1

JTE9 Chapter 8 ~ Introduction to Statistical Inferences

Level of

Confidence

1

Maximum

Error

E

Sample

Size

n

En

z

2

JTE9 Chapter Goals

• Learn the basic concepts of estimation and hypothesis testing

• Consider questions about a population mean using two methods that assume the population standard deviation is known

• Consider: what value or interval of values can we use to estimate a population mean?

• Consider: is there evidence to suggest the hypothesized mean is incorrect?

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JTE9 8.1 ~ The Nature of Estimation

• Discuss estimation more precisely

• What makes a statistic good ?

• Assume the population standard deviation, , is known throughout this chapter

• Concentrate on learning the procedures for making statistical inferences about a population mean

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JTE9 Point Estimate for a Parameter

Point Estimate for a Parameter: The value of the corresponding statistic

Note: The quality of an estimation procedure is enhanced if the sample statistic is both less variable and unbiased

How good is the point estimate? Is it high? Or low? Would another sample yield the same result?

Example: is a point estimate (single number value) for the mean of the sampled populationx 14 7.

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JTE9 Unbiased Statistic

Unbiased Statistic: A sample statistic whose sampling distribution has a mean value equal to the value of the population parameter being estimated. A statistic that is not unbiased is a biased statistic.

Example: The figures on the next slide illustrate the concept of being unbiased and the effect of variability on a point estimate

Assume A is the parameter being estimated

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JTE9 Illustrations

A

Positive biasOver estimateLow variability

A

UnbiasedOn target estimate

A

Negative biasUnder estimateHigh variability

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JTE9 Notes

2. Sample means vary from sample to sample. We don’t expect the sample mean to be exactly equal the population mean .

3. We do expect the sample mean to be close to the population mean

4. Since closeness is measured in standard deviations, we expect the sample mean to be within 2 standard deviations of the population mean

x

1. The sample mean, ,is an unbiased statistic because the mean value of the sampling distribution is equal to the population mean:

x

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JTE9 Important Definitions

Interval Estimate: An interval bounded by two values and used to estimate the value of a population parameter. The values that bound this interval are statistics calculated from the sample that is being used as the basis for the estimation.

Level of Confidence 1 - : The probability that the sample to be selected yields an interval that includes the parameter being estimated

Confidence Interval: An interval estimate with a specified level of confidence

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• To construct a confidence interval for a population mean , use the CLT

• Use the point estimate as the central value of an intervalx

Summary

• The level of confidence for the resulting interval is approximately 95%, or 0.95

• We can be more accurate in determining the level of confidence

x

• Since the sample mean ought to be within 2 standard deviations of the population mean (95% of the time), we can find the bounds to an interval centered at :

x xx x 2 2( ) ( ) to

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JTE9 Illustration

x x x2( )x x 2( )

Distribution of x

• The interval is an approximate 95% confidence interval for the population mean based on this x

x xx x 2 2 to

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JTE9 8.2 ~ Estimation of Mean ( Known)

• Formalize the interval estimation process as it applies to estimating the population mean based on a random sample

• Assume the population standard deviation is known

• The assumptions are the conditions that need to exist in order to correctly apply a statistical procedure

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JTE9 The Assumption...

Assumption satisfied by:

1. Knowing that the sampled population is normally distributed, or

2. Using a large enough random sample (CLT)Note: The CLT may be applied to smaller samples (for examplen = 15) when there is evidence to suggest a unimodal distribution that is approximately symmetric. If there is evidence of skewness, the sample size needs to be much larger.

x

The assumption for estimating the mean using a known :

The sampling distribution of has a normal distribution

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2. z: confidence coefficient, the number of multiples of the standard error needed to construct an interval estimate of the correct width to have a level of confidence 1-

The 1- Confidence Interval of

• A 1- confidence interval for is found by

0 z

1 / 2 / 2

- z z

xn

xn

to z z

Notes:

1. is the point estimate and the center point of the confidence intervalx

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3. : standard error of the mean

The standard deviation of the distribution of

/ n

x

Notes Continued

4. : maximum error of estimate E

One-half the width of the confidence interval (the product of the confidence coefficient and the standard error)

n( / )z

5. : lower confidence limit (LCL)

