one population variance or standard deviation
Assumes population is approximately normally distributed
Null hypothesis is H0: 2 = 02
4. Test statistic
Hypothesized pop. variance
Sample variance   2 1)    (n S

95. Chi-Square (2) Distribution
Select simple random
Population
sample, size n.
Sampling Distributions
Compute s 2
for Different Sample s
Sizes
The sampling distribution is a function of the sample sizes upon which the sample variances are based.
Hint: Recall the formula for variance!
s2 = (x -x)2/(n-1) m
Compute c 2 =
(n-1)s 2 / s 2 1 2 3 c 2
Astronomical number of c 2
values 65

96. Finding Critical Value Example
What is the critical 2 value given: Ha: 2 > 0.7 n = 3  =.05? c 2
Upper Tail Area DF
.995 …
.95
.05 1
...
0.004
3.841
0.010
0.103
5.991
2 Table (Portion)
Reject
 = .05
df = n - 1 = 2
5.991 26

97. Finding Critical Value Example
What is the critical 2 value given: Ha: 2 < 0.7 n = 3  =.05?
What do you do if the rejection region is on the left? 26

98. Finding Critical Value Example
What is the critical 2 value given: Ha: 2 < 0.7 n = 3  =.05? c 2
Upper Tail Area DF
.995 …
.95
.05 1
...
0.004
3.841
0.010
0.103
5.991
2 Table (Portion)
Upper Tail Area for Lower Critical Value = = .95
 = .05
Reject H0
df = n - 1 = 2
.103 26

99. Chi-Square (2) Test Example
Is the variation in boxes of cereal, measured by the variance, equal to 15 grams? A random sample of 25 boxes had a standard deviation of 17.7 grams. Test at the .05 level of significance.

100. Chi-Square (2) Test Solution
Ha:
 =
df =
Critical Value(s):
2 = 15
2  15
Test Statistic:
Decision:
Conclusion:
= 33.42
.05
= 24  2
 /2 = .025
Do not reject at  = .05
There is no evidence 2 is not 15
39.364
12.401

101. Conclusion Distinguished Types of Hypotheses
Described Hypothesis Testing Process
Explained p-Value Concept
Solved Hypothesis Testing Problems Based on a Single Sample
Explained Power of a Test
As a result of this class, you will be able to ...