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  • Statