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Statistics

Hypothesis Tests

- Developing Null and Alternative Hypotheses
- Type I and Type II Errors
- Population Mean: σ Known
- Population Mean: σUnknown
- Population Proportion
- Hypothesis Testing and Decision Making
- Calculating the Probability of Type II Errors
- Determining the Sample Size for
- Hypothesis Tests About a Population Mean

- Population Proportion
- Hypothesis Testing and Decision Making
- Calculating the Probability of Type II Errors
- Determining the Sample Size for
- Hypothesis Tests About a Population Mean

- John Morrell & Company is considered
the oldest continuously operating meat

manufacturer in the United States.

- Market research at Morrell provides management with up to- date information on the company’s various products and how the products compare with competing brands of similar products.

- One research question concerned whether Morrell’s Convenient Cuisine Beef Pot Roast was the preferred choice of more than 50% of the consumer population. The hypothesis test for the research question is Ho: p ≤ .50 vs. Ha: p > .50
- In this chapter we will discuss how to formulate hypotheses and how to conduct tests like the one used by Morrell.

- Hypothesis testing can be used to determine
- whether a statement about the value of a
- population parameter should or should not be
- rejected.

- The null hypothesis, denoted by Ho , is a
- tentative assumption about a population
- parameter.

- The alternative hypothesis, denoted by Ha,
- is the opposite of what is stated in the null
- hypothesis.

- The alternative hypothesis is what the test is
attempting to establish.

- Testing Research Hypotheses
- Testing the Validity of a Claim
- Testing in Decision-Making Situations

- Testing Research Hypotheses

- The research hypothesis should be expressed
- as the alternative hypothesis.

- The conclusion that the research hypothesis
- is true comes from sample data that contradict
- the null hypothesis.

- Testing Research Hypotheses

Example:

Consider a particular automobile model that

currently attains an average fuel efficiency of 24 miles per gallon.

A product research group developed a new

fuel injection system specifically designed to

increase the miles-per-gallon rating.

The appropriate null and alternative hypotheses

for the study are:

Ho: μ ≤ 24

Ha: μ > 24

- Testing Research Hypotheses

- In research studies such as these, the null and
- alternative hypotheses should be formulated
- so that the rejection of H0 supports the research conclusion. The research hypothesis therefore should be expressed as the alternative hypothesis.

- Testing the Validity of a Claim

- Manufacturers’ claims are usually given the
- benefit of the doubt and stated as the null
- hypothesis.

- The conclusion that the claim is false comes
- from sample data that contradict the null
- hypothesis.

- Testing the Validity of a Claim

Example:

Consider the situation of a manufacturer of soft

drinks who states that it fills two-liter containers

of its products with an average of at least 67.6

fluid ounces.

- Testing the Validity of a Claim

A sample of two-liter containers will be selected,

and the contents will be measured to test the

manufacturer’s claim.

The null and alternative hypotheses as follows.

Ho: μ ≥ 67. 6

Ha: μ < 67. 6

- Testing the Validity of a Claim

- In any situation that involves testing the validity
- of a claim, the null hypothesis is generally based
- on the assumption that the claim is true. The
- alternative hypothesis is then formulated so that
- rejection of H0will provide statistical evidence
- that the stated assumption is incorrect.
- Action to correct the claim should be considered
- whenever H0 is rejected.

- Testing in Decision-Making Situations

- A decision maker might have to choose
- betweentwo courses of action, one
- associated with the null hypothesis and
- another associated with the alternative
- hypothesis.

- Testing in Decision-Making Situations

Example:

Accepting a shipment of goods from a supplier

or returning the shipment of goods to the supplier.

Assume that specifications for a particular part

require a mean length of two inches per part.

The null and alternative hypotheses would be

formulated as follows.

H0 : μ = 2

Ha: μ≠ 2

- Testing in Decision-Making Situations

- If the sample results indicate H0 cannot be rejected and the shipment will be accepted.
- If the sample results indicate H0 should be rejected the parts do not meet specifications. The quality control inspector will have sufficient evidence to return the shipment to the supplier.

