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Chapter 8 Introduction to Statistical Inferences

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?

8.1 The Nature of Estimation

- Discuss estimation more precisely.
- What makes a statistic good?
- Assume the population standard deviation, s, is

known throughout this chapter. - Concentrate on learning the procedures for making

statistical inferences about a population mean m.

- Point Estimate for a Parameter
- The value of the corresponding statistic.
- Example
- is a point estimate (single number

value) for the mean m of the sampled population. - Problem
- How good is the point estimate? Is it high? Or

low? Would another sample yield the same

result? - Note
- The quality of an estimation procedure is

enhanced if the sample statistic is both less

variable and unbiased.

- 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.

Negative bias Under estimate High variability

Unbiased On target estimate

Positive bias Over estimate Low variability

- Note
- 1. The sample mean, , is an unbiased statistic

because the mean value of the sampling

distribution is equal to the population mean - 2. Sample means vary from sample to sample. We

dont expect the sample mean to exactly equal the

population mean m. - 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.

- 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 - a
- 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.

- Summary
- To construct a confidence interval for a

population mean m, use the CLT. - Use the point estimate as the central value

of an interval. - 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 - 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.

- Illustration
- The interval
- is an approximate 95 confidence interval for the

population mean m.

Distribution of

8.2 Estimation of Mean m (s Known)

- Formalize the interval estimation process as it

applies to estimating the population mean m based

on a random sample. - Assume the population standard deviation s is

known. - The assumptions are the conditions that need to

exist in order to correctly apply a statistical

procedure.

- The assumption for estimating the mean m using a

known s - The sampling distribution of has a normal

distribution. - Assumption satisfied by 1. Knowing that the

sampling population is normally distributed, or - 2. Using a large enough random sample (CLT).
- Note The CLT may be applied to smaller samples

(for example n 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.

- A (large sample) 1 - a confidence interval for m

is found by - Note
- 1. is the point estimate and the center point

of the confidence interval. - 2. z(a/2) 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 - a.

- 3. standard error of the mean.
- The standard deviation of the distribution of
- 4. maximum error of

estimate E. - One-half the width of the confidence interval

(the product of the confidence coefficient and

the standard error). - 5. lower

confidence limit (LCL). - upper

confidence limit (UCL).

- The Confidence Interval A Five-Step Model
- 1. Describe the population parameter of concern.
- 2. Specify the confidence interval criteria.
- a. Check the assumptions.
- b. Identify the probability distribution and the

formula to - be used.
- c. Determine the level of confidence, 1 - a.
- 3. Collect and present sample information.
- 4. Determine the confidence interval.
- a. Determine the confidence coefficient.
- b. Find the maximum error of estimate.
- c. Find the lower and upper confidence limits.
- 5. State the confidence interval.

- 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 concern.
- The mean, m, weight of all boxes of this cereal.
- 2. Specify the confidence interval criteria.
- a. Check the assumptions.
- The weights are normally distributed. The

distribution of is normal. - b. Identify the probability distribution and

formula to be used. - Use the standard normal variable z with s

0.27.

- c. Determine the level of confidence, 1 - a.
- The question asks for 95 confidence 1 - a

0.95. - 3. Collect and present information.
- The sample information is given in the statement

of the - problem.
- Given
- 4. Determine the confidence interval.
- a. Determine the confidence coefficient.
- The confidence coefficient is found using Table

4B

- b. Find the maximum error of estimate.
- Use the maximum error part of the formula for a

CI. - c. Find the lower and upper confidence limits.
- Use the sample mean and the maximum error
- 5. State the confidence interval.
- 9.75 to 10.00 is a 95 confidence interval for

the true mean weight, m, of cereal boxes.

- 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, m,

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 s 10.5.

- c. The level of confidence 99, or 1 - a

0.99. - 3. Sample information.
- Given n 100 and 72.6.
- 4. The confidence interval.
- a. Confidence coefficient
- b. Maximum error
- c. The lower and upper limits
- 5. Confidence interval With 99 confidence we

can say, - The mean test score is between 69.9 and 75.3.