: upper confidence limit (UCL)

x

x n( / )z

n( / )z

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JTE9 The Confidence IntervalA Five-Step Model:

1. Describe the population parameter of concern2. Specify the confidence interval criteria

a. Check the assumptionsb. Identify the probability distribution and the formula to be usedc. Determine the level of confidence, 1 -

3. Collect and present sample information

4. Determine the confidence intervala. Determine the confidence coefficientb. Find the maximum error of estimatec. Find the lower and upper confidence limits

5. State the confidence interval

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2. Specify the confidence interval criteria

a. Check the assumptionsThe weights are normally distributed, the distribution of is normal

b. Identify the probability distribution and formula to be usedUse the standard normal variable z with = 0.27

c. Determine the level of confidence, 1 - The question asks for 95% confidence: 1 - = 0.95 x

Example Example: The weights of full boxes of a certain kind of cereal are normally

distributed with a standard deviation of 0.27 oz. A sample of 18 randomly selected boxes produced a mean weight of 9.87 oz. Find a 95% confidence interval for the true mean weight of a box of this

cereal.Solution:

1. Describe the population parameter of concernThe mean, , weight of all boxes of this cereal

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3. Collect and present informationThe sample information is given in the statement of the problem

Given: n x 18 9 87; .

Solution Continued

1.15 1.28 1.65 1.96 2.33 2.580.75 0.80 0.90 0.95 0.98 0.991

4. Determine the confidence interval

a. Determine the confidence coefficient

The confidence coefficient is found using Table 4B:

z

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JTE9 Solution Continued

b. Find the maximum error of estimateUse the maximum error part of the formula for a CI

5. State the confidence interval

9.75 to 10.00 is a 95% confidence interval for the true mean weight, , ofcereal boxes

En

1960 27

1801247.

..= z

c. Find the lower and upper confidence limitsUse the sample mean and the maximum error:

xn

9 87 01247 9 87 01247

9 7453 9 9947

9 75 10 00

to

to

to

to

. . . .

. .

. .

z xn

+ z

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JTE9 Example Example: A random sample of the test scores of 100 applicants for clerk-

typist positions at a large insurance company showed a mean score of 72.6. Determine a 99% confidence interval for the mean

score of all applicants at the insurance company. Assume the standard deviation of test scores is 10.5.

Solution:

1. Parameter of concern

The mean test score, , of all applicants at the insurance company

2. Confidence interval criteria

a. Assumptions: The distribution of the variable, test score, is not known. However, the sample size is large enough (n = 100) so that the CLT applies

b. Probability distribution: standard normal variable z with = 10.5

c. The level of confidence: 99%, or 1 - = 0.99

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3. Sample informationGiven: n = 100 and = 72.6x

Solution Continued

5. Confidence intervalWith 99% confidence we say, “The mean test score is between 69.9 and 75.3”, or “69.9 to 75.3 is a 99% confidence interval for the true mean test score”

Note: The confidence is in the process. 99% confidence means: if we conduct the experiment over and over, and construct lots of confidence intervals, then 99% of the confidence intervals will contain the true mean value .

72 6 2 709 69 891 72 6 2 709 75 309. . . . . . to

4. The confidence intervala. Confidence coefficient:

b. Maximum error:

c. The lower and upper limits:

z(0.005) .2 58 z

E n ( / ) ( . )( . / ) . 2 58 10 5 100 2 709z

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JTE9 Sample Size

• Problem: Find the sample size necessary in order to obtain a specified maximum error and level of confidence (assume the standard deviation is known)

En

z

Solve this expression for n:

nE

2z

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JTE9 Example

Example: Find the sample size necessary to estimate a population mean to within 0.5 with 95% confidence if the standard deviation is 6.2

Note: When solving for sample size n, always round up to the next largest integer (Why?)

Solution:

Therefore, n = 591

n

( . )( . )0.