- Testing in Decision-Making Situations

- We see that for these types of situations, action is taken both when H0cannot be rejected and when H0 can be rejected.

Summary of Forms for Null and Alternative Hypotheses about a Population Mean

- The equality part of the hypotheses always
- appears in the null hypothesis.(Follows Minitab)

- In general, a hypothesis test about the value of
- apopulation mean must take one of the
- followingthree forms (where 0is the
- hypothesized value of the population mean).

Summary of Forms for Null and Alternative Hypotheses about a Population Mean

One-tailed

(lower-tail)

One-tailed

(upper-tail)

Two-tailed

- Example: Metro EMS

A major west coast city provides

one of the most comprehensive

emergency medical services in

the world.

Operating in a multiple

hospital system with

approximately 20 mobile medical

units, the service goal is to respond to medical

emergencies with a mean time of 12 minutes or less.

Null and Alternative Hypotheses

- Example: Metro EMS

The director of medical services

wants to formulate a hypothesis

test that could use a sample of

emergency response times to

determine whether or not the

service goal of 12 minutes or less

is being achieved.

Null and Alternative Hypotheses

The emergency service is meeting

the response goal; no follow-up

action is necessary.

H0:

The emergency service is not

meeting the response goal;

appropriate follow-up action is

necessary.

Ha:

where: μ = mean response time for the population

of medical emergency requests

- Because hypothesis tests are based on sample
data, we must allow for the possibility of errors.

- A Type I error is rejecting H0 when it is true.

- Applications of hypothesis testing that only
control the Type I error are often called

significance tests.

Type I error

- The probability of making a Type I error when the null hypothesis is true as an equality is called the level of significance.
- The Greek symbol α(alpha) is used to denote the level of significance, and common choices for α are .05 and .01.

Type I error

- If the cost of making a Type I error is high, small values of α are preferred. If the cost of making a Type I error is not too high, larger values of α are typically used.

Type I and Type II Errors

Type II error

- A Type II error is accepting H0 when it is false.

- It is difficult to control for the probability of
- making a Type II error. Most applications of
- hypothesis testing control for the probability
- of making a Type I error.

- Statisticians avoid the risk of making a Type II
error by using “do not reject H0” and not

“accept H0”.

Type I and Type II Errors

Population Condition

H0 True

(m< 12)

H0 False

(m > 12)

Conclusion

Not rejectH0

(Concludem< 12)

Correct

Decision

Type II Error

Type I Error

Correct

Decision

RejectH0

(Conclude m > 12)

Steps of Hypothesis Testing

Step 1.Develop the null and alternative hypotheses.

Step 2.Specify the level of significance α .

Step 3.Collect the sample data and compute

the test statistic.

p-Value Approach

Step 4.Use the value of the test statistic to compute

the p-value.

Step 5.Reject H0 if p-value <α.

Steps of Hypothesis Testing

Critical Value Approach

Step 4. Use the level of significance to determine the critical value and the rejection rule.

Step 5. Use the value of the test statistic and the rejection rule to determine whether to

reject H0.

p-Value Approach to

One-Tailed Hypothesis Testing

- The p-value is the probability, computed using
the test statistic, that measures the support

(or lack of support) provided by the sample

for the null hypothesis. The p-value is also called

the observed level of significance.

- If the p-value is less than or equal to the level
of significance α , the value of the test statistic

is in the rejection region.

- Reject H0 if the p-value <α .

Critical Value Approach to

One-Tailed Hypothesis Testing

- The test statistic z has a standard normal
probability distribution.

- We can use the standard normal probability
- distribution table to find the z-value with an
- area of α in the lower (or upper) tail of the
- distribution.

Critical Value Approach to

One-Tailed Hypothesis Testing

- The value of the test statistic that established
- theboundary of the rejection region is called
- thecritical value for the test.

