chapter 9: statistical inference: hypothesis testing for single populations

Chapter 9:Statistical Inference:

Hypothesis Testing for Single Populations

LO1 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.

LO2 Reach a statistical conclusion in hypothesis testing problems about a population mean with a known population standard deviation using the z statistic.

LO3 Reach a statistical conclusion in hypothesis testing problems about a population mean with an unknown population standard deviation using the t statistic.

Learning Objectives

LO4 Reach a statistical conclusion in hypothesis testing problems about a population proportion using the z statistic.

LO5 Reach a statistical conclusion in hypothesis testing problems about a population variance using the chi-square statistic.

LO6 Solve for possible Type II errors when failing to reject the null hypothesis.

Learning Objectives

• It is one of the most important mechanisms for decision making

• It reduces the risk of intuitive guessing by using evidence based decision making

• Hypothesis testing is central to inferential statistics• With this tool business researchers can structure problems in

order to be able to use statistical evidence to test various theories about business phenomena

• It is widely used in business to determine if production processes are functioning efficiently and effectively (or are out of control); or if new management approaches are making a significantly positive difference

Importance of Hypothesis Testing in Decision Making

• Research Hypothesis– A statement of what the researcher believes will be the outcome of an

experiment or a study

• Statistical Hypotheses– A formal structure used to scientifically test the research hypothesis

• Substantive Hypotheses– Conclusion reached on the basis of the data obtained and used in the

study. – It may merely be a statistical outcome strongly and logically suggested

by the data available and the technique (models) used.– Thus, a statistically significant difference may not imply a material, or

substantive difference for a business situation.

• Warning– Since what is mathematically significant may not be so in the business sense,

business decision makers must exercise care in interpreting statistical results.

Types of Hypotheses

• Older workers are more loyal to a company.

• Companies with more than $1 billion of assets spend a higher percentage of their annual budget on advertising than do companies with less than $1 billion of assets.

• The price of scrap metal is a good indicator of the industrial production index six months later.

Example of Research Hypotheses

HTAB System to Test Hypotheses

Step 1. Establish a null and an alternative hypothesis.Step 2. Determine the appropriate statistical test.Step 3. Set the value of alpha, the level of significance.Step 4. Establish the decision rule.Step 5. Gather sample data.Step 6. Analyze the data.Step 7. Reach a statistical conclusion.Step 8. Make a business decision.

Statistical Procedures for Hypothesis Testing

• Type I Error– Committed by a decision to reject a true null hypothesis– This can happen when by chance the sample is not representative of

the population – The probability of committing a Type I error is called α (alpha), the

level of significance.

• Type II Error– Committed by failing to reject a false null hypothesis– In this case the null hypothesis is really false but a decision is made

not to reject it. – The probability of committing a Type II error is called (beta).

Type I and Type II Errors

Decision Table for Hypothesis Testing

Step 1: Establish a null and an alternative hypothesis.

– A Null hypothesis: The null condition assumed to be true.– An Alternative Hypothesis: Declares something new is

happening.

– Notation• Null: H0

• Alternative: Ha

Statistical Hypotheses

• The Null and Alternative Hypotheses are mutually exclusive. Only one of them can be true.

• The Null and Alternative Hypotheses are collectively exhaustive. They are stated to include all possibilities. (An abbreviated form of the null hypothesis is often used.)

• The Null Hypothesis is assumed to be true.

• The burden of proof falls on the Alternative Hypothesis.

Null and Alternative Hypotheses

• Note that the Null Hypothesis (H0) always has an equal sign in the algebraic form of the expression

• The Alternative Hypothesis (Ha) always has the algebraic sign of less than, greater than or not equal

Examples:

Algebraic Expressions of the Null Hypothesis

H0: µ ≤ 2Ha: µ > 2

H0: µ = 2Ha: µ ≠ 2

H0: µ ≥ 2Ha: µ ˂ 2

or or

• A manufacturer is filling 2 kg packages with flour.

• The company wants the package contents to average 2 kilograms.

Null and Alternative Hypotheses: Example

• Because of an increased marketing effort, company officials believe the company’s market share is now greater than 18%, and the officials would like to prove it.

Null and Alternative Hypotheses: Example

HTAB Paradigm: Task 2

• Conceptually and graphically, statistical outcomes that result in the rejection of the null hypothesis lie in what is termed the rejection region.

• Statistical outcomes that fail to result in the rejection of the null hypothesis lie in what is termed the nonrejection region.

Rejection and Nonrejection Regions

• There are three possibilities which can be stipulated in the alternative hypothesis.

• The three possibilities are: >, <, or ≠.

