Business Statistics For Contemporary Decision Making 8th Edition Black
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Chapter 9 Statistical Inference: Hypothesis Testing for Single Populations
LEARNING OBJECTIVES
The main objective of Chapter 9 is to help you to learn how to test Hypotheses on single populations, thereby enabling you to:
1. Develop both one- and two-tailed null and alternative hypotheses that can be tested in a business setting by examining the rejection and non-rejection regions in light of Type I and Type II errors.
2. Reach a statistical conclusion in hypothesis testing problems about a population mean with a known population standard deviation using the z statistic.
3. Reach a statistical conclusion in hypothesis testing problems about a population mean with an unknown population standard deviation using the t statistic.
4. Reach a statistical conclusion in hypothesis testing problems about a population proportion using the z statistic.
5. Reach a statistical conclusion in hypothesis testing problems about a population variance using the chi-square statistic.
6. Solve for possible Type II errors when failing to reject the null hypothesis.
CHAPTER TEACHING STRATEGY
For some instructors, this chapter is the cornerstone of the first statistics course. Hypothesis testing presents the logic in which ideas, theories, etc., are scientifically examined. The student can be made aware that much of the development of concepts to this point including sampling, level of data measurement, descriptive tools such as mean and standard deviation, probability, and distributions pave the way for testing hypotheses. Often students (and instructors) will say "Why do we need to test this hypothesis when we can make a decision by examining the data?" Sometimes it is true that examining the data could allow hypothesis decisions to be made. However, by using the methodology and structure of hypothesis testing even in "obvious" situations, the researcher has added credibility and rigor to his/her findings. Some statisticians actually report findings in a court of law as an expert witness. Others report their findings in a journal, to the public, to the corporate board, to a client, or to their manager. In each case, by using the hypothesis testing method rather than a "seat of the pants" judgment, the researcher stands on a much firmer foundation by using the principles of hypothesis testing and random sampling. Chapter 9 brings together many of the tools developed to this point and formalizes a procedure for testing hypotheses.
The statistical hypotheses are set up as to contain all possible decisions. The two-tailed test always has = and in the null and alternative hypothesis. One-tailed tests are presented with = in the null hypothesis and either > or < in the alternative hypothesis. If in doubt, the researcher should use a two-tailed test. Chapter 9 begins with a two-tailed test example. Often that which the researcher wants to demonstrate true or prove true is set up as the alternative hypothesis. The null hypothesis is that the new theory or idea is not true, the status quo is still true, or that there is no difference. The null hypothesis is assumed to be true before the process begins. Some researchers liken this procedure to a court of law where the defendant is presumed innocent (assume null is true - nothing has happened). Evidence is brought before the judge or jury. If enough evidence is presented, then the null hypothesis (defendant innocent) can no longer be accepted or assume true. The null hypothesis is rejected as not true and the alternate hypothesis is accepted as true by default. Emphasize that the researcher needs to make a decision after examining the observed statistic.
Some of the key concepts in this chapter are one-tailed and two-tailed test and Type I and Type II error. In order for a one-tailed test to be conducted, the problem must include some suggestion of a direction to be tested. If the student sees such words as greater, less than, more than, higher, younger, etc., then he/she knows to use a one-tail test. If no direction is given (test to determine if there is a "difference"), then a two-tailed test is called for. Ultimately, students will see that the only effect of using a one-tailed test versus a two-tailed test is on the critical table value. A one-tailed test uses all of the value of alpha in one tail. A two-tailed test splits alpha and uses alpha/2 in each tail thus creating a critical value that is further out in the distribution. The result is that (all things being the same) it is more difficult to reject the null hypothesis with a two-tailed test. Many computer packages such as Minitab include in the results a p-value. If you designate that the hypothesis test is a two-tailed test, the computer will double the p-value so that it can be compared directly to alpha.