- The rejection rule is:
- Lower tail: Reject H0 if z< - z α
- Upper tail: Reject H0 if z>zα

- Population Mean: σ Known
- One-tailed testsabout a population mean take one of the following two forms.
Lower Tail Test Upper Tail Test

Sampling

distribution

of

Lower-Tailed Test About a Population Mean:sKnown

p-Value <a,

so reject H0.

- p-Value Approach

a = .10

p-value

72

z

z =

-1.46

-za =

-1.28

0

Sampling

distribution

of

Lower-Tailed Test About a Population Mean : sKnown

p-Value <a,

so reject H0.

- p-Value Approach

a = .04

p-Value

11

z

za =

1.75

z =

2.29

0

Sampling

distribution

of

Lower-Tailed Test About a Population Mean : sKnown

- Critical Value Approach

Reject H0

a 1

Do Not Reject H0

z

-za = -1.28

0

Sampling

distribution

of

Upper-Tailed Test About a Population Mean : sKnown

- Critical Value Approach

Reject H0

Do Not Reject H0

z

za = 1.645

0

Example:

- The label on a large can of Hilltop Coffee states that the can contains 3 pounds of coffee. The Federal Trade Commission (FTC) knows that Hilltop’s production process cannot place exactly 3 pounds of coffee in each can, even if the mean filling weight for the population of all cans filled is 3 pounds per can.

Example:

- However, as long as the population mean filling weight is at least 3 pounds per can, the rights of consumers will be protected.

Step 1. Develop the null and alternative hypotheses.

denoting the population mean filling weight and

the hypothesized value of the population mean is

μ0 = 3. The null and alternative hypotheses are

H0 : μ ≥ 3

Ha: μ < 3

Step 2. Specify the level of significance α.

we set the level of significance for the hypothesis

test at α= .01

Step 3. Collect the sample data and compute the

test statistic.

Test statistic:

Sample data: σ = .18 and sample size n=36,

= 2.92 pounds.

Step 3. Collect the sample data and compute the

test statistic.

We have z = -2.67

- p-Value Approach
Step 4. Use the value of the test statistic to compute the p-value.

Using the standard normal distribution table, the

area between the mean and z = - 2.67 is .4962.

The p-value is .5000 - .4962 = .0038,

Step 5. p-value = .0038 < α = .01

Reject H0.

- p-value for The Hilltop Coffee Study when
sample mean = 2.92 and z = -2 .67.

Critical Value Approach

Step 4. Use the level of significance to determine the critical value and the rejection rule.

The critical value is z = -2.33.

Step 5. Use the value of the test statistic and the rejection rule to determine whether to reject H0.

We will reject H0 if z <- 2.33.

Because z = - 2.67 < 2.33, we can reject H0 and conclude that Hilltop Coffee is under filling cans.

One-Tailed Tests About a Population Mean : sKnown

- Example: Metro EMS

The response times for a random

sample of 40 medical emergencies

were tabulated. The sample mean

is 13.25 minutes. The population

standard deviation is believed to

be 3.2 minutes.

One-Tailed Tests About a Population Mean : sKnown

- Example: Metro EMS

The EMS director wants to

perform a hypothesis test, with a

.05 level of significance, to determine

whether the service goal of 12 minutes or less

is being achieved.

One-Tailed Tests About a Population Mean: sKnown

- p -Value and Critical Value Approaches

1. Develop the hypotheses.

H0:

Ha:

a = .05

2. Specify the level of significance.

3. Compute the value of the test statistic.

One-Tailed Tests About a Population Mean: sKnown

- p –Value Approach

4. Compute the p –value.

For z = 2.47, cumulative probability = .9932.

p–value = 1 - .9932 = .0068

5. Determine whether to reject H0.

Because p–value = .0068 <α = .05, we reject H0.

We are at least 95% confident that Metro EMS is not meeting the response goal of 12 minutes.

Sampling

distribution

of

One-Tailed Tests About a Population Mean: sKnown

- p –Value Approach

a = .05

p-value

z

za =

1.645

z =

2.47

0

One-Tailed Tests About a Population Mean: sKnown

- Critical Value Approach

4. Determine the critical value and rejection rule.

For α = .05, z.05 = 1.645

Reject H0 if z> 1.645

5. Determine whether to reject H0.

Because 2.47 > 1.645, we reject H0.

