Chapter 8: Introduction to Statistical Inferences - PowerPoint PPT Presentation

Loading...

PPT – Chapter 8: Introduction to Statistical Inferences PowerPoint presentation | free to download - id: 7c0d3c-NmIwO



Loading


The Adobe Flash plugin is needed to view this content

Get the plugin now

View by Category
About This Presentation
Title:

Chapter 8: Introduction to Statistical Inferences

Description:

Chapter 8: Introduction to ... Find the lower and upper confidence limits. 5. ... Example: A company advertises the net weight of its cereal is 24 ounces. – PowerPoint PPT presentation

Number of Views:70
Avg rating:3.0/5.0
Slides: 84
Provided by: Author544
Learn more at: http://facweb.bhc.edu
Category:

less

Write a Comment
User Comments (0)
Transcript and Presenter's Notes

Title: Chapter 8: Introduction to Statistical Inferences


1
Chapter 8 Introduction to Statistical Inferences
2
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?

3
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.

4
  • 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.

5
  • 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.

6
Negative bias Under estimate High variability
Unbiased On target estimate
Positive bias Over estimate Low variability
7
  • 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.

8
  • 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.

9
  • 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.

10
  • Illustration
  • The interval
  • is an approximate 95 confidence interval for the
    population mean m.

Distribution of
11
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.

12
  • The assumption for estimating the mean m using a
    known s
  • The sampling distribution of has a normal
    distribution.
  • Assumption satisfied by1. 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.

13
  • 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.

14
  • 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).

15
  • 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.

16
  • 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.

17
  • 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

18
  • 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.

19
  • 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.

20
  • 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.

21
  • 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

22
  • 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?)

23
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.

24
  • 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.

25
  • 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.

26
  • 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.

27
  • 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.

28
  • 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.

29
  • 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.

30
  • 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.

31
  • 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.
32
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.

33
  • 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.

34
  • 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.

35
  • Interrelationship between the probability of a
    type I error (a), the probability of a type II
    error (b), and the sample size (n).

36
  • 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.

37
  • 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.

38
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.

39
  • 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.

40
  • 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.

41
  • 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.

42
  • 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.

43
  • 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.

44
  • 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

45
  • Common Phrases and Their Negations

46
  • 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.

47
  • 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.

48
  • 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
49
  • 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.

50
  • 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.

51
  • 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)

52
  • 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

53
  • 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

54
  • 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.

55
  • 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.

56
  • 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.

57
  • 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

58
  • 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

59
  • 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?

60
  • 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.

61
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.

62
  • 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.

63
  • 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.

64
  • 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.

65
  • 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.

66
  • 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.

67
  • 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

68
  • Common Phrases and Their Negations

69
  • 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.

70
  • 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)

71
  • 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).

72
  • Illustration
  • Critical Region and Critical Value(s).

Critical Value
Critical Region
73
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.
74
  • 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.

75
  • 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

76
  • 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

77
  • 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

78
  • 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.

79
  • 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.

80
  • 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

81
  • 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

82
  • 4. The Probability Distribution
  • The calculated value of z, z 5.44, is in the
    critical region.

83
  • 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?
About PowerShow.com