- 69.9 to 75.3 is a 99 confidence interval for

the true - mean test score.

- Note The confidence is in the process.
- 95 confidence means if we conduct the

experiment over and over, and construct lots of

confidence intervals, then 95 of the confidence

intervals will contain the true mean value m. - 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). - Solve this expression for n

- Example Find the sample size necessary to

estimate a population mean to within .5 with 95

confidence if the standard deviation is 6.2. - Solution
- Therefore, n 591.
- Note When solving for sample size n, always

round up to the next largest integer. (Why?)

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.

- 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 on 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.

- There are two hypotheses involved in making a

decision. - Null Hypothesis, H0
- 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.

- Note
- 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.

- 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. - Solution
- H0 The new pain reliever is no better than the

leading pain reliever. - Ha The new pain reliever lasts longer than the

leading pain reliever.

- Example 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. - Solution
- H0 The mean level of radon for homes in the

development is 4 (or less). - Ha The mean level of radon for homes in the

development is greater than 4.

- 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.

- Example 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. - Solution
- H0 The proportion of defective parts is 0.01 (or

less). - Ha The proportion of defective parts is greater

than 0.01.

- Fail to Reject H0
- 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.

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 H0
- 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

unacceptable.

- Note
- 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 a.
- Probability of a type II error b.

- Note
- 1. Would like a and b to be as small as possible.
- 2. a and b are inversely related.
- 3. Usually set a (and dont worry too much about

b. Why?) - 4. Most common values for a and b are 0.01 and

0.05. - 5. 1 - b 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.

- Interrelationship between the probability of a

type I error (a), the probability of a type II

error (b), and the sample size (n).

- Level of Significance a
- The probability of committing the type 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 H0 or reject H0. - Note
- 1. The value of the test statistic is used in

conjunction with a decision rule to determine

fail to reject H0 or reject H0. - 2. The decision rule is established prior to

collecting the data and specifies how you will

reach the decision.

- The Conclusion
- a. If the decision is reject H0, then the

conclusion should be worded something like,

There is sufficient evidence at the a level of

significance to show that . . . (the meaning of

the alternative hypothesis). - b. If the decision is fail to reject H0, then the

conclusion should be worded something like,

There is not sufficient evidence at the a level

of significance to show that . . . (the meaning

of the alternative hypothesis). - Note
- 1. The decision is about H0.
- 2. The conclusion is a statement about Ha.
- 3. There is always the chance of making an error.

8.4 Hypothesis Test of Mean m (s 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 m of a

population (s known) a probability-value

approach.

- The assumption for hypothesis tests about a mean

m using a known s - The sampling distribution of has a normal

distribution. - Recall
- 1. The distribution of has mean m.
- 2. The distribution of has standard deviation
- 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 Probability-Value Hypothesis Test A

Five-Step Procedure - 1. The Set-Up
- a. Describe the population parameter of concern.
- b. State the null hypothesis (H0) and the

alternative hypothesis (Ha). - 2. The Hypothesis Test Criteria
- a. Check the assumptions.
- b. Identify the probability distribution and the

test statistic formula to be used. - c. Determine the level of significance, a.
- 3. The Sample Evidence
- a. Collect the sample information.
- b. Calculate the value of the test statistic.
- 4. The Probability Distribution
- a. Calculate the p-value for the test statistic.
- b. Determine whether or not the p-value is

smaller than a. - 5. The Results
- a. State the decision about H0.
- b. State a conclusion about Ha.

- Example A company advertises the net weight of

its cereal is 24 ounces. A consumer group would

like to check this claim. 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 groups null and alternative

hypotheses. Assume s .2. - Solution
- 1. The Set-Up
- a. Describe the population parameter of concern.
- The population parameter of interest is the

mean m, the mean weight of the cereal boxes.

- b. State the null hypothesis (H0) and the

alternative hypothesis (Ha). - Formulate two opposing statements concerning m.
- H0 m 24 ( ) (the mean is at least 24)
- Ha m lt 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 lt, , or gt. 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.