. .196 6 2

5[24 304] 590 684

22

n E

2z

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JTE9 8.3 ~ The Nature of Hypothesis Testing

• Formal process for making an inference

• Consider many of the concepts of a hypothesis test and look at several decision-making situations

• The entire process starts by identifying something of concern and then formulating two hypotheses about it

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JTE9 Hypothesis

Hypothesis: A statement that something is true

Statistical Hypothesis Test: A process by which a decision is made between two opposing hypotheses. The two opposing hypotheses are formulated so that each hypothesis is the negation of the other. (That way one of them is always true, and the other one is always false). Then one hypothesis is tested in hopes that it can be shown to be a very improbable occurrence thereby implying the other hypothesis is the likely truth.

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JTE9 Null & Alternative Hypothesis

There are two hypotheses involved in making a decision:

Null Hypothesis, Ho: The hypothesis to be tested. Assumed to be true. Usually a statement that a population parameter has a specific value. The “starting point” for the investigation.

Alternative Hypothesis, Ha: A statement about the same population parameter that is used in the null hypothesis. Generally this is a statement that specifies the population parameter has a value different, in some way, from the value given in the null hypothesis. The rejection of the null hypothesis will imply the likely truth of this alternative hypothesis.

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JTE9 Notes

1. Basic idea: proof by contradiction

Assume the null hypothesis is true and look for evidence to suggest that it is false

2. Null hypothesis: the status quo

A statement about a population parameter that is assumed to be true

3. Alternative hypothesis: also called the research hypothesis

Generally, what you are trying to prove?

We hope experimental evidence will suggest the alternative hypothesis is true by showing the unlikeliness of the truth of the null hypothesis

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JTE9 Example

Example: Suppose you are investigating the effects of a new pain reliever. You hope the new drug relieves minor

muscle aches and pains longer than the leading pain reliever. State the null and alternative hypotheses.

Solutions:

• Ho: The new pain reliever is no better than the leading pain reliever

• Ha: The new pain reliever lasts longer than the leading pain reliever

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JTE9 ExampleExample: You are investigating the presence of radon in homes

being built in a new development. If the mean level of radon is greater than 4 then send a warning to all home owners in the development. State the null and alternative hypotheses.

Solutions:

• Ho: The mean level of radon for homes in the development is4 (or less)

• Ha: The mean level of radon for homes in the development is greater than 4

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JTE9

Null Hypothesis

Decision True FalseFail to reject Ho Type A correct decision Type II error

Reject Ho Type I error Type B correct decision

Hypothesis Test Outcomes

Type A correct decision: Null hypothesis true, decide in its favor

Type B correct decision: Null hypothesis false, decide in favor of alternative hypothesis

Type I error: Null hypothesis true, decide in favor of alternative hypothesis

Type II error: Null hypothesis false, decide in favor of null hypothesis

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JTE9 ExampleExample: A calculator company has just received a large shipment of parts used to

make the screens on graphing calculators. They consider the shipment acceptable if the proportion of defective parts is 0.01 (or less). If the proportion of defective parts is greater than 0.01 the shipment is unacceptable and returned to the manufacturer. State the null and alternative hypotheses, and describe the four possible outcomes

and the resulting actions that would occur for this test.

Solutions:

• Ho: The proportion of defective parts is 0.01 (or less)

• Ha: The proportion of defective parts is greater than 0.01

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JTE9

Null Hypothesis Is True:

Type A correct decision

Truth of situation: The proportion of defective parts is 0.01 (or less)

Conclusion: It was determined that the proportion of defective parts is 0.01 (or less)

Action: The calculator company received parts with an acceptable proportion of defectives

Null Hypothesis Is False:

Type II error

Truth of situation: The proportion of defective parts is greater than 0.01

Conclusion: It was determined that the proportion of defective parts is 0.01 (or less)

Action: The calculator company received parts with an unacceptable proportion of defectives

Fail To Reject Ho

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Null hypothesis is true:

Type I error

Truth of situation: The proportion of defectives is 0.01 (or less)

Conclusion: It was determined that the proportion of defectives is greater than 0.01

Action: Send the shipment back to the manufacturer. The proportion of defectives is acceptable

Null hypothesis is false:

