1. © 2011 Pearson Education, Inc

2. Statistics for Business and Economics
Chapter 6
Inferences Based on a Single Sample: Tests of Hypotheses
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3. © 2011 Pearson Education, Inc
Content
The Elements of a Test of Hypothesis
Formulating Hypotheses and Setting Up the Rejection Region
Test of Hypothesis about a Population Mean: Normal (z) Statistic
Observed Significance Levels: p-Values
Test of Hypothesis about a Population Mean: Student’s t-Statistic
As a result of this class, you will be able to ...
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4. © 2011 Pearson Education, Inc
Content
Large-Sample Test of Hypothesis about a Population Proportion
Calculating Type II Error Probabilities: More about *
Test of Hypothesis about a Population Variance
As a result of this class, you will be able to ...
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5. © 2011 Pearson Education, Inc
Learning Objectives
Test a specific value of a population parameter (mean or proportion), called a test of hypothesis
Provide a measure of reliability for the hypothesis test, called the significance level of the test
As a result of this class, you will be able to ...
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6. The Elements of a Test of Hypothesis
6.1
The Elements of a Test of Hypothesis
:1, 1, 3
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7. 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 
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8. What’s a Hypothesis?
A statistical hypothesis is a statement about the numerical value of a population parameter.
I believe the mean GPA of this class is 3.5!
© T/Maker Co.
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Null Hypothesis
The null hypothesis, denoted H0, represents the hypothesis that will be accepted unless the data provide convincing evidence that it is false. This usually represents the “status quo” or some claim about the population parameter that the researcher wants to test.
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10. Alternative Hypothesis
The alternative (research) hypothesis, denoted Ha, represents the hypothesis that will be accepted only if the data provide convincing evidence of its truth. This usually represents the values of a population parameter for which the researcher wants to gather evidence to support.
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11. Alternative Hypothesis
Opposite of null hypothesis
The hypothesis that will be accepted only if the data provide convincing evidence of its truth
Designated Ha
Stated in one of the following forms
Ha:   some value)
Ha:   some value)
Ha:   some value)
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12. Identifying Hypotheses
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)
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13. 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
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14. 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
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15. 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
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16. 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
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Test Statistic
The test statistic is a sample statistic, computed from information provided in the sample, that the researcher uses to decide between the null and alternative hypotheses.
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18. Test Statistic - Example
The sampling distribution of assuming µ = 2,400. the chance of observing more than standard deviations above 2,400 is only .05 – if in fact the true mean µ is 2,400.
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Type I Error
A Type I error occurs if the researcher rejects the null hypothesis in favor of the alternative hypothesis when, in fact, H0 is true. The probability of committing a Type I error is denoted by .
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Rejection Region
The rejection region of a statistical test is the set of possible values of the test statistic for which the researcher will reject H0 in favor of Ha.
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Type II Error
A Type II error occurs if the researcher accepts the null hypothesis when, in fact, H0 is false. The probability of committing a Type II error is denoted by .
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22. Conclusions and Consequences for a Test of Hypothesis
True State of Nature
Conclusion
H0 True
Ha True
Accept H0 (Assume H0 True)
Correct decision
Type II error (probability )
Reject H0 (Assume Ha True)
Type I error (probability )
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23. Elements of a Test of Hypothesis
1. Null hypothesis (H0): A theory about the specific values of one or more population parameters. The theory generally represents the status quo, which we adopt until it is proven false.
2. Alternative (research) hypothesis (Ha): A theory that contradicts the null hypothesis. The theory generally represents that which we will adopt only when sufficient evidence exists to establish its truth.
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24. Elements of a Test of Hypothesis
3. Test statistic: A sample statistic used to decide whether to reject the null hypothesis.
4. Rejection region: The numerical values of the test statistic for which the null hypothesis will be rejected. The rejection region is chosen so that the probability is  that it will contain the test statistic when the null hypothesis is true, thereby leading to a Type I error. The value of  is usually chosen to be small (e.g., .01, .05, or .10) and is referred to as the level of significance of the test.
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25. Elements of a Test of Hypothesis
5. Assumptions: Clear statement(s) of any assumptions made about the population(s) being sampled.
6. Experiment and calculation of test statistic: Performance of the sampling experiment and determination of the numerical value of the test statistic.
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26. Elements of a Test of Hypothesis
7. Conclusion:
a. If the numerical value of the test statistic falls in the rejection region, we reject the null hypothesis and conclude that the alternative hypothesis is true. We know that the hypothesis-testing process will lead to this conclusion incorrectly (Type I error) only 100% of the time when H0 is true.
