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Statistics. Hypothesis Tests. Contents. Developing Null and Alternative Hypotheses Type I and Type II Errors Population Mean: σ Known Population Mean: σ Unknown Population Proportion Hypothesis Testing and Decision Making

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Statistics

Statistics

Hypothesis Tests


Contents

Contents

  • 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


Contents1

Contents

  • 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


Statistics in practice

STATISTICSin PRACTICE

  • 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.


Statistics in practice1

STATISTICSin PRACTICE

  • 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.


Developing null and alternative hypotheses

Developing Null and Alternative Hypotheses

  • 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.


Developing null and alternative hypotheses1

Developing Null and Alternative Hypotheses

  • 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.


Developing null and alternative hypotheses2

Developing Null and Alternative Hypotheses

  • Testing Research Hypotheses

  • Testing the Validity of a Claim

  • Testing in Decision-Making Situations


Developing null and alternative hypotheses3

Developing Null and Alternative Hypotheses

  • 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.


Developing null and alternative hypotheses4

Developing Null and Alternative Hypotheses

  • 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.


Developing null and alternative hypotheses5

Developing Null and Alternative Hypotheses

The appropriate null and alternative hypotheses

for the study are:

Ho: μ ≤ 24

Ha: μ > 24


Developing null and alternative hypotheses6

Developing Null and Alternative Hypotheses

  • 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.


Developing null and alternative hypotheses7

Developing Null and Alternative Hypotheses

  • 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.


Developing null and alternative hypotheses8

Developing Null and Alternative Hypotheses

  • 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.


Developing null and alternative hypotheses9

Developing Null and Alternative Hypotheses

  • 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


Developing null and alternative hypotheses10

Developing Null and Alternative Hypotheses

  • 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.


Developing null and alternative hypotheses11

Developing Null and Alternative Hypotheses

  • 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.


Developing null and alternative hypotheses12

Developing Null and Alternative Hypotheses

  • 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


Developing null and alternative hypotheses13

Developing Null and Alternative Hypotheses

  • 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.


Developing null and alternative hypotheses14

Developing Null and Alternative Hypotheses

  • 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.


Statistics

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).


Statistics

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

One-tailed

(lower-tail)

One-tailed

(upper-tail)

Two-tailed


Null and alternative hypotheses

Null and Alternative Hypotheses

  • 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.


Statistics

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.


Statistics

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


Type i and type ii errors

Type I and Type II Errors

  • 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 and type ii errors1

Type I and Type II Errors

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 and type ii errors2

Type I and Type II Errors

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.


Statistics

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”.


Statistics

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)


Statistics

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.


Statistics

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.


Statistics

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 .


Statistics

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.


Statistics

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 


One tailed hypothesis testing

One-Tailed Hypothesis Testing

  • Population Mean: σ Known

  • One-tailed testsabout a population mean take one of the following two forms.

    Lower Tail Test Upper Tail Test


Statistics

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


Statistics

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


Statistics

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


Statistics

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


One tailed hypothesis testing1

One-Tailed Hypothesis Testing

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.


One tailed hypothesis testing2

One-Tailed Hypothesis Testing

Example:

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


One tailed hypothesis testing3

One-Tailed Hypothesis Testing

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


One tailed hypothesis testing4

One-Tailed Hypothesis Testing

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.


One tailed hypothesis testing5

One-Tailed Hypothesis Testing

Step 3. Collect the sample data and compute the

test statistic.

We have z = -2.67


One tailed hypothesis testing6

One-Tailed Hypothesis Testing

  • 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.


One tailed hypothesis testing7

One-Tailed Hypothesis Testing

  • p-value for The Hilltop Coffee Study when

    sample mean = 2.92 and z = -2 .67.


One tailed hypothesis testing8

One-Tailed Hypothesis Testing

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.


One tailed hypothesis testing9

One-Tailed Hypothesis Testing

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.


Statistics

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.


Statistics

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.


Statistics

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.


Statistics

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.


Statistics

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


Statistics

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.


Two tailed hypothesis testing

Two-Tailed Hypothesis Testing

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


Statistics

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.


Statistics

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 <.


Statistics

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


Two tailed hypothesis testing1

Two-Tailed Hypothesis Testing

  • 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.


Two tailed hypothesis testing2

Two-Tailed Hypothesis Testing

  • 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.


Two tailed hypothesis testing3

Two-Tailed Hypothesis Testing

  • 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


Two tailed hypothesis testing4

Two-Tailed Hypothesis Testing

  • If the sample mean is significantly less than

    295 yards or significantly greater than 295 yards, we will reject H0.


Two tailed hypothesis testing5

Two-Tailed Hypothesis Testing

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


Two tailed hypothesis testing6

Two-Tailed Hypothesis Testing

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


Two tailed hypothesis testing7

Two-Tailed Hypothesis Testing

  • 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.


Two tailed hypothesis testing8

Two-Tailed Hypothesis Testing

  • p-Value for The Maxflight Hypothesis Test


Two tailed hypothesis testing9

Two-Tailed Hypothesis Testing

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.


Two tailed hypothesis testing10

Two-Tailed Hypothesis Testing

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.