Type B correct decision

Truth of situation: The proportion of defectives is greater than 0.01

Conclusion: It was determined that the proportion of defectives is greater than 0.01

Action: Send the shipment back to the manufacturer. The proportion of defectives is unacceptable

Reject Ho

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Correct Decision Type ProbabilityFailure to reject a true Ho A 1 - Rejection of a false Ho B 1 -

ErrorsNotes:

1. The type II error sometimes results in what represents a lost opportunity

2. Since we make a decision based on a sample, there is always the chance of making an error

Probability of a type I error = Probability of a type II error =

Error in Decision Type ProbabilityRejection of a true Ho I Failure to reject a false Ho II

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JTE9 Notes

1. Would like and to be as small as possible

2. and are inversely related

3. Usually set (and don’t worry too much about Why?)

4. Most common values for and are 0.01 and 0.05

5. 1 - : the power of the statistical test

A measure of the ability of a hypothesis test to reject a false null hypothesis

6. Regardless of the outcome of a hypothesis test, we never really know for sure if we have made the correct decision

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error) I (typeP

error) II (typeP

Sample

Size

n

Interrelationship

Interrelationship between the probability of a type I error (), the probability of a type II error (), and the sample size (n)

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JTE9 Level of Significance & Test Statistic

Level of Significance, : The probability of committing thetype I error

Test Statistic: A random variable whose value is calculated from the sample data and is used in making the decision fail to reject Ho or reject Ho

Notes: The value of the test statistic is used in conjunction with a decision rule to determine fail to reject Ho or reject Ho

The decision rule is established prior to collecting the data and specifies how you will reach the decision

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JTE9 The Conclusion

a. If the decision is reject Ho, then the conclusion should be worded something like, “There is sufficient evidence at the level of significance to show that . . . (the meaning of the alternative hypothesis)”

b. If the decision is fail to reject Ho, then the conclusion should be worded something like, “There is not sufficient evidence at the level of significance to show that . . . (the meaning of the alternative hypothesis)”

Notes: The decision is about Ho

The conclusion is a statement about Ha

There is always the chance of making an error

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JTE9 8.4 ~ Hypothesis Test of Mean

( known): A Probability-Value Approach

• The concepts and much of the reasoning behind hypothesis tests are given in the previous sections

• Formalize the hypothesis test procedure as it applies to statements concerning the mean of a population ( known): a probability-value approach

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JTE9 The Assumption...

Hypothesis test:

1. A well-organized, step-by-step procedure used to make a decision

2. Probability-value approach (p-value approach): a procedure that has gained popularity in recent years. Organized into five steps.

The assumption for hypothesis tests about a mean using a known : The sampling distribution of has a normal distributionx

Recall:

1. The distribution of has mean 2. The distribution of has standard deviationx

nx

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JTE9 The Probability-Value Hypothesis Test

A Five-Step Procedure:1. The Set-Up

a. Describe the population parameter of concernb. State the null hypothesis (Ho) and the alternative hypothesis (Ha)

2. The Hypothesis Test Criteriaa. Check the assumptionsb. Identify the probability distribution and the test statistic formula to be usedc. Determine the level of significance,

3. The Sample Evidencea. Collect the sample informationb. Calculate the value of the test statistic

4. The Probability Distributiona. Calculate the p-value for the test statisticb. Determine whether or not the p-value is smaller than

5. The Resultsa. State the decision about Ho

b. State a conclusion about Ha

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JTE9 Example

Example: A company advertises the net weight of its cereal is 24 ounces. A consumer group suspects the boxes are

underfilled. They cannot check every box of cereal, so a sample of cereal boxes will be examined. A decision will be made about the true mean weight based on the sample mean. State the consumer group’s null and

alternative hypotheses. Assume = 0.2

Solution:

1. The Set-Up

a. Describe the population parameter of concern

The population parameter of interest is the mean , the mean weight of the cereal boxes

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b. State the null hypothesis (Ho) and the alternative hypothesis (Ha)

Formulate two opposing statements concerning

Ho: = 24 ( ) (the mean is at least 24)

Ha: < 24 (the mean is less than 24)

Note: The trichotomy law from algebra states that two numerical values must be related in exactly one of three possible relationships: <, =, or >. All three of these possibilities must be accounted for between the two opposing hypotheses in order for the hypotheses to be negations of each other.