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27. Elements of a Test of Hypothesis
7. Conclusion:
b. If the test statistic does not fall in the rejection region, we do not reject H0. Thus, we reserve judgment about which hypothesis is true. We do not conclude that the null hypothesis is true because we do not (in general) know the probability  that our test procedure will lead to an incorrect acceptance of H0 (Type II error).
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28. Determining the Target Parameter
Key Words or Phrases
Type of Data
Mean; average
Quantitative p
Proportion; percentage; fraction; rate
Qualitative 2
Variance; variability; spread
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29. Formulating Hypotheses and Setting Up the Rejection Region
6.2
Formulating Hypotheses and Setting Up the Rejection Region
:1, 1, 3
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30. Steps for Selecting the Null and Alternative Hypotheses
1. Select the alternative hypothesis as that which the sampling experiment is intended to establish. The alternative hypothesis will assume one of three forms:
a. One-tailed, upper-tailed (e.g., Ha: µ > 2,400)
b. One-tailed, lower-tailed (e.g., Ha: µ < 2,400)
c. Two-tailed (e.g., Ha: µ ≠ 2,400)
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31. Steps for Selecting the Null and Alternative Hypotheses
2. Select the null hypothesis as the status quo, that which will be presumed true unless the sampling experiment conclusively establishes the alternative hypothesis. The null hypothesis will be specified as that parameter value closest to the alternative in one-tailed tests and as the complementary (or only unspecified) value in two-tailed tests.
(e.g., H0: µ = 2,400)
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One-Tailed Test
A one-tailed test of hypothesis is one in which the alternative hypothesis is directional and includes the symbol “ < ” or “ >.”
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Two-Tailed Test
A two-tailed test of hypothesis is one in which the alternative hypothesis does not specify departure from H0 in a particular direction and is written with the symbol “ ≠.”
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34. 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
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35. Rejection Region (One-Tail Test)Ho
Value
Critical a
Sample Statistic
Rejection
Region
Fail to Reject
Sampling Distribution
1 – 
Level of Confidence
Rejection region does NOT include critical value.
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36. Rejection Regions (Two-Tailed Test)Ho
Value
Critical
1/2 a
Sample Statistic
Rejection
Region
Fail to Reject
Sampling Distribution
1 – 
Level of Confidence
Rejection region does NOT include critical value.
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37. Rejection Regions Alternative Hypotheses Lower-Tailed Upper-Tailed
Two-Tailed
 = .10
z < –1.28
z > 1.28
z < –1.645 or z > 1.645
 = .05
z < –1.645
z > 1.645
z < –1.96 or z > 1.96
 = .01
z < –2.33
z > 2.33
z < –2.575 or z > 2.575
Rejection region does NOT include critical value.
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38. Test of Hypotheses about a Population Mean: Normal (z) Statistic
6.3
Test of Hypotheses about a Population Mean: Normal (z) Statistic
:1, 1, 3
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39. Large-Sample Test of Hypothesis about µ
One-Tailed Test Two-Tailed Test
H0: µ = µ0 H0: µ = µ0
Ha: µ < µ0 Ha: µ ≠ µ0
(or Ha: µ > µ0)
Test Statistic: Test Statistic:
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40. Large-Sample Test of Hypothesis about µ
One-Tailed Test
Rejection region:
z < –z
(or z > zwhen Ha: µ > µ0)
where z is chosen so that
P(z > z) = 
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41. Large-Sample Test of Hypothesis about µ
Two-Tailed Test
Rejection region:
|z| > z
where z is chosen so that
P(|z| > z) = /2
Note: µ0 is the symbol for the numerical value assigned to µ under the null hypothesis.
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42. Conditions Required for a Valid Large-Sample Hypothesis Test for µ
1. A random sample is selected from the target population.
2. The sample size n is large (i.e., n ≥ 30). (Due to the Central Limit Theorem, this condition guarantees that the test statistic will be approximately normal regardless of the shape of the underlying probability distribution of the population.)
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43. Possible Conclusions for a Test of Hypothesis
1. If the calculated test statistic falls in the rejection region, reject H0 and conclude that the alternative hypothesis Ha is true. State that you are rejecting H0 at the  level of significance. Remember that the confidence is in the testing process, not the particular result of a single test.
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44. Possible Conclusions for a Test of Hypothesis
2. If the test statistic does not fall in the rejection region, conclude that the sampling experiment does not provide sufficient evidence to reject H0 at the  level of significance. [Generally, we will not “accept” the null hypothesis unless the probability  of a Type II error has been calculated.]
