1. Chapter 9 Hypothesis Testing: Single Population
Statistics for
Business and Economics 8th Edition
Chapter 9
Hypothesis Testing:
Single Population
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What is a Hypothesis?
9.1
A hypothesis is a claim
(assumption) about a
population parameter:
population mean
population proportion
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 = .68
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The Null Hypothesis, H0
States the assumption (numerical) to be tested
Example: The average number of TV sets in U.S. Homes is equal to three ( )
A hypothesis about a parameter that will be maintained unless there is strong evidence against the null hypothesis.
Is always about a population parameter, not about a sample statistic
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4. 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
A null hypothesis is a claim (or statement) about a population parameter that is assumed to be true until it is declared false.

5. The Alternative Hypothesis, H1
Is the opposite of the null hypothesis
e.g., The average number of TV sets in U.S. homes is not equal to 3 ( H1: μ ≠ 3 )
Never contains the “=” , “≤” or “” sign
May or may not be supported
If we reject the null hypothesis, then the second hypothesis, named the “alternative hypothesis” will be accepted.
Is generally the hypothesis that the researcher is trying to support
An alternative hypothesis is a claim about a population parameter that will be true if the null hypothesis is false.
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6. Hypothesis Testing 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
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7. Rejection and Nonrejection Regions
Figure 9.1 Nonrejection and rejection regions for the court case.

8. Four Possible Outcomes for a Test of Hypothesis (Two Types of Errors)
Actual Situation
The person is not guilty
The person is guilty
Court's Decision
Correct Decision
Type 2 error
The person is guilty
Type 1 error
Table: Four possible outcomes for a court case.

9. Two Types of Errors Definition
A Type I error occurs when a true null hypothesis is rejected. The value of α represents the probability of committing this type of error; that is,
α = P(H0 is rejected | H0 is true)
The value of α represents the significance level of the test.
A Type II error occurs when a false null hypotheses is not rejected. The value of β represents the probability of committing a Type II error; that is,
β = P (H0 is not rejected | H0 is false)
The value of 1 – β is called the power of the test. It represents the probability of not making a Type II error.

10. Level of Significance, 
Defines the unlikely values of the sample statistic if the null hypothesis is true
Defines rejection region of the sampling distribution
Is designated by  , (level of significance)
Typical values are .01, .05, or .10
Is selected by the researcher at the beginning
Provides the critical value(s) of the test
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11. Level of Significance and the Rejection Region
Represents
critical value a a
H0: μ = 3 H1: μ ≠ 3 /2 /2
Rejection region is shaded
Two-tail test
H0: μ ≤ 3 H1: μ > 3 a
Definition
A two-tailed test has rejection regions in both tails, a left-tailed test has the rejection region in the left tail, and a right-tailed test has the rejection region in the right tail of the distribution curve.
Upper-tail test
H0: μ ≥ 3 H1: μ < 3 a
Lower-tail test
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12. Signs in H0 and H1 and Tails of a Test
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13. Hypothesis Tests for the Mean
 Known
 Unknown
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14. Test of Hypothesis for the Mean (σ Known)
9.2
Convert sample result ( ) to a z value
Hypothesis
Tests for 
σ Known
σ Unknown
Consider the test
The decision rule is:
(Assume the population is normal)
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Decision Rule
H0: μ = μ0
H1: μ > μ0
Alternate rule: a
Do not reject H0
Reject H0 Z zα μ0
Critical value
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16. Example 1: 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 test:
H0: μ ≤ the average is not over $52 per month
H1: μ > the average is greater than $52 per month
(i.e., sufficient evidence exists to support the
manager’s claim)
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17. Example: Find Rejection Region
(continued)
Suppose that  = .10 is chosen for this test
Find the rejection region:
Reject H0
 = .10
Do not reject H0
Reject H0
1.28
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18. Example: Sample Results
(continued)
Obtain sample and compute the test statistic
Suppose a sample is taken with the following results: n = 64, x = ( = 10 was assumed known)
Using the sample results,
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Example: Decision
(continued)
Reach a decision and interpret the result:
Reject H0
 = .10
Do not reject H0
Reject H0
1.28
z = 0.88
Do not reject H0 since z = 0.88 < 1.28
(So, Accept H0)
i.e.: there is not sufficient evidence that the
mean bill is over $52,
OR, the average bill is not over $52 per month
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One-Tail Tests
In many cases, the alternative hypothesis focuses on one particular direction
This is an upper-tail test since the alternative hypothesis is focused on the upper tail above the mean of 3
H0: μ ≤ 3
H1: μ > 3
This is a lower-tail test since the alternative hypothesis is focused on the lower tail below the mean of 3
H0: μ ≥ 3
H1: μ < 3
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Upper-Tail Tests
H0: μ ≤ 3
H1: μ > 3
There is only one critical value, since the rejection area is in only one tail a
Do not reject H0
Reject H0 zα Z μ
Critical value
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Lower-Tail Tests
H0: μ ≥ 3
H1: μ < 3
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
Critical value
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Two-Tail Tests
In some settings, the alternative hypothesis does not specify a unique direction
H0: μ = 3 H1: μ ¹ 3
/2
/2
There are two critical values, defining the two regions of rejection x 3
Reject H0
Do not reject H0
Reject H0 z
-z/2
+z/2
Lower critical value
Upper
critical value
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24. Assume n=25 and α=0.05. Also assume that the sample mean was 83.
Example 9.1 (B.p-336) Evaluating a new production process (hypothesis test: Upper tail Z test)
The production manager of Northern Windows Inc. has asked you to evaluate a proposed new procedure for producing its Regal line of double-hung windows. The present process has a mean production of 80 units per hour with a population standard deviation of σ = 8. the manager indicates that she does not want to change a new procedure unless there is strong evidence that the mean production level is higher with the new process.
Assume n=25 and α=0.05. Also assume that the sample mean was 83.
What decision would you recommend based on hypothesis testing?
Statistics for Business and Economics, 6e © 2007 Pearson Education, Inc.