Example glow toothpaste

oz.

Glow

Example: Glow Toothpaste

  • 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.


Example glow toothpaste1

oz.

Glow

Example: Glow Toothpaste

  • 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.


Statistics

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.


Statistics

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.


Statistics

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.


Statistics

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


Statistics

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.


Statistics

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


Summary of hypothesis tests about a population mean known case

Summary of Hypothesis Tests about a Population Mean: σ Known Case


Confidence interval approach to two tailed tests about a population meant

Confidence Interval Approach toTwo-Tailed Tests About a Population Meant

  • 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.


Confidence interval approach to two tailed tests about a population mean

Glow

Confidence Interval Approach toTwo-Tailed Tests About a Population Mean

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.


Tests about a population mean s unknown

Tests About a Population Mean:s Unknown

  • Test Statistic

This test statistic has a t distribution

with n - 1 degrees of freedom.


Statistics

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 < - tor t >t


P values and the t distribution

p -Values and the t Distribution

  • 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.


Tests about a population mean s unknown1

Tests About a Population Mean:s Unknown

  • 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.


Tests about a population mean s unknown2

Tests About a Population Mean:sUnknown

  • The null and alternative hypotheses for this upper tail test are as follows:

    H0 : μ ≤ 7

    Ha : μ > 7


Tests about a population mean s unknown3

Tests About a Population Mean:sUnknown

  • 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


Statistics

  • 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.


Statistics

  • 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.


Statistics

  • 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.


Example highway patrol

Example: Highway Patrol

  • 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.


Statistics

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.


Statistics

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.


Statistics

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


Statistics

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.


Statistics

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


Statistics

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.


Statistics

One-Tailed Test About a Population Mean: s Unknown

Reject H0

Do Not Reject H0



t

ta =

1.669

0


Tests about a population mean s unknown4

Tests About a Population Mean:s Unknown

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


Statistics

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).


Statistics

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

One-tailed

(lower tail)

One-tailed

(upper tail)

Two-tailed


Statistics

Tests About a Population Proportion

  • Test Statistic

where:

assuming np> 5 and n(1 – p)> 5


Statistics

Tests About a Population Proportion

  • Rejection Rule: p –Value Approach

Reject H0 if p –value < α

  • Rejection Rule: Critical Value Approach

H0: pp

Reject H0 if z>z

H0: pp

Reject H0 if z< -z

H0: pp

Reject H0 if z< -z or z>z


Upper tail test about a population proportion

Upper Tail Test About aPopulation Proportion

  • 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


Upper tail test about a population proportion1

Upper Tail Test About aPopulation Proportion

  • level of significance of α = .05 be used.

  • Test statistics


Upper tail test about a population proportion2

Upper Tail Test About aPopulation Proportion

  • 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


Upper tail test about a population proportion3

Upper Tail Test About aPopulation Proportion

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.


Upper tail test about a population proportion4

Upper Tail Test About aPopulation Proportion

  • 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.


Statistics

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.


Statistics

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.


Statistics

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


Statistics

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.


Statistics

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.


Statistics

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.


Tests about a population proportion

Tests About a Population Proportion

  • Summary of Hypothesis Tests about a Population Proportion


Statistics

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.


Hypothesis testing and decision making

Hypothesis Testing and Decision Making

  • 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


Hypothesis testing and decision making1

Hypothesis Testing and Decision Making

  • 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.


Statistics

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.


Statistics

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.


Calculating the probability of a type ii error in hypothesis tests about a population mean

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

  • 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.


Calculating the probability of a type ii error in hypothesis tests about a population mean1

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

  • The critical value approach and z.05 = 1.645, the rejection rule is

    Reject H0 if z ≤ -1.645


Calculating the probability of a type ii error in hypothesis tests about a population mean2

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

  • The rejection rule indicates that we will reject H0 if

    That indicates that we will reject H0if

    and accept the shipment whenever > 116.71.


Calculating the probability of a type ii error in hypothesis tests about a population mean3

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

  • 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.)


Calculating the probability of a type ii error in hypothesis tests about a population mean4

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

  • 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 .


Calculating the probability of a type ii error in hypothesis tests about a population mean5

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

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


Calculating the probability of a type ii error in hypothesis tests about a population mean6

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

  • 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


Statistics

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.


Statistics

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.


Statistics

4. We will acceptH0when < 12.8323

Calculating the Probability of a Type II Error

1.Hypotheses are:H0:μandHa:μ

2. Rejection rule is: RejectH0 if z>1.645

  • 3.Value of the sample mean that identifies

  • the rejection region:


  • Statistics

    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


    Statistics

    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.


    Statistics

    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.


    Statistics

    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.)


    Statistics

    Power Curve

    H0 False

    m


    Statistics

    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.


    Determining the sample size for a hypothesis test about a population mean

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

    • 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.


    Determining the sample size for a hypothesis test about a population mean1

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


    Determining the sample size for a hypothesis test about a population mean2

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

    • we compute c using the following formulas


    Determining the sample size for a hypothesis test about a population mean3

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

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


    Statistics

    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


    Statistics

    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


    Statistics

    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.


    Statistics

    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.


    Statistics

    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


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