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45. Two-Tailed z Test Example
Does an average box of cereal contain 368 grams of cereal? A random sample of 25 boxes had x = The company has specified  to be 25 grams. Test at the .05 level of significance.
368 gm.
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46. 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
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47. 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?
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48. 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
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49. 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.
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50. 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
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51. 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 less than 32?
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52. 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
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53. Observed Significance Levels: p-Values
6.4
Observed Significance Levels: p-Values
:1, 1, 3
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54. © 2011 Pearson Education, Inc
p-Value
The observed significance level, or p-value, for a specific statistical test is the probability (assuming H0 is true) of observing a value of the test statistic that is at least as contradictory to the null hypothesis, and supportive of the alternative hypothesis, as the actual one computed from the sample data.
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55. © 2011 Pearson Education, Inc
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
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56. Steps for Calculating the p-Value for a Test of Hypothesis
1. Determine the value of the test statistic z corresponding to the result of the sampling experiment.
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57. Steps for Calculating the p-Value for a Test of Hypothesis
2a. If the test is one-tailed, the p-value is equal to the tail area beyond z in the same direction as the alternative hypothesis. Thus, if the alternative hypothesis is of the form > , the p-value is the area to the right of, or above, the observed z-value. Conversely, if the alternative is of the form < , the p-value is the area to the left of, or below, the observed z-value.
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58. Steps for Calculating the p-Value for a Test of Hypothesis
2b. If the test is two-tailed, the p-value is equal to twice the tail area beyond the observed z-value in the direction of the sign of z – that is, if z is positive, the p-value is twice the area to the right of, or above, the observed z-value. Conversely, if z is negative, the p-value is twice the area to the left of, or below, the observed z-value.
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59. Reporting Test Results as p-Values: How to Decide Whether to Reject H0
1. Choose the maximum value of  that you are willing to tolerate.
2. If the observed significance level (p-value) of the test is less than the chosen value of , reject the null hypothesis. Otherwise, do not reject the null hypothesis.
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60. 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 15 grams. Find the p-Value.
368 gm.
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61. Two-Tailed z Test p-Value Solution
1.50 
z value of sample statistic (observed)
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62. Two-Tailed Z Test p-Value Solution
p-value is P(z  –1.50 or z  1.50)
.5000 –  z
1.50
–1.50
1/2 p-Value
From z table: lookup 1.50
.4332  
z value of sample statistic (observed)
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63. Two-Tailed z Test p-Value Solution
p-value is P(z  –1.50 or z  1.50) = .1336
1/2 p-Value
1/2 p-Value
.0668
.0668 z
–1.50
1.50
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64. Two-Tailed z Test p-Value Solution
p-Value =   = .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
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65. 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 15 grams. Find the p-Value.
368 gm.
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66. One-Tailed z Test p-Value Solution
1.50 
z value of sample statistic
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67. One-Tailed z Test p-Value Solution
p-Value is P(z 1.50) 
p-Value z
1.50
.5000 – 
Use alternative hypothesis to find direction
From z table: lookup 1.50
.4332  
z value of sample statistic
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68. 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
.5000 –
.4332 
From z table: lookup 1.50 
z value of sample statistic
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69. 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
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70. p-Value Thinking Challenge
You’re an analyst for Ford. You want to find out if the average miles per gallon of Escorts is less than 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)? .
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71. © 2011 Pearson Education, Inc
p-Value Solution*
p-Value is P(z  -2.65) = p-Value < ( = .01). Reject H0.
p-Value
.004
.5000 – 
Use alternative hypothesis to find direction 
From z table: lookup 2.65
.4960  z
–2.65 
z value of sample statistic
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72. © 2011 Pearson Education, Inc
Converting a Two-Tailed p-Value from a Printout to a One-Tailed p-Value
if Ha is of the form > and z is positive
or Ha is of the form < and z is negative
if Ha is of the form > and z is negative
Ha is of the form < and z is positive
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73. Test of Hypothesis about a Population Mean: Student’s t-Statistic
6.5
Test of Hypothesis about a Population Mean: Student’s t-Statistic
:1, 1, 3
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74. Small-Sample Test of Hypothesis about µ
One-Tailed Test
H0: µ = µ0
Ha: µ < µ0 (or Ha: µ > µ0)
Test statistic:
Rejection region: t < –t
(or t > t when Ha: µ > µ0)
where t and t are based on (n – 1) degrees of freedom
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75. Small-Sample Test of Hypothesis about µ
Two-Tailed Test
H0: µ = µ0
Ha: µ ≠ µ0
Test statistic:
Rejection region: |t| > t
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76. Conditions Required for a Valid Small-Sample Hypothesis Test for µ
1. A random sample is selected from the target population.
2. The population from which the sample is selected has a distribution that is approximately normal.
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77. 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.