We are at least 95% confident that Metro EMS is not meeting the response goal of 12 minutes.

- In hypothesis testing, the general form for a two-tailed test about a population mean is as follows:

p-Value Approach to

Two-Tailed Hypothesis Testing

- Compute the p-value using the following three
steps:

1. Compute the value of the test statistic z.

2. If z is in the upper tail (z > 0), find the area

under the standard normal curve to the right

of z. If z is in the lower tail (z < 0), find the

area under the standard normal curve to the

left of z.

p-Value Approach to

Two-Tailed Hypothesis Testing

3. Double the tail area obtained in step 2 to obtain

the p –value.

- The rejection rule:
- Reject H0 if the p-value <.

Critical Value Approach to

Two-Tailed Hypothesis Testing

- The critical values will occur in both the lower
and upper tails of the standard normal curve.

- Use the standard normal probability
distribution table to find za/2 (the z-value

with an area of a/2 in the upper tail of

the distribution).

- The rejection rule is:
- Reject H0 if z< -za/2 or z>za/2

- Example: The U.S. Golf Association (USGA) establishes rules that manufacturers of golf equipment must meet if their products are to be acceptable for use in USGA events.
- MaxFlight produce golf balls with an average distance of 295 yards.

- When the average distance falls below 295 yards, the company worries about losing sales because the golf balls do not provide as much distance as advertised.

- When the average distance passes 295 yards, MaxFlight’s golf balls may be rejected by the
USGA for exceeding the overall distance

standard concerning carry and roll.

- A hypothesized value of μ0= 295 and the null and alternative hypotheses test are as follows:
H0 : μ = 295 VS Ha : μ≠ 295

- If the sample mean is significantly less than
295 yards or significantly greater than 295 yards, we will reject H0.

Step 1. Develop the null and alternative

hypotheses.

A hypothesized value of μ0 = 295 and

H0 : μ = 295

Ha : μ≠ 295

Step 2. Specify the level of significance α.

we set the level of significance for the hypothesis test at α = .05

Step 3. Collect the sample data and compute the

test statistic.

Test statistic:

Sample data: σ = 12 and sample size n=50,

= 297.6 yards.

We have

- p-Value Approach
Step 4. Use the value of the test statistic to compute the p-value.

The two-tailed p-value in this case is given by

P(z< -1.53) + P(z> 1.53)= 2(.0630) = .1260.

Step 5.We do not reject H0 because the

p-value = .1260 > .05. No action will be taken

to adjust manufacturing process.

- p-Value for The Maxflight Hypothesis Test

Critical Value Approach

Step 4. Use the level of significance to determine the critical value and the rejection rule. The critical values are - z.025 = -1.96 and z.025 = 1.96.

Step 5. Use the value of the test statistic and the rejection rule to determine whether to reject H0.

We will reject H0 if z < - 1.96 or if z > 1.96

Because the value of the test statistic is z=1.53, the statistical evidence will not permit us to reject the null hypothesis

at the .05 level of significance.

oz.

Glow

- Two-Tailed Test About a Population Mean: s Known

The production line for Glow toothpaste

is designed to fill tubes with a mean weight

of 6 oz. Periodically, a sample of 30 tubes

will be selected in order to check the

filling process.

oz.

Glow

- Two-Tailed Test About a Population Mean: s Known

Quality assurance procedures call for

the continuation of the filling process if the

sample results are consistent with the assumption

that the mean filling weight for the population of

toothpaste

tubes is 6 oz.; otherwise the process will be

adjusted.

oz.

Glow

Example: Glow Toothpaste

- Two-Tailed Test About a Population Mean: s Known

Assume that a sample of 30 toothpaste tubes

provides a sample mean of 6.1 oz. The population

standard deviation is believed to be 0.2 oz.