25. Example 9.2 Lower Tail Test
The production manager of Twin Forks ball bearing has asked your assistance in evaluating a modified ball bearing production process. When the process is operating properly the process produces ball bearings whose weights are normally distributed with a population mean of 5 ounces and a population standard deviation of 0.1 ounce. The sample mean was and n= 16.
A new raw material supplier was used for a recent production run, and the manager wants to know if that change has resulted in lowering of the mean weight of bearings.
What will be your conclusion for a lower tail test?
Statistics for Business and Economics, 6e © 2007 Pearson Education, Inc.

26. Example 9.3 Two tailed test
The production manager of Circuits Unlimited has asked for your assistance in analyzing a production process. The process involves drilling holes whose diameters are normally distributed with population mean 2 inches and population standard deviation 0.06 inch. A random sample of nine measurements had a sample mean of 1.95 inches.
Use a significance level of 0.05 to determine if the observed sample mean is unusual and suggests that the machine should be adjusted.
Statistics for Business and Economics, 6e © 2007 Pearson Education, Inc.

27. t Test of Hypothesis for the Mean (σ Unknown)
Convert sample result ( ) to a t test statistic
Hypothesis
Tests for 
σ Known
σ Unknown
Consider the test
The decision rule is:
(Assume the population is normal)
Statistics for Business and Economics, 6e © 2007 Pearson Education, Inc.

28. t Test of Hypothesis for the Mean (σ Unknown)
(continued)
For a two-tailed test:
Consider the test
(Assume the population is normal, and the population variance is unknown)
The decision rule is:
Statistics for Business and Economics, 6e © 2007 Pearson Education, Inc.

29. Example 2: Two-Tail Test ( Unknown)
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 = H1: μ ¹ 168
Statistics for Business and Economics, 6e © 2007 Pearson Education, Inc.

30. Example Solution: Two-Tail Test
H0: μ = H1: μ ¹ 168
a/2=.025
a/2=.025
a = 0.05
n = 25
 is unknown, so
use a t statistic
Critical Value:
t24 , .025 = ±
Reject H0
Do not reject H0
Reject H0
t n-1,α/2
-t n-1,α/2
2.0639
1.46
Do not reject H0: not sufficient evidence that true mean cost is different than $168
Statistics for Business and Economics, 6e © 2007 Pearson Education, Inc.