Solution Continued

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JTE9

Null Hypothesis Alternative Hypothesis

1. greater than or equal to less than

2. less than or equal to greater than

3. equal to not equal to

)()(

)(

)()()(

Possible Statements of Null & Alternative Hypotheses

Notes: The null hypothesis will be written with just the equal sign (a value is assigned) When equal is paired with less than or greater than, the combined symbol is written beside the

null hypothesis as a reminder that all three signs have been accounted for in these two opposing statements.

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JTE9

Example: A freezer is set to cool food to . If the temperature is higher, the food could spoil, and if the temperature is lower, the freezer is wasting energy. Random freezers are selected and tested as they come off the assembly line. The assembly line is stopped if there is any evidence to suggest improper cooling. State the null and alternative hypotheses.

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Example: An automobile manufacturer claims a new model gets at least 27 miles per gallon. A consumer groups

disputes this claim and would like to show the mean miles per gallon is lower. State the null and alternative hypotheses.

Solution: Ho: = 27 () and Ha: < 27

Solution: Ho: = 10 and Ha: 10

Examples

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JTE9

)( :o H

)( :o H

)( :o H

)( : aH

)( : aH

)( : aH

at least less thanno less than less thannot less than less than

at most more thanno more than more thannot greater than greater than

is is notnot different from different fromsame as not same as

Common Phrases & Their Negations

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2. The Hypothesis Test Criteria

a. Check the assumptions

The weight of cereal boxes is probably mound shaped. A sample size of 40 should be sufficient for the CLT to apply. The sampling distribution of the sample mean can be expected to be normal.

Example Continued: Weight of cereal boxes

Recall: Ho: = 24 () (at least 24) Ha: < 24 (less than 24)

Example Continued

b. Identify the probability distribution and the test statistic to be used

To test the null hypothesis, ask how many standard deviations away from is the sample mean

n

xz

* :statistictest

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4. The Probability Distribution

a. Calculate the p-value for the test statistic

c. Determine the level of significance

Let = 0.05

3. The Sample Evidence

a. Collect the sample information

A random sample of 40 cereal boxes is examined

40 and 95.23 nx

5811.14020.

2495.23*

n

xz

b. Calculate the value of the test statistic ( = 0.2)

Solution Continued

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Probability-Value, or p-Value: The probability that the test statistic could be the value it is or a more extreme value (in the direction of the alternative hypothesis) when the null hypothesis is true (Note: the symbol P will be used to represent the p-value, especially in algebraic situations)

0571.04429.00.5000

)58.1()58.1(*)(P

zPzPzzP

058.1 z

P

Probability-Value or p-Value

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JTE9 Solution Continued

b. Determine whether or not the p-value is smaller than The p-value (0.0571) is greater than (0.05)

5. The Results

Decision Rule:

a. If the p-value is less than or equal to the level of significance , then the decision must be to reject Ho

b. If the p-value is greater than the level of significance , then the decision must be to fail to reject Ho

a. State the decision about Ho

Decision about Ho : Fail to reject Ho

b. Write a conclusion about Ha

There is not sufficient evidence at the 0.05 level of significance to show that the mean weight of cereal boxes is less than 24 ounces

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JTE9 Notes

If we fail to reject Ho, there is no evidence to suggest the null hypothesis is false. This does not mean Ho is true.

The p-value is the area, under the curve of the probability distribution for the test statistic, that is more extreme than the calculated value of the test statistic.

There are 3 separate cases for p-values. The direction (or sign) of the alternative hypothesis is the key.

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JTE9 Finding p-Values

0 z|*| z|*| z

3. Ha contains (Two-tailed)

p-value = P(z < |z*|) + P(z > |z*|)

0 z*z

1. Ha contains > (Right tail)

p-value = P(z > z*)

0 z*z

2. Ha contains < (Left tail)

p-value = P(z < z*)

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JTE9

Example: The mean age of all shoppers at a local jewelry store is 37 years (with a standard deviation of 7 years). In an attempt to attract older adults with more disposable income, the store launched a new advertising campaign. Following the advertising, a random sample of 47 shoppers showed a mean age of 39.3. Is there sufficient evidence to suggest the

advertising campaign has succeeded in attracting older customers?