• The rejection regions for these possibilities, if a standard normal distribution is used for the test statistic, are shown on the next slide.

Possible Rejection and Nonrejection Regions


Rejection regionfor hypothesisthat involve thestandard normal distribution andthe > symbol (right–tailed test)


Rejection regionfor hypothesisthat involve thestandard normal distribution andthe < symbol (left–tailed test)


Rejection regionfor hypothesisthat involve thestandard normal distribution andthe symbol (two–tailed test)

Example:

•A survey of chartered accountants, done more than 15 years ago, found that their average salary was $74,914. An accounting researcher would like to test whether this average has changed over the years. A sample of 112 CAs produced a mean salary of $78,695. Assume that the population standard deviation of salaries = $14,530.

Testing Hypotheses about a Population Mean Using the z Statistic ( Known)

• Step 1: Hypothesize

• Step 2: Test


• Step 3: Specify the Type I error rate

– = 0.05 z/2 = 1.96

• Step 4: Establish the decision rule

– Reject H0 if the test statistic < -1.96 or it the test statistic > 1.96.


• Step 5: Gather sample data

x-bar = $78,695, n = 112, = $14,530, hypothesized = $74,914.

• Step 6: Compute the test statistic.


• Step 7: Reach a statistical conclusion

– Since z = 2.75 > 1.96, reject H0.

• Step 8: Business decision

– Statistically, the researcher has enough evidence to reject the figure of $74,914 as the true average salary for CAs. In addition, based on the evidence gathered, it may suggest that the average has increased over the 10-year period.


• Test statistic:

Testing Hypotheses about a Population Mean Using the z Statistic ( Known) from a Finite Population

• Another way to reach a statistical conclusion in hypothesis testing problems is to use the p-value, sometimes referred to as the observed significance level.

p-value < , reject H0

p-value , do not reject H0

Using the p-Value to Test Hypotheses

• One should be careful when using p-values from statistical software outputs.

• Both Minitab and Excel report the actual p-values for hypothesis tests.

• Minitab doubles the p-value for a two-tailed test so you can compare with .

• Excel does not double the p-value for a two-tailed test. So when using the p-value from Excel, you may multiply the value by 2 and then compare with .


Demonstration Problem: Minitab

• The critical value method determines the critical mean value required for z to be in the rejection region and uses it to test the hypotheses.

Critical Value Method to Test Hypotheses

• For the previous example,


• Thus, a sample mean greater than $77,605 or less than $72,223 will result in the rejection of the null hypothesis.

• This method is particularly attractive in industrial settings where standards can be set ahead of time and then quality control technicians can gather data and compare actual measurements of products to specifications.


• In this case, the test statistic will be

Testing Hypotheses about a Population Mean Using the t Statistic ( Unknown)

Example: Masses in Kilograms of a Sample of 20 Plates

22.6 22.2 23.2 27.4 24.527.0 26.6 28.1 26.9 24.926.2 25.3 23.1 24.2 26.125.8 30.4 28.6 23.5 23.6

Two-tailed Test: Unknown, = .05

Two-tailed Test: Unknown, = .05

-2.093 < t= 1.04 <2.093, therefor the statistical conclusion is to fail to reject the null hypothesis. It appears that the machine is not out of control.

MINITAB Computer Printout for the Machine Plate Example

Machine Plate Example: Excel(Part 1)

Machine Plate Example: Excel(Part 2)

z Test of Population Proportion

• A manufacturer believes exactly 8% of its products contain at least one minor flaw. Suppose the company wants to test this belief. A sample of 200 products resulted in 33 items have at least one minor flaw. Use a probability of a Type I error of 0.10.

z Test of Population Proportion

Testing Hypotheses about a Proportion: Manufacturer Example (Part 2)

Minitab Computer Printout for the Minor Flaw Example

Using the Critical Value Method

• The test statistic for this test is

Testing Hypotheses About a Variance

Testing Hypotheses About a Variance: Demonstration Problem 9.4

A small business has 37 employees. Because of the uncertain demand for its product, the company usually pays overtime in any given week. The company assumed that about 50 total hours of overtime per week are required and that the variance on this figure is about 25. Company officials want to know whether the variance of overtime hours has changed. Given here is a sample of 16 weeks of overtime data (in hours per week). Assume hours of overtime are normally distributed. Use these data to test the null hypothesis that the variance of overtime data is 25. Let α = 0.10.

57 56 52 44

46 53 44 44

48 51 55 48

63 53 51 50

• Step 1:

• Step 2: Test statistic


• Step 3: Because this is a two-tailed test, = 0.10 and /2 = 0.05.