- Possible Statements of Null and Alternative

Hypotheses - Note
- 1. The null hypothesis will be written with just

the equal sign (a value is assigned). - 2. 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.

- 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 H0 m 27 (³) and Ha m lt 27
- 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. - Solution H0 m 10 and Ha m ¹ 10

- Common Phrases and Their Negations

- Example (continued) Weight of cereal boxes.
- Recall H0 m 24 (³) (at least 24) Ha m lt

24 (less than 24) - 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. - b. Identify the probability distribution and the

test statistic to be used. - To test the null hypothesis, ask how many

standard deviations away from m is the sample

mean.

- c. Determine the level of significance.
- Consider the four possible outcomes and their

consequences. Let a 0.05. - 3. The Sample Evidence
- a. Collect the sample information.
- A random sample of 40 cereal boxes is examined.
- b. Calculate the value of the test statistic.

(s .2) - 4. The Probability Distribution
- a. Calculate the p-value for the test statistic.

- 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.)

P

- b. Determine whether or not the p-value is

smaller than a. - The p-value (0.0571) is greater than a (0.05).
- 5. The Results
- Decision Rule
- a. If the p-value is less than or equal to the

level of significance a, then the decision must

be to reject H0. - b. If the p-value is greater than the level of

significance a, then the decision must be to fail

to reject H0. - a. State the decision about H0.
- Decision about H0 Fail to reject H0.
- 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.

- Note
- 1. If we fail to reject H0, there is no evidence

to suggest the null hypothesis is false. This

does not mean H0 is true. - 2. 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. - 3. There are 3 separate cases for p-values. The

direction (or sign) of the alternative hypothesis

is the key.

- Finding p-values
- 1. Ha contains gt (Right tail)
- p-value P(z gt z)
- 2. Ha contains lt (Left tail)
- p-value P(z lt z)
- 3. Ha contains (Two-tailed)
- p-value
- P(z lt -z) P(z gt z)

- 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, m, of all

shoppers. - b. The hypotheses
- H0 m 37 ()
- Ha m gt 37

- 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 given. We

will find a p- value. - 3. The Sample Evidence
- a. Sample information
- b. Calculated test statistic

- 4. The Probability Distribution
- a. The p-value
- b. Determine whether or not the p-value is

smaller than a. - A comparison is not possible, no a given.
- 5. The Results
- Because the p-value is so small (P lt 0.05),

there is evidence to suggest the mean age of

shoppers at the jewelry store is greater than 37.

- 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 H0.
- 2. When the p-value is fairly small (like 0.01),

the evidence - against H0 is quite strong and H0 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 H0 were

true, and the - rejection of H0 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 H0 is true, and no

one will reject - H0.

- 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
- Tendency for people to put off determining the

level of significance.

- Example The active ingredient for a drug is

manufactured using fermentation. The standard

process yields a mean of 26.5 grams (assume s

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, m. - b. The null and alternative hypotheses
- H0 m 26.5
- Ha m ¹ 26.5

- 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.
- 3. The Sample Evidence
- a. From the sample
- b. The calculated test statistic

- 4. The Probability Distribution
- a. The p-value
- b. The p-value is large. There is no a given in

the statement of the problem. - Note Suppose we took repeated samples of size

32. - 1. What results would you expect?
- 2. What does the p-value really measure?

- 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.

8.5 Hypothesis Test of mean m (s known) A

Classical Approach

- Concepts and reasoning behind hypothesis testing

given in previous section. - Formalize the hypothesis test procedure as it

applies to statements concerning m of a

population with known s a classical approach.

- The assumption for hypothesis tests about mean m

using a known s - The sampling distribution of has a normal

distribution. - Recall
- 1. The distribution of has mean m.
- 2. The distribution of has standard deviation

- Hypothesis test
- 1. A well-organized, step-by-step procedure used

to make a decision. - 2. The classical approach is the hypothesis test

process that has enjoyed popularity for many

years.