Solution:

1. The Set-Up

a. Parameter of concern: the mean age, , of all shoppers

b. The hypotheses:

Ho: = 37 ()

Ha: > 37

Example

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2. The Hypothesis Test Criteria

a. The assumptions: The distribution of the age of shoppers is unknown. However, the sample size is large enough for the CLT to apply.

b. The test statistic: The test statistic will be z*

c. The level of significance: none givenWe will find a p-value

Solution Continued

3.39 ,47 xn

25.2477

373.39*

nx

z

3. The Sample Evidence

a. Sample information:

b. Calculated test statistic:

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4. The Probability Distribution

a. The p-value:

b. Determine whether or not the p-value is smaller than A comparison is not possible, no given

0122.0

4878.05000.0

)25.2(

*)(value-

zP

zzPp

0 25.2 z

Solution Continued

5. The Results

Because the p-value is so small (P < 0.05), there is evidence to suggest the mean age of shoppers at the jewelry store is greater than 37

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JTE9 p-Value

The idea of the p-value is to express the degree of belief in the null hypothesis:

1. When the p-value is minuscule (like 0.0001), the null hypothesis would be rejected by everyone because the sample results are very unlikely for a true Ho

2. When the p-value is fairly small (like 0.01), the evidence against Ho is quite strong and Ho will be rejected by many

3. When the p-value begins to get larger (say, 0.02 to 0.08), there is too much probability that data like the sample involved could have occurred even if Ho were true, and the rejection of Ho is not an easy decision

4. When the p-value gets large (like 0.15 or more), the data is not at all unlikely if the Ho is true, and no one will reject Ho

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JTE9 p-Value Advantages & Disadvantage

Advantages of p-value approach:

1. The results of the test procedure are expressed in terms of a continuous probability scale from 0.0 to 1.0, rather than simply on a reject or fail to reject basis

2. A p-value can be reported and the user of the information can decide on the strength of the evidence as it applies to his/her own situation

3. Computers can do all the calculations and report the p-value, thus eliminating the need for tables

Disadvantage:

1. Tendency for people to put off determining the level of significance

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JTE9

Example: The active ingredient for a drug is manufactured using fermentation. The standard process yields a mean of 26.5 grams (assume = 3.2). A new mixing technique during fermentation is implemented. A random sample of 32 batches showed a sample mean 27.1. Is there any evidence to suggest the new mixing technique has changed the yield?

Solution:

1. The Set-Up

a. The parameter of interest is the mean yield of active ingredient, b. The null and alternative hypotheses:

H0: = 26.5

Ha: 26.5

Example

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JTE9

2 The Hypothesis Test Criteria

. a. Assumptions: A sample of size 32 is large enough to satisfy the CLT

b. The test statistic: z*

c. The level of significance: find a p-value

Solution Continued

06.1322.3

5.261.27*

nx

z

3. The Sample Evidence

a. From the sample:

b. The calculated test statistic:

1.27 ,32 xn

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JTE9 Solution Continued

2892.01446.02

)3554.05000.0(2

)06.1(2

|)*|(2value-

zP

zzPp

4. The Probability Distribution

a. The p-value:

b. The p-value is largeThere is no given in the statement of the problem

0 06.1 z

5. The Results

Because the p-value is large (P = 0.2892), there is no evidence to suggest the new mixing technique has changed the mean yield

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JTE9 8.5 ~ Hypothesis Test of mean ( known):

A Classical Approach

• Concepts and reasoning behind hypothesis testing given in Section 8.3

• Formalize the hypothesis test procedure as it applies to statements concerning of a population with known : a classical approach

61

JTE9 The Assumption...