Perform a hypothesis test, at the .03

level of significance, to help determine

whether the filling process should continue

operating or be stopped and corrected.

Glow

1. Determine the hypotheses.

Two-Tailed Tests About a Population Mean:sKnown

- p –Value and Critical Value Approaches

α = .03

2. Specify the level of significance.

3. Compute the value of the test statistic.

Glow

Two-Tailed Tests About a Population Mean : sKnown

- p –Value Approach

4. Compute the p –value.

For z = 2.74, cumulative probability = .9969

p–value = 2(1 - .9969) = .0062

5. Determine whether to reject H0.

Because p–value = .0062 < α = .03, we reject H0.

We are at least 97% confident that the mean filling weight of the toothpaste tubes is not 6 oz.

Glow

Two-Tailed Tests About a Population Mean : sKnown

- p-Value Approach

1/2

p -value

= .0031

1/2

p -value

= .0031

a/2 =

.015

a/2 =

.015

z

z = -2.74

0

z = 2.74

-za/2 = -2.17

za/2 = 2.17

Glow

Two-Tailed Tests About a Population Mean : sKnown

- Critical Value Approach

4. Determine the critical value and rejection rule.

For a/2 = .03/2 = .015, z.015 = 2.17

Reject H0 if z< -2.17 or z> 2.17

5. Determine whether to reject H0.

Because 2.47 > 2.17, we reject H0.

We are at least 97% confident that the mean filling weight of the toothpaste tubes is not 6 oz.

Glow

Sampling

distribution

of

Two-Tailed Tests About a Population Mean : σKnown

- Critical Value Approach

Reject H0

Reject H0

Do Not Reject H0

a/2 = .015

a/2 = .015

z

0

2.17

-2.17

- Select a simple random sample from the
population and use the value of the sample

mean to develop the confidence interval for

the population mean .

(Confidence intervals are covered in Chapter 8.)

- If the confidence interval contains the
hypothesized value 0do not reject H0.

Otherwise, reject H0.

Glow

The 97% confidence interval for is

or 6.02076 to 6.17924

Because the hypothesized value for the

population mean,0=6, is not in this interval,

the hypothesis-testing conclusion is that the

null hypothesis,H0: = 6, can be rejected.

- Test Statistic

This test statistic has a t distribution

with n - 1 degrees of freedom.

Tests About a Population Mean:s Unknown

- Rejection Rule: p -Value Approach

Reject H0 if p –value <a

- Rejection Rule: Critical Value Approach

Reject H0 if t<-t

H0:

H0:

Reject H0 if t>t

H0:

Reject H0 if t < - tor t >t

- The format of the t distribution table provided in
most statistics textbooks does not have sufficient

detail to determine the exactp-value p-value for

a hypothesis test.

- However, we can still use the t distribution table
to identify a range for the p-value.

- An advantage of computer software packages is
that the computer output will provide the p-value

for the t distribution.

- Example: A business travel magazine wants to classify transatlantic gateway airports according to the mean rating for the population of business travelers.
- A rating scale with a low score of 0 and a high score of 10 will be used, and airports with a population mean rating greater than 7 will be designated as superior service airports.

- The null and alternative hypotheses for this upper tail test are as follows:
H0 : μ ≤ 7

Ha : μ > 7

- Test Statistic
The sampling distribution

of t has n – 1 df.

We have =7.25, s = 1.052, and n = 60, and the value of the test statistic is

- p-Value Approach
Using Table 2 in Appendix B, the t distribution with 59 degrees of freedom.

- We see that t 1.84 is between 1.671 and 2.001. Although the table does not provide the exact p-value, the values in the “Area in Upper Tail” row show that the p-value must be less than .05 and greater than .025.

- MINITAB OUTPUT
- The test statistic t = 1.84, and the exact
p-value is .035 for the Heathrow rating hypothesis test.

- A p-value = .035 < .05 leads to the rejection of the null hypothesis.

- Critical Value Approach
The rejection rule is thus

Reject H0 if t >1.671

With the test statistic t = 1.84 > 1.671, H0is rejected.