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78. Two-Tailed t Test Solution
H0:
Ha:
 =
df =
Critical Value(s):
 = 368
  368
Test Statistic:
Decision:
Conclusion:
.05
36 – 1 = 35
Reject at  = .05 t
2.030
-2.030
.025
Reject H
There is evidence population average is not 368
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79. 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.
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80. Two-Tailed t Test Solution*
H0:
Ha:
 
df 
Critical Value(s):
 = 3.25
  3.25
Test Statistic:
Decision:
Conclusion:
.01
64 – 1 = 63
Do not reject at  = .01 t
2.656
-2.656
.005
Reject H
There is no evidence average is not 3.25
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81. One-Tailed t Test Example
Is the average capacity of batteries less than 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.
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82. One-Tailed t Test Solution
H0:
Ha:
 =
df =
Critical Value(s):
 = 140
 < 140
Test Statistic:
Decision:
Conclusion:
.05
20 – 1 = 19
Reject at  = .05 t
-1.729
.05
Reject H0
There is evidence population average is less than 140
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83. 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.
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84. One-Tailed t Test Solution*
H0:
Ha:
 =
df =
Critical Value(s):
 = 5
 > 5
Test Statistic:
Decision:
Conclusion:
.05
10 – 1 = 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
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85. Large-Sample Test of Hypothesis about a Population Proportion
6.6
Large-Sample Test of Hypothesis about a Population Proportion
:1, 1, 3
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86. Large-Sample Test of Hypothesis about p
One-Tailed Test
H0: p = p0
Ha: p < p0 (or Ha: p > p0)
Test statistic:
Rejection region:
z < –z(or z > z when Ha: p > p0)
Note: p0 is the symbol for the numerical value of p assigned in the null hypothesis
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87. Large-Sample Test of Hypothesis about p
Two-Tailed Test
H0: p = p0
Ha: p ≠ p0
Test statistic:
Rejection region: |z| < z
Note: p0 is the symbol for the numerical value of p assigned in the null hypothesis
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88. Conditions Required for a Valid Large-Sample Hypothesis Test for p
1. A random sample is selected from a binomial population.
2. The sample size n is large. (This condition will be satisfied if both np0 ≥ 15 and nq0 ≥ 15.)
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89. 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.
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90. 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
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91. 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?
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92. 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%
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93. Calculating Type II Error Probabilities: More about 
6.7
Calculating Type II Error Probabilities: More about 
:1, 1, 3
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94. © 2011 Pearson Education, Inc
Type II Error
The Type II error probability  is calculated assuming that the null hypothesis is false because it is defined as the probability of accepting H0 when it is false.The situation corresponding to accepting the null hypothesis, and thereby risking a Type II error, is not generally as controllable. For that reason, we adopted a policy of nonrejection of H0 when the test statistic does not fall in the rejection region, rather than risking an error of unknown magnitude.
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95. Steps for Calculating  for a Large-Sample Test about µ
1. Calculate the value(s) of corresponding to the border(s) of the rejection region. There will be one border value for a one-tailed test and two for a two-tailed test. The formula is one of the following, corresponding to a test with level of significance :
Upper-tailed test:
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96. Steps for Calculating  for a Large-Sample Test about µ
Lower-tailed test:
Two-tailed test:
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97. Steps for Calculating  for a Large-Sample Test about µ
2. Specify the value of µa in the alternative hypothesis for which the value of  is to be calculated. Then convert the border value(s) of to z-value(s) using the alternative distribution with mean µa. The general formula for the z-value is
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98. Steps for Calculating  for a Large-Sample Test about µ
Sketch the alternative distribution (centered at µa) and shade the area in the acceptance (nonrejection) region. Use the z-statistic(s) and Table IV in Appendix B to find the shaded area, which is .
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99. © 2011 Pearson Education, Inc
Power of Test
Probability of rejecting false H0
Correct decision
Equal to 1 – 
Used in determining test adequacy
Affected by
True value of population parameter
Significance level 
Standard deviation & sample size n
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100. Two-Tailed z Test Example
Does an average box of cereal contain 368 grams of cereal? A random sample of 25 boxes had x = The company has specified  to be 15 grams. Test at the .05 level of significance.