• Step 4: The degrees of freedom are 16 – 1 = 15. The two critical chi-square values are 2

(1 – 0.05), 15 = 2 0.95, 15 = 7.26093

and 2 0.05, 15 = 24.9958.

• Step 5: The data are listed in the text & previous slide.

• Step 6: The sample variance is s2 = 28.06. The observed chi-square value is calculated as 2 = 16.84.


• Step 7: The observed chi-square value is in the non-rejection region because

• 2 0.05, 15 = 7.26094 < 2

observed = 16.84 < 2 0.05, 15 = 24.9958

• Step 8: This result indicates to the company managers that the variance of weekly overtime hours is about what they expected.


• A Type II error can be committed only when the researcher fails to reject the null hypothesis and the null hypothesis is false.

• A Type II error, β, varies with possible values of the alternative parameter.

• Often, when the null hypothesis is false, the value of the alternative mean is unknown, so the researcher will compute the probability of committing Type II errors for several possible values. How can the probability of committing a Type II error be computed for a specific alternative value of the mean?

Solving for Type II Errors

• Suppose a test is conducted on the following hypotheses: H0: = 12 ounces vs. Ha: < 12 g when the sample size is 60 with mean of 11.985 and a standard deviation of 0.10g.

• The first step in determining the probability of a Type II error: calculate a critical value for the sample mean for this case.

• For =0.05, the critical value for the sample mean is

Solving for Type II Errors for Small Machine Parts Example

Graphic for Type II Error:Machine Parts Example

• The decision to accept or reject the null hypothesis is being made on the assumption that the critical region is to the left of z = - 1.645 where µ = 12, σ = 10.

• However, the distribution of the population from which the samples come may be different.

Solving for Type II Errors: Small Machine Parts

• Assume that the alternative mean µ = 11.99 . Since a Type II error, , varies with possible values of the alternative parameter, then for an alternative mean of 11.99 (< 12) the corresponding z-value is -0.85

• The probability of committing a Type II error is all the area to the right of 11.979 in distribution (b), or 0.3023 + 0.5000 = 0.8023.

• Hence, there is an 80.23% chance of committing a Type II error if the alternative mean is 11.99 g.

Solving for Type II Errors: Small Machine Parts

• Let the alternative mean now be 11.96, to the left of 11.979.

• Note that the new Critical Z1 is 1.47 and that β = 0.0798 < the previous value of β = 0.8023.

Type II Error When the Difference Between the Alternative Mean and the Hypothesized Mean Increases

Demonstration Problem 9.5.

• Thus the probability of committing a type II error is only 0.0708.

• Suppose you are conducting a two-tailed hypothesis test of proportions. The null hypothesis is that the population proportion is 0.40. The alternative hypothesis is that the population proportion is not 0.40. A random sample of 250 produces a sample proportion of 0.44. With alpha of 0.05, the table z value for α/2 is ±1.96. The observed z from the sample information is:

Demonstration Problem 9.6 Proportions Example

The critical values are 0.34 on the lower end and 0.46 on the upper end.

Demonstration Problem 9.6 Proportions Example

• Suppose the alternative population proportion is really 0.36. What is the probability of committing a Type II error?

• Solving for the area between

• and p1 = 0.36 yields:

• The area to the right of -0.66 is 0.2454 + 0.4995 = 0.7449, the probability of committing a type II error.

• Because the probability of committing a Type II error changes for each possible different value of the alternative parameter, it is best to examine a series of possible alternative values.

• The power of a test is (1 - ) It is the probability of rejecting the null hypothesis when it is false.

• The series of values of (1-β) resulting from changes in the alternative value of the parameter are called the power curve

• The series of values of resulting from the changes in the alternate value of the parameters are called the operating curve.

Operating Characteristic and Power Curve

Operating Characteristic and Power Curves for Small Parts Machines Example

Note that as the alternative means get closer to the hypothesized mean of µ =12g that the probability of a type II error is greater: it is difficult to discriminate between distributions with means of 12 and 11.999. The reverse is true when the alternative mean is vey different, for example 11.95. The calculations below show power curves and operating curves for µ = 12g and α = 0.05.

Operating Characteristic Curve Small Parts Machines Problem

12.0011.9911.9811.9711.9611.95

0.9

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

0.0

Alternative Means

Type II Er

ror

The Power Curve for Small Parts Machines Problem

12.0011.9911.9811.9711.9611.95

1.0

0.8

0.6

0.4

0.2

0.0

Alternative Means

Pow

er

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chapter 9: statistical inference: hypothesis testing for single populations

Documents

statistical conclusion

hypothesis testing problems

statistical evidence

statistical outcome

statistical inference

statistical results

statistical procedures

alternative hypothesis