In discussing Type I and Type II errors, there are a few things to consider. Once a decision is made regarding the null hypothesis, there is a possibility that the decision is correct or that an error has been made. Since the researcher virtually never knows for certain whether the null hypothesis was actually true or not, a probability of committing one of these errors can be computed. Emphasize with the students that a researcher can never commit a Type I error and a Type II error at the same time. This is so because a Type I error can only be committed when the null hypothesis is rejected and a Type II error can only be committed when the decision is to not reject the null hypothesis. Type I and Type II errors are important concepts for managerial students to understand even beyond the realm of statistical hypothesis testing. For example, if a manager decides to fire or not fire an employee based on some evidence collected, he/she could be committing a Type I or a Type II error depending on the decision. If the production manager decides to stop the production line because of evidence of faulty raw materials, he/she might be committing a Type I error. On the other hand, if the manager fails to shut the production line down even when faced with evidence of faulty raw materials, he/she might be committing a Type II error.
The student can be told that there are some widely accepted values for alpha (probability of committing a Type I error) in the research world and that a value is usually selected before the research begins. On the other hand, since the value of Beta (probability of committing a Type II error) varies with every possible alternate value of the parameter being tested, Beta is usually examined and computed over a range of possible values of that parameter. As you can see, the concepts of hypothesis testing are difficult and represent higher levels of learning (logic, transfer, etc.). Student understanding of these concepts will improve as you work your way through the techniques in this chapter and in chapter 10.
CHAPTER OUTLINE
9.1 Introduction to Hypothesis Testing Types of Hypotheses
Research Hypotheses
Statistical Hypotheses
Substantive Hypotheses
Eight-Step Process for Testing Hypotheses
Rejection and Non-rejection Regions
Type I and Type II errors
Comparing Type I and Type II Errors
9.2 Testing Hypotheses About a Population Mean Using the z Statistic (known) An Example Using the Eight-Step Approach
Using the p-Value to Test Hypotheses
Testing the Mean with a Finite Population
Using the Critical Value Method to Test Hypotheses
Using the Computer to Test Hypotheses about a Population Mean Using the z Statistic
9.3 Testing Hypotheses About a Population Mean Using the t Statistic (unknown) Using the Computer to Test Hypotheses about a Population Mean Using the t Test
9.4 Testing Hypotheses About a Proportion Using the Computer to Test Hypotheses about a Population Proportion
9.5 Testing Hypotheses About a Variance
9.6 Solving for Type II Errors
Some Observations About Type II Errors
Operating Characteristic and Power Curves
Effect of Increasing Sample Size on the Rejection Limits
KEY TERMS
Alpha()
Alternative Hypothesis
Beta()
Critical Value
Critical Value Method
Hypothesis
Hypothesis Testing
Level of Significance
Nonrejection Region
Null Hypothesis
Observed Significance Level
Observed Value
One-tailed Test
Operating-Characteristic Curve (OC)
p-Value
Power
Power Curve
Rejection Region
Research Hypothesis
Statistical Hypothesis
Substantive Result
Two-Tailed Test
Type I Error
Type II Error
SOLUTIONS TO PROBLEMS IN CHAPTER 9
9.1 a) Two-Tailed
b) One-Tailed
c) One-Tailed
d) Two-Tailed
9.2 a) Correct Decision
b) Type II error
c) Correct Decision
d) Type I error
Reject the null hypothesis
b) from Table A.5, inside area between z = 0 and z = 2.77 is .4972
p-value = .5000 - .4972 = .0028
Since the p-value of .0028 is less than /2 = .005, the decision is to: Reject the null hypothesis
c) critical mean values:
Fail to reject the null hypothesis
b) Probability > observed z = 1.59 is .5000 - .4441 = .0559 (the p-value) which is less than = .10.
c) Critical mean value:
Since the observed x = 1,215 is greater than the critical x = 1212.04, the decision is to reject the null hypothesis.