- One-Tailed Test About a Population Mean: s Unknown

A State Highway Patrol periodically samples

vehicle speeds at various locations

on a particular roadway.

The sample of vehicle speeds

is used to test the hypothesis

H0: m< 65

The locations where H0 is rejected are deemed

the best locations for radar traps.

Example: Highway Patrol

- One-Tailed Test About a Population Mean: s Unknown

At Location F, a sample of 64 vehicles shows a

mean speed of 66.2 mph with a

standard deviation of

4.2 mph. Use a = .05 to

test the hypothesis.

One-Tailed Test About a Population Mean : s Unknown

- p –Value and Critical Value Approaches

1. Determine the hypotheses.

H0: < 65

Ha: m> 65

a= .05

2. Specify the level of significance.

3. Compute the value of the test statistic.

One-Tailed Test About a Population Mean : s Unknown

- p –Value Approach

4. Compute the p –value.

For t = 2.286, the p–value must be less than .025 (for t = 1.998) and greater than .01 (for t = 2.387).

.01 < p–value < .025

One-Tailed Test About a Population Mean: s Unknown

- p –Value Approach

5. Determine whether to reject H0.

Because p–value <a = .05, we reject H0.

We are at least 95% confident that the mean speed of vehicles at Location F is greater than 65 mph.

One-Tailed Test About a Population Mean: s Unknown

- Critical Value Approach

4. Determine the critical value and rejection rule.

For a = .05 and d.f. = 64 – 1 = 63, t.05 = 1.669

Reject H0 if t> 1.669

One-Tailed Test About a Population Mean: s Unknown

- Critical Value Approach

5. Determine whether to reject H0.

Because 2.286 > 1.669, we reject H0.

We are at least 95% confident that the mean speed of vehicles at Location F is greater than 65 mph. Location F is a good candidate for a radar trap.

One-Tailed Test About a Population Mean: s Unknown

Reject H0

Do Not Reject H0

t

ta =

1.669

0

- Summary of Hypothesis Tests about a Population Mean: s Unknown Case

A Summary of Forms for Null and Alternative Hypotheses About a Population Proportion

- The equality part of the hypotheses always
- appearsin the null hypothesis.

- In general, a hypothesis test about the value of
- a population proportion pmust take one of the
following three forms (where p0 is the

hypothesizedvalue of the population

proportion).

A Summary of Forms for Null and Alternative Hypotheses About a Population Proportion

One-tailed

(lower tail)

One-tailed

(upper tail)

Two-tailed

Tests About a Population Proportion

- Test Statistic

where:

assuming np> 5 and n(1 – p)> 5

Tests About a Population Proportion

- Rejection Rule: p –Value Approach

Reject H0 if p –value < α

- Rejection Rule: Critical Value Approach

H0: pp

Reject H0 if z>z

H0: pp

Reject H0 if z< -z

H0: pp

Reject H0 if z< -z or z>z

- Example: Pine Creek golf course
Over the past year, 20% of the players at Pine Creek were women.

A special promotion designed to attract women golfers. One month after the promotion was implemented.

- The null and alternative hypotheses are as
follows: H0: p ≤ .20

Ha: p > .20

- level of significance of α = .05 be used.
- Test statistics

- level of significance of α = .05 be used.
- Suppose a random sample of 400 players was
selected, and that 100 of the players were women.

The proportion of women golfers in the

sample is

and the value of the test statistic is

p-Value Approach

- the p-value is the probability that z is greater than or equal to z = 2.50. Thus, the p-value for the test is .5000 – .4938 = .0062.

- A p-value = .0062 < .05 gives sufficient statistical evidence to reject H0 at the .05 level of significance.
Critical Value Approach

- The critical value corresponding to an area of .05 in the upper tail of a standard normal distribution is z.05 = 1.645. To reject H0 if
z ≥ 1.645.

- Because z = 2.50 > 1.645, H0 is rejected.

Two-Tailed Test About a

Population Proportion

- Example: National Safety Council

For a Christmas and New Year’s week, the

National Safety Council estimated that

500 people would be killed and 25,000

injured on the nation’s roads. The

NSC claimed that 50% of the

accidents would be caused by

drunk driving.

Two-Tailed Test About a

Population Proportion

- Example: National Safety Council

A sample of 120 accidents showed that

67 were caused by drunk driving. Use

these data to test the NSC’s claim with

α = .05.

Two-Tailed Test About a

Population Proportion

- p –Value and Critical Value Approaches

1. Determine the hypotheses.

2. Specify the level of significance.

a = .05

a common

error is using in this formula

Two-Tailed Test About a

Population Proportion

- p –Value and Critical Value Approaches

3. Compute the value of the test statistic.

Two-Tailed Test About a

Population Proportion

- p -Value Approach

4. Compute the p -value.

For z = 1.28, cumulative probability = .8997

p–value = 2(1 - .8997) = .2006

5. Determine whether to reject H0.

Because p–value = .2006 > a = .05, we cannot reject H0.

Two-Tailed Test About a

Population Proportion

- Critical Value Approach

4. Determine the critical value and rejection rule.

Fora/2 = .05/2 = .025, z.025 = 1.96

Reject H0 if z < -1.96 or z > 1.96

5. Determine whether to reject H0.

Because 1.278 > -1.96 and < 1.96, we cannot reject H0.

- Summary of Hypothesis Tests about a Population Proportion

Hypothesis Testing and Decision Making

- In many decision-making situations the
- decision maker may want, and in some cases
- may be forced, to take action with both the
- conclusion do not rejectH0 and the conclusion rejectH0.

- In such situations, it is recommended that
- the hypothesis-testing procedure be extended
- to include consideration of making a Type II
- error.

- Example:
A quality control manager must decide to accept a shipment of batteries from a supplier or to return the shipment because of poor quality.

Suppose the null and alternative hypotheses about the population mean follow.

H0 : μ ≥ 120

Ha : μ < 120

- If H0 is rejected, the appropriate action is to return the shipment to the supplier.
- If H0 is not rejected, the decision maker must
still determine what action should be taken.

- In such decision-making situations, it is recommended to control the probability of making a Type II error. Because knowledge
of the probability of making a Type II error

will be helpful.

Calculating the Probability of a Type II Error in Hypothesis Tests About a Population Mean

1. Formulate the null and alternative hypotheses.

- Using the critical value approach, use the level
- of significance to determine the critical value
- and the rejection rule for the test.

- Using the rejection rule, solve for the value of
- the sample mean corresponding to the critical
- value of the test statistic.

5. Using the sampling distribution offor a value

of μ satisfying the alternative hypothesis, and

the acceptance region from step 4, compute the probability that the sample mean will be in the acceptance region. (This is the probability of making a Type II error at the chosen

level of μ.)

Calculating the Probability of a Type II Error in Hypothesis Tests About a Population Mean

4. Use the results from step 3 to state the values of the sample mean that lead to the acceptance of H0; this defines the acceptance region.

- Example: Batteries’ Quality
- Suppose the null and alternative hypotheses are
H0 : μ ≥ 120

Ha : μ < 120

- level of significance of α = .05.
- Sample size n=36 and σ = 12 hours.

- The critical value approach and z.05 = 1.645, the rejection rule is
Reject H0 if z ≤ -1.645

- The rejection rule indicates that we will reject H0 if
That indicates that we will reject H0if

and accept the shipment whenever > 116.71.

- Compute probabilities associated with
making a Type II error (whenever the true

mean is less than 120 hours and we make

the decision to accept H0: μ ≥120.)

- If μ =112 is true, the probability of making a Type II error is the probability that the sample mean is greater than 116.71 when μ = 112,that is, P( ≥ 116.71 | μ = 112) =?
- The probability of making a Type II error (β ) when μ = 112 is .0091 .

- We can repeat these calculations for other values of μ less than 120.