31. Tests of the Population Proportion
Involves categorical variables
Two possible outcomes
“Success” (a certain characteristic is present)
“Failure” (the characteristic is not present)
Fraction or proportion of the population in the “success” category is denoted by P
Assume sample size is large
Statistics for Business and Economics, 6e © 2007 Pearson Education, Inc.

32. Proportions Sample proportion in the success category is denoted by
(continued)
Sample proportion in the success category is denoted by
When nP(1 – P) > 9, can be approximated by a normal distribution with mean and standard deviation
Statistics for Business and Economics, 6e © 2007 Pearson Education, Inc.

33. Hypothesis Tests for Proportions
The sampling distribution of is approximately normal, so the test statistic is a z value:
Hypothesis
Tests for P
nP(1 – P) > 9
nP(1 – P) < 9
Not discussed in this chapter
Statistics for Business and Economics, 6e © 2007 Pearson Education, Inc.

34. Example 3: 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  = .05 significance level.
Check:
Our approximation for P is
= 25/500 = .05
nP(1 - P) = (500)(.05)(.95)
= > 9 
Statistics for Business and Economics, 6e © 2007 Pearson Education, Inc.

35. Z Test for Proportion: Solution
Test Statistic:
H0: P = H1: P ¹ .08
a = .05
n = 500, = .05
Decision:
Critical Values: ±1.96
Reject H0 at  = .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
Statistics for Business and Economics, 6e © 2007 Pearson Education, Inc.

36. Two Procedures Two procedures to make tests of hypothesis
1. The p-value approach
2. The critical-value approach
Prem Mann, Introductory Statistics, 7/E Copyright © 2010 John Wiley & Sons. All right reserved

37. p-Value 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
Decision rule: compare the p-value to 
If p-value <  , reject H0
If p-value   , do not reject H0
Statistics for Business and Economics, 6e © 2007 Pearson Education, Inc.

38. HYPOTHESIS TESTS ABOUT μ: σ KNOWN
Definition
Assuming that the null hypothesis is true, the p-value can be defined as the probability that a sample statistic (such as the sample mean) is at least as far away from the hypothesized value in the direction of the alternative hypothesis as the one obtained from the sample data under consideration.
Note that the p–value is the smallest significance level at which the null hypothesis is rejected.
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39. Figure 9.5 The p–value for a right-tailed test.
Prem Mann, Introductory Statistics, 7/E Copyright © 2010 John Wiley & Sons. All right reserved

40. Figure 9.6 The p–value for a two-tailed test.
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41. Calculating the z Value for x
When using the normal distribution, the value of z for x for a test of hypothesis about μ is computed as follows:
The value of z calculated for x using this formula is also called the observed value of z.
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42. Steps to Perform a Test of Hypothesis Using the p–Value Approach
State the null and alternative hypothesis.
Select the distribution to use.
Calculate the p–value.
Make a decision.
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43. Example 9-1 (p–Value Approach)
At Canon Food Corporation, it used to take an average of 90 minutes for new workers to learn a food processing job. Recently the company installed a new food processing machine. The supervisor at the company wants to find if the mean time taken by new workers to learn the food processing procedure on this new machine is different from 90 minutes.
A sample of 20 workers showed that it took, on average, 85 minutes for them to learn the food processing procedure on the new machine. It is known that the learning times for all new workers are normally distributed with a population standard deviation of 7 minutes.
Find the p–value for the test that the mean learning time for the food processing procedure on the new machine is different from 90 minutes. What will your conclusion be if α = .01?
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44. Example 9-1: Solution Step 1: H0: μ = 90 H1: μ ≠ 90
Step 2: The population standard deviation σ is known, the sample size is small (n < 30), but the population distribution is normal. We will use the normal distribution to find the p–value and make the test.
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45. Example 9-1: Solution Step 3:
see 3.19 from z table, we find so =
p-value = 2(.0007) = .0014
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46. Figure 9: The p-value for a two-tailed test.
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47. Example 9-1: Solution
Step 4: Because α = .01 is greater than the p-value of .0014, we reject the null hypothesis at this significance level.
Therefore, we conclude that the mean time for learning the food processing procedure on the new machine is different from 90 minutes.
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