Statistically, we can conclude that urban air soot is significantly lower. From a business and community point-of-view, assuming that the sample result is representative of how the air actually is now, is a reduction of suspended particles from 82 to 78.125 really an important reduction in air pollution (is it substantive)? Certainly it marks an important first step and perhaps a significant start. Whether or not it would really make a difference in the quality of life for people in the city of St. Louis remains to be seen. Most likely, politicians and city chamber of commerce folks would jump on such results as indications of improvement in city conditions.
the decision is to fail to reject the null hypothesis.
Since the observed z = 2.64 > z 01 = 2.33, the decision is to Reject the null hypothesis
The table value for z = 2.64 is .4959. The p-value is .5000 - .4959 = .0041. Since this is less than = .01, the decision using the p-value is to reject the null hypothesis.
Whether or not the result is substantive is somewhat dependent on the business decision maker. However, a difference of 2.4 years from the value from just a couple of years ago would seem to be considerable and therefore this may be a substantive increase.
9.11 Ho: µ = $50 Ha: µ > $50
x = $52.17 n = 23 = $3.49 = .10
For one-tailed test, = .10, z.10 = 1.28
Since the observed z = 2.98 > z 10 = 1.28, the decision is to Reject the null Hypothesis.
The table value for z = 2.98 is .4986. The p-value is .5000 - .4986 = .0014. Since this is less than = .10, the decision using the p-value is to reject the null hypothesis.
In our decision to reject the hypothesized mean of $50, we are saying that we have concluded that the mean is greater than $50. However, our sample mean is only $2.17 more than $50. While this may be an indication of carpet cleaning inflation, for many customers, an additional $2.17 may not be substantial nor cause them to forgo the cleaning.
9.12 Ho: µ = 123
Ha: µ > 123
= .05 n = 40 40 people were sampled x = 132.36 = 27.68
This is a one-tailed test. Since the p-value = .019, we reject the null hypothesis at = .05. The average water usage per person is greater than 123 gallons.
The decision is to Fail to reject the null hypothesis
The decision is to Fail to reject the null hypothesis
Two-tailed
The decision is to fail to reject the null hypothesis.
Since the Excel p-value = .256 > /2 = .025 and MINITAB p-value =.513 > .05, the decision is to fail to reject the null hypothesis. She would not conclude that her city is any different from the ones in the national survey.
9.24
9.25
The decision is to Fail to reject the null hypothesis
The decision is to Fail to reject the null hypothesis
The decision is to Fail to reject the null hypothesis
p-Value Method:
z = -0.60
from Table A.5, area = .2257
Area in tail = .5000 - .2257 = .2743 which is the p-value
Since the p-value = .2743 > /2 = .025, the decision is to Fail to reject the null hypothesis
Solving
.257 and .323 are the critical values
Since p ˆ = .28 is not outside critical values in tails, the decision is to Fail to reject the null hypothesis
Since the observed z = -1.89 is greater than z.005= -2.575, The decision is to fail to reject the null hypothesis. There is not enough evidence to declare that the proportion is any different than .48. 9.28
Since the observed z = -3.00 is less than z.01= -2.33, The decision is to reject the null hypothesis.
Since the observed z = 1.23 is less than z.005= 1.645, The decision is to fail to reject the null hypothesis. There is not enough evidence to declare that the proportion is any different than .31.
Since the observed z = 1.34 is less than z.05= 1.645, The decision is to fail to reject the null hypothesis. There is not enough evidence to declare that the proportion is less than .24.
Since the observed
= 2.02 is less than z.01= 2.33, The decision is to fail to reject the null hypothesis. There is not enough evidence to declare that the proportion is greater than .18.
decision is to reject the null hypothesis.
9.32 Ho: p = .47
Ha: p .47
n = 67 x = 40 = .05 /2 = .025
For a two-tailed test, z.025 = +1.96
Since the observed z = 2.08 is greater than z.025= 1.96, The decision is to reject the null hypothesis.
Ha: 2 > 20
2 .05,14 = 23.6848
2 = 20 1)(32) (15 = 22.4
Since 2 = 22.4 < 2 .05,14 = 23.6848, the decision is to fail to reject the null hypothesis.