Recall:

1. The distribution of has mean 2. The distribution of has standard deviation

nx

x

Hypothesis Test: A well-organized, step-by-step procedure used to make a decision. The classical approach is the hypothesis test process that has enjoyed popularity for many years.

x

The assumption for hypothesis tests about mean using a known : The sampling distribution of has a normal distribution

62

JTE9 The Classical Hypothesis Test

1. The Set-Upa. Describe the population parameter of concernb. State the null hypothesis (Ho) and the alternative hypothesis (Ha)

2. The Hypothesis Test Criteriaa. Check the assumptionsb. Identify the probability distribution and the test statistic to be usedc. Determine the level of significance,

3. The Sample Evidencea. Collect the sample informationb. Calculate the value of the test statistic

4. The Probability Distributiona. Determine the critical region(s) and critical value(s)b. Determine whether or not the calculated test statistic is in the critical region

5. The Resultsa. State the decision about Ho

b. State the conclusion about Ha

A Five-Step Procedure:

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JTE9

Example: A company advertises the net weight of its cereal is 24 ounces. A consumer group suspects the boxes are underfilled. They cannot check every box of cereal, so a sample of cereal boxes will be examined. A decision will be made about the true mean weight based on the sample mean. State the consumer group’s null and alternative hypotheses. Assume = 0.2

Solution:

1. The Set-Up

a. Describe the population parameter of concern

The population parameter of interest is the mean, , the mean weight of the cereal boxes

Example

64

JTE9

b. State the null hypothesis (Ho) and the alternative hypothesis (Ha)

Formulate two opposing statements concerning the Ho: = 24 ( ) (the mean is at least 24)Ha: < 24 (the mean is less than 24)

Solution Continued

Note: The trichotomy law from algebra states that two numerical values must be related in exactly one of three possible relationships: <, =, or >. All three of these possibilities must be accounted for between the two opposing hypotheses in order for the hypotheses to be negations of each other.

65

JTE9

Null Hypothesis Alternative Hypothesis

1. greater than or equal to less than

2. less than or equal to greater than

3. equal to not equal to

)( )(

)( )()( )(

Possible Statements of Null & Alternative Hypotheses

Notes: The null hypothesis will be written with just the equal sign

(a value is assigned) When equal is paired with less than or greater than, the combined symbol is written beside the null hypothesis as a reminder that all three signs have been accounted for in these two opposing statements

66

JTE9

Example: An automobile manufacturer claims a new model gets at least 27 miles per gallon. A consumer groups disputes

this claim and would like to show the mean miles per gallon is lower. State the null and alternative hypotheses.

Solution: Ho: = 27 () and Ha: < 27

Solution: Ho: = 10 and Ha: 10

Examples

Example: A freezer is set to cool food to . If the temperature is higher, the food could spoil, and if the temperature is lower, the freezer is wasting energy. Random freezers are selected and tested as they come off the assembly line. The assembly line is stopped if there is any evidence to suggest improper cooling. State the null and alternative hypotheses.

10

67

JTE9

)( :o H

)( :o H

)( :o H

)( : aH

)( : aH

)( : aH

at least less thanno less than less thannot less than less than

at most more thanno more than more thannot greater than greater than

is is notnot different from different fromsame as not same as

Common Phrases & Their Negations

68

JTE9 Example Continued Example (continued): Weight of cereal boxes

n

xz

* :statistictest

2. The Hypothesis Test Criteria

a. Check the assumptions

The weight of cereal boxes is probably mound shaped. A sample size of 40 should be sufficient for the CLT to apply. The sampling distribution of the sample mean can be expected to be normal.

Solution Continued:

b. Identify the probability distribution and the test statistic to be used

To test the null hypothesis, ask how many standard deviations away from is the sample mean

Recall: Ho: = 24 (>) (at least 24) Ha: < 24 (less than 24)

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c. Determine the level of significance

Consider the four possible outcomes and their consequencesLet = 0.05

Solution Continued

3. The Sample Evidencea. Collect the sample information

A random sample of 40 cereal boxes is examined

b. Calculate the value of the test statistic ( = 0.2)

40 and 95.23 nx

5811.14020.