- The powerof the test = the probability of correctly rejecting H0 when it is false. For any particular value of μ, the power is 1-β.
- Power Curve for The Lot-Acceptance Hypothesis Test

Calculating the Probability of a Type II Error

- Example: Metro EMS (revisited)

Recall that the response times for

a random sample of 40 medical

emergencies were tabulated. The

sample mean is 13.25 minutes.

The population standard deviation

is believed to be 3.2 minutes.

Calculating the Probability of a Type II Error

- Example: Metro EMS (revisited)

The EMS director wants to

perform a hypothesis test, with a

.05 level of significance, to determine

whether or not the service goal of 12 minutes or less is being achieved.

4. We will acceptH0when < 12.8323

Calculating the Probability of a Type II Error

1.Hypotheses are:H0:μandHa:μ

2. Rejection rule is: RejectH0 if z>1.645

- 3.Value of the sample mean that identifies

Values of m b 1-b

Calculating the Probability of a Type II Error

5.Probabilities that the sample mean will be

in the acceptance region:

14.0 -2.31 .0104 .9896

13.6 -1.52 .0643 .9357

13.2 -0.73 .2327 .7673

12.8323 0.00 .5000 .5000

12.8 0.06 .5239 .4761

12.4 0.85 .8023 .1977

12.0001 1.645 .9500 .0500

Example: m = 12.0001, b= .9500

Calculating the Probability of a Type II Error

- Calculating the Probability of a Type II Error

Observations about the preceding table:

- When the true population mean m is close to
- the null hypothesis value of 12, there is a high
- probability that we will make a Type II error.

Example: m = 14.0, b = .0104

Calculating the Probability of a Type II Error

- Calculating the Probability of a Type II Error

- When the true population mean m is far above the null hypothesis value of 12, there is a low probability that we will make a Type II error.

Power of the Test

- The probability of correctly rejecting H0
- when it is false is called the power of the test.

- For any particular value of μ, the power is
- 1 – β.

- We can show graphically the power
- associated with each value of μ; such a graph
- is called a power curve. (See next slide.)

Power Curve

H0 False

m

Determining the Sample Size for a Hypothesis Test About a Population Mean

- The specified level of significance determines the probability of making a Type I error.

- By controlling the sample size, the probability
- of making a Type II error is controlled.

- Example: how a sample size can be determined for the lower tail test about a population mean.
H0 : μ ≥ μ0

Ha : μ < μ0

- Let α be the probability of a Type I error and zαand zβare the z value corresponding to an area of α and β, respectively in the upper tail of the standard normal distribution.

- we compute c using the following formulas

- To determine the required sample size, we solve for the n as follows.

c

c

Determining the Sample Size for a Hypothesis Test About a Population Mean

H0: m0

Ha:m0

Reject H0

a

m0

Sampling

Distribution of when

H0 is true

and m = m0

Sampling

distribution

of when

H0 is false

and ma > m0

b

ma

Determining the Sample Size for a Hypothesis Test About a Population Mean

where

z= z value providing an area of in the tail

z= z value providing an area of in the tail

= population standard deviation

0 = value of the population mean in H0

a = value of the population mean used for theType II error

Note: In a two-tailed hypothesis test, use z /2 not z

Relationship Among a, b, and n

- Once two of the three values are known, the other can be computed.
- For a given level of significance a, increasing the sample size n will reduce b.
- For a given sample size n, decreasing a will increaseb, whereas increasing a will decrease b.

Determining the Sample Size for a Hypothesis Test About a Population Mean

- Let’s assume that the director of medical
services makes the following statements

about the allowable probabilities for the Type I and Type II errors:

- If the mean response time is m = 12 minutes, I am willing to risk an a = .05 probability of rejecting H0.

- If the mean response time is 0.75 minutes over the specification (m = 12.75), I am willing to risk a b= .10 probability of not rejecting H0.

Determining the Sample Size for a Hypothesis Test About a Population Mean

Given

a = .05, b = .10

z= 1.645, z= 1.28

0 = 12, a = 12.75

= 3.2