2 = 8.5 1)(17) (22 = 42
Since 2 = 42 > 2 .05,21 = 32.6706, the decision is to reject the null hypothesis
45 2 .01,7 = 18.4753
2 = 45 1)(4.12) (8 2 = 2.64
Since 2 = 2.64 < 2 .01,7 = 18.4753, the decision is to fail to reject the null hypothesis.
= 14 0833) 1)(30 (12 = 23.64
= 21.9200, the decision is to reject the null hypothesis
2 .01,15 = 30.5780
2 = .001 1)(.00144667) (16 = 21.7
Since 2 = 21.7 < 2 .01,15 = 30.5780, the decision is to fail to reject the null hypothesis
Since 2 = 49.93 > 2 .05,12 = 21.0261, the decision is to reject the null hypothesis. The variance has changed.
16.8119, the decision is to reject the null hypothesis
from Table A.5, area for z = -1.84 is .4671 = .4671 + .5000 = .9671
d) As gets smaller (other variables remaining constant), gets larger. Decreasing the probability of committing a Type I error increases the probability of committing a Type II error if other variables are held constant.
from Table A.5, area for z = -0.90 is .3159
= .3159 + .5000 = .8159
b)
d)
= .0636 + .5000 = .5636
= .5000 - .1331 = .3669
e) As the alternative value gets farther from the null hypothesized value, the probability of committing a Type II error reduces (all other variables being held constant).
Since this is two-tailed, /2 = .005 z.005 = ±2.575
from Table A.5 for z = 0.04, area = .0160
-5.11 Area associated with z = -5.11 is .5000
= .5000 + .0160 = .5160
from Table A.5, the area for z = -4.14 is .5000 = .5000 - .5000 = .0000
from Table A.5, the area for z = 1.08 is .3599
= .5000 + .3599 = .8599
Power = 1 - = 1 - .8599 = .1401
For
From Table A.5, the area for z = 0.21 is .0832
= .5000 + .0832 = .5832
Power = 1 - = 1 - .5832 = .4168
For 47 years:
z = 58 7 8 47 46.239 = -0.67
From Table A.5, the area for z = -0.67 is .2486
= .5000 - .2486 = .2514
Power = 1 - = 1 - .2514 = .7486
For 48 years:
z = 58 7 8 48 46.239 = 1.54
From Table A.5, the area for z = 1.54 is .4382
= .5000 - .4382 = .0618
Power = 1 - = 1 - .0618 = .9382
Since the observed z = -0.48 > z.10 = -1.28, the decision is to fail to reject the null hypothesis
Type II error: Solving for the critical proportion,
p ˆ = .683
For pa = .69
.69 .683 = -0.33
z = 463 (.69)(.31)
From Table A.5, the area for z = -0.33 is .1293
The probability of committing a Type II error = .1293 + .5000 = .6293
For pa = .66
66 683 = 1.04
z = 463 (.66)(.34)
From Table A.5, the area for z = 1.04 is .3508
The probability of committing a Type II error = .5000 - .3508 = .1492
For pa = .60
60 683 = 3.65
z = 463 (.60)(.40)
From Table A.5, the area for z = 3.65 is essentially, .5000
The probability of committing a Type II error = .5000 - .5000 = .0000
9.44 8 steps:
1) Ho: µ = 36
x
2) z = n
3) = .01
4) two-tailed test, /2 = .005, z.005 = + 2.575 If the observed value of z is greater than 2.575 or less than -2.575, the decision will be to reject the null hypothesis.
5) n = 63, x = 38.4, = 5.93
7) Since the observed value of z = 3.21 is greater than z.005 = 2.575, the decision is to reject the null hypothesis.
8) The mean is likely to be greater than 36.
9.45 8 steps:
1) Ho: µ = 7.82 Ha: µ < 7.82
2) The test statistic is
= n s x 3) = .05
4) df = n - 1 = 16, t.05,16 = -1.746. If the observed value of t is less than -1.746, then the decision will be to reject the null hypothesis.