- The Classical Hypothesis Test A Five-Step

Procedure - 1. The Set-Up
- a. Describe the population parameter of concern.
- b. State the null hypothesis (H0) and the

alternative hypothesis (Ha). - 2. The Hypothesis Test Criteria
- a. Check the assumptions.
- b. Identify the probability distribution and the

test statistic to be used. - c. Determine the level of significance, a.
- 3. The Sample Evidence
- a. Collect the sample information.
- b. Calculate the value of the test statistic.
- 4. The Probability Distribution
- a. 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 Results
- a. State the decision about H0.
- b. State the conclusion about Ha.

- Example A company advertises the net weight of

its cereal is 24 ounces. A consumer group would

like to check this claim. 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 groups null and alternative

hypotheses. Assume s .2. - Solution
- 1. The Set-Up
- a. Describe the population parameter of concern.
- The population parameter of interest is the

mean, m, the mean weight of the cereal boxes.

- b. State the null hypothesis (H0) and the

alternative hypothesis (Ha). - Formulate two opposing statements concerning

the m. - H0 m 24 ( ) (the mean is at least 24)
- Ha m lt 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 lt, , or gt. 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.

- Possible Statements of Null and Alternative

Hypotheses - Note
- 1. The null hypothesis will be written with just

the equal sign (a value is assigned). - 2. 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.

- 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 H0 m 27 (³) and Ha m lt 27
- 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. - Solution H0 m 10 and Ha m ¹ 10

- Common Phrases and Their Negations

- Example (continued) Weight of cereal boxes.
- Recall H0 m 24 (³) (at least 24) Ha m lt

24 (less than 24) - 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. - b. Identify the probability distribution and the

test statistic to be used. - To test the null hypothesis, ask how many

standard deviations away from m is the sample

mean.

- c. Determine the level of significance.
- Consider the four possible outcomes and their

consequences. Let a 0.05. - 3. The Sample Evidence
- a. Collect the sample information.
- A random sample of 40 cereal boxes is examined.
- b. Calculate the value of the test statistic.

(s .2)

- 4. The Probability Distribution
- a. Determine the critical region(s) and 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).

- Illustration
- Critical Region and Critical Value(s).

Critical Value

Critical Region

b. Determine whether or not the calculated test

statistic is in the critical region.

Location of z

The calculated value of z, z -1.58, is in

the noncritical region. 5. The Results We

need a decision rule.

- Decision Rule
- a. If the test statistic falls within the

critical region, we will reject H0. (The

critical value is part of the critical region.) - b. If the test statistic is in the noncritical

region, we will fail to reject H0. - a. State the decision about H0.
- Decision Fail to reject H0.
- 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.

- Note
- 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). - 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 a

- 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 s 7.) Is there any evidence

to suggest the mean water pressure is different

from 56? Use a 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 hypotheses.
- H0 m 56
- Ha m ¹ 56

- 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 statistic.
- The test statistic is z.
- c. Determine the level of significance a 0.01

(given) - 3. The Sample Evidence
- a. The sample information
- b. Calculate the value of the test statistic

- 4. The Probability Distribution
- a. Determine the critical regions and the

critical values. - b. Determine whether or not the calculated test

statistic is in the critical region. - The calculated value of z, z 1.077, is in

the noncritical region.

- 5. The Results
- a. State the decision about H0.
- Fail to reject H0.
- c. State the conclusion about Ha.
- There is no evidence to suggest the water

pressure is different from 56.

- 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 s 7.5). Is there

any evidence to suggest the mean time spent on

homework is greater than 30 minutes? Use a

0.05. - Solution
- 1. The parameter of concern m, the mean time

spent doing homework each night. - H0 m 30 ()
- Ha m gt 30

- 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 a 0.01.
- 3. The Sample Evidence

- 4. The Probability Distribution
- The calculated value of z, z 5.44, is in the

critical region.

- 5. The Results
- Decision Reject H0.
- 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?