368 gm.
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101. © 2011 Pearson Education, Inc
Finding Power Step 1  x 
= 368
Reject H0
Do Not
Hypothesis: H0: 0  368 Ha: 0 < 368
 = .05 
Draw
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102. © 2011 Pearson Education, Inc
Finding Power Steps 2 & 3  x  a
= 360
‘True’ Situation: a = 360 (Ha)  
Draw
Specify 
1–
= 368
Reject H0
Do Not
Hypothesis: H0: 0  368 Ha: 0 < 368
 = .05 
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103. © 2011 Pearson Education, Inc
Finding Power Step 4  x 
= 368
Reject H0
Do Not
Hypothesis: H0: 0  368 Ha: 0 < 368
 = .05 
Draw 
‘True’ Situation: a = 360 (Ha)
Draw  
1–
Specify 
= 360  x
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104. © 2011 Pearson Education, Inc
Finding Power Step 5   x  a
= 360
‘True’ Situation: a = 360 (Ha)  
Draw
Specify
= 368
Reject H0
Do Not
Hypothesis: H0: 0  368 Ha: 0 < 368
 = .05 
 = .154 
1– =.846
z Table
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105. Properties of  and Power
1. For fixed n and , the value of  decreases, and the power increases as the distance between the specified null value µ0 and the specified alternative value µa increases.
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106. Properties of  and Power
2. For fixed n and values of µ0 and µa, the value of  increases, and the power decreases as the value of  is decreased.
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107. Properties of  and Power
3. For fixed n and values of µ0 and µa, the value of  decreases, and the power increases as the sample size n is increased.
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108. Test of Hypothesis about a Population Variance
6.8
Test of Hypothesis about a Population Variance
:1, 1, 3
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109. © 2011 Pearson Education, Inc
Variance
Although many practical problems involve inferences about a population mean (or proportion), it is sometimes of interest to make an inference about a population variance, 2.
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110. Test of a Hypothesis about 2
One-Tailed Test
H0:  = 0
Ha:  < 0(or Ha:  > 0)
Test statistic:
Rejection region:
(or  >  when Ha:  > 0)
where 0 is the hypothesized variance and the distribution of  is based on (n – 1) degrees of freedom.
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111. Test of a Hypothesis about 2
Two-Tailed Test
H0:  = 0
Ha:  ≠ 0
Test statistic:
Rejection region:
where 0 is the hypothesized variance and the distribution of  is based on (n – 1) degrees of freedom.
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112. Conditions Required for a Valid Hypothesis Test for s2
1. A random sample is selected from the target population.
2. The population from which the sample is selected has a distribution that is approximately normal.
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113. Several 2 probability Distributions
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114. Part of Table VI: Critical Values of Chi Square
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115. 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
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116. 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?
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117. 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 = 1–.05 = .95
 = .05
Reject H0
df = n - 1 = 2
.103
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118. 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.
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119. Chi-Square (2) Test Solution
Ha:
 =
df =
Critical Value(s):
2 = 15
2  15
Test Statistic:
Decision:
Conclusion:
= 33.42
.05
25 – 1 = 24  2
/2 = .025
Do not reject at  = .05
There is no evidence 2 is not 15
39.364
12.401
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120. © 2011 Pearson Education, Inc
Key Ideas
Key Words for Identifying the Target Parameter
 – Mean, Average
p – Proportion, Fraction, Percentage, Rate, Probability
2 – Variance, Variability, Spread
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121. © 2011 Pearson Education, Inc
Key Ideas
Elements of a Hypothesis Test
1. Null hypothesis (H0)
2. Alternative hypothesis (Ha)
3. Test statistic (z, t, or 2)
4. Significance level ()
5. p-value
6. Conclusion
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122. © 2011 Pearson Education, Inc
Key Ideas
Errors in Hypothesis Testing
Type I Error = Reject H0 when H0 is true (occurs with probability )
Type II Error = Accept H0 when H0 is false (occurs with probability )
Power of Test = P(Reject H0 when H0 is false) = 1 – 
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123. © 2011 Pearson Education, Inc
Key Ideas
Forms of Alternative Hypothesis
Lower-tailed : Ha :  < 50
Upper-tailed : Ha :  > 50
Two-tailed : Ha :  ≠ 50
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124. © 2011 Pearson Education, Inc
Key Ideas
Using p-values to Decide
1. Choose significance level ()
2. Obtain p-value of the test
3. If  > p-value, reject H0
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