5) n = 17 x = 7.01 s = 1.69
7) Since the observed t = -1.98 is less than the table value of t = -1.746, the decision is to reject the null hypothesis
8) The population mean is significantly less than 7.82.
9.46 8 steps:
3)
4) This is a one-tailed test, z.10 = 1.28. If the observed value of z is greater than 1.28, the decision will be to reject the null hypothesis.
5) n = 783 x = 230
7) Since z = 0.85 is less than z.10 = 1.28, the decision is to fail to reject the null hypothesis
8) There is not enough evidence to declare that p > .28.
b.
1) Ho: p = .61
Ha: p .61
ˆ
2) z = n q p p p
3) = .05
4) This is a two-tailed test, z.025 = + 1.96. If the observed value of z is greater than 1.96 or less than -1.96, then the decision will be to reject the null hypothesis.
5) n = 401 p ˆ = .56
6) z = 401 (.61)(.39) .61 .56 = -2.05
7) Since z = -2.05 is less than z.025 = -1.96, the decision is to reject the null hypothesis.
8) The population proportion is not likely to be .61.
9.47 8 steps:
7) Since the observed 2 = 32.675 is less than 33.4087, the decision is to fail to reject the null hypothesis.
8) The population variance is not significantly more than 15.4.
from table A.5, area for z = -1.28 is .3997
= .5000 - .3997 = .1003 b)
Since
9.55 H0: 2 = 4 n = 8 s = 7.80 = .10 df = 8 – 1 = 7
Ha: 2 > 4
2 .10,7 = 12.0170
2 = 4 1)(7.80) (8 2 = 106.47
Since observed 2 = 106.47 > 2 .10,7 = 12.017, the decision is to reject the null hypothesis.
9.56 H0: p = .46 Ha: p > .46
Using a one-tailed test, z.05 = 1.645 z = 125
Since the observed value of z = 1.53 < z.05 = 1.645, the decision is to fail to reject the null hypothesis Solving
from Table A.5, the area for z = 0.74 is .2704 = .5000 + .2704 = .7704
9.57 n = 16 x = 175 s = 14.28286 df = 16 - 1 = 15 = .05
H0: µ = 185
Ha: µ < 185
t.05,15 = - 1.753
t = n
= 16 28286 14 185 175 = -2.80
Since observed t = - 2.80 < t.05,15 = - 1.753, the decision is to reject the null hypothesis
9.58 H0: p = .16
Ha: p > .16
n = 428 x = 74 = .01 ��= �� �� = 74 428 = .1729
For a one-tailed test, z.01 = 2.33
Since the observed z = 0.73 < z.01 = 2.33, the decision is to fail to reject the null hypothesis. The probability of committing a Type I error is .01.
To determine the probability of a Type II error:
from Table A.5, the areas for z = 0.48 and z = 4.41 are .1844 and .5000
= .5000 - .1844 = .3156
The upper tail has no effect on .
= .05 df = 11 two-tailed test, /2 = .025
2 .025,11 = 21.9200
2 ..975,11 = 3.81574
If the observed 2 is greater than 21.9200 or less than 3.81574, the decision is to reject the null hypothesis.
Since the observed 2 = 45.866 is greater than 2 .025,11 = 21.92, the decision is to reject the null hypothesis. The population variance is significantly more than
9.64
9.65 The sample size is 22. x is 3.969 s = 0.866 df = 21
The test statistic is:
The observed t = -2.33. The p-value is .015.
The results are statistical significant at = .05.
The decision is to reject the null hypothesis.
9.66 H0: p = .25
Ha: p .25
This is a two-tailed test with = .05. n = 384.
Since the p-value = .045 < = .05, the decision is to reject the null hypothesis.
The sample proportion, p ˆ = .205729 which is less than the hypothesized p = .25. One conclusion is that the population proportion is lower than .25.