2495.23*

n

xz

4. The Probability Distribution

a. Determine the critical region(s) and critical value(s)

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JTE9 Critical Region & Critical Value(s)

Critical Region: The set of values for the test statistic that will cause us to reject the null hypothesis. The set of values that are not in the critical region is called the noncritical region (sometimes called the acceptance region).

Critical Value(s): The first or boundary value(s) of the critical region(s)

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065.1 z

05.0

Critical Region

Critical Value

Critical Region & Critical Value(s)

Illustration:

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5. The Results

We need a decision rule

Solution Continued4. The Probability Distribution (Continued)

b. Determine whether or not the calculated test statistic is in the critical region

065.1 z

The calculated value of z, z* = 1.58, is in the noncritical region

*

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JTE9 Decision Rule

Decision Rule:

a. If the test statistic falls within the critical region, we will reject Ho

(the critical value is part of the critical region)

b. If the test statistic is in the noncritical region, we will fail to reject Ho

a. State the decision about Ho

Decision: Fail to reject Ho

b. State the conclusion about Ha

Conclusion: There is not enough evidence at the 0.05 level of significance to show that the mean weight of cereal boxes is less than 24

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JTE9 Notes

1. The null hypothesis specifies a particular value of a population parameter

2. The alternative hypothesis can take three forms. Each form dictates a specific location of the critical region(s)

Sign in theAlternativeHypothesisCritical Region One region Two regions One region

Left side Half on each side Right sideOne-tailed test Two-tailed test One-tailed test

3. For many hypothesis tests, the sign in the alternative hypothesis points in the direction in which the critical region is located

4. Significance level:

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Example: The mean water pressure in the main water pipe from a town well should be kept at 56 psi. Anything less and several homes will have an insufficient supply, and anything greater could burst the pipe. Suppose the water pressure is checked at 47 random times. The sample mean is 57.1. (Assume

= 7). Is there any evidence to suggest the mean water pressure is different from 56? Use = 0.01

Solution:1. The Set-Up

a. Describe the parameter of concern:The mean water pressure in the main pipe

b. State the null and alternative hypothesesHo: = 56Ha: 56

Example

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2. The Hypothesis Test Criteria

a. Check the assumptions:A sample of n = 47 is large enough for the CLT to apply

b. Identify the test statisticThe test statistic is z*

c. Determine the level of significance: = 0.01 (given)

3. The Sample Evidence

a. The sample information:

b. Calculate the value of the test statistic:

47 ,1.57 nx

077.1477561.57

* n

xz

Solution Continued

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JTE9 Solution Continued

4. The Probability Distribution

a. Determine the critical regions and the critical values

0 58.258.2 z

005.0 005.0

b. Determine whether or not the calculated test statistic is in thecritical region

The calculated value of z, z* = 1.077, is in the noncritical region

*

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5. The Results

a. State the decision about Ho:

Fail to reject Ho

b. State the conclusion about Ha:

There is no evidence to suggest the water pressure is differentfrom 56 at the 0.01 level of significance

Solution Continued

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Example: An elementary school principal claims students receive no more than 30 minutes of homework each night. A random sample of 36 students showed a sample mean of 36.8 minutes spent doing homework (assume = 7.5). Is there any evidence to suggest the mean time spent on homework is greater than 30 minutes? Use = 0.05

Solution:

1. The Set-Up

The parameter of concern: , the mean time spent doing homework each night

Ho: = 30 ()Ha: > 30

Example

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2. The Hypothesis Test Criteria

a. The sample size is n = 36, the CLT applies

b. The test statistic is z*

c. The level of significance is given: = 0.01

Solution Continued

36 ,8.36 nx

44.5365.7

308.36*

n

xz

3. The Sample Evidence

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4. The Probability Distribution

Solution Continued

0 33.2 z

01.0

*

The calculated value of z, z* = 5.44, is in the critical region

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JTE9 Solution Continued

5. The Results

Decision: Reject Ho

Conclusion: There is sufficient evidence at the 0.01 level of significance to conclude the mean time spent on homework by the elementary students is more than 30 minutes

Note: Suppose we took repeated sample of size 36. What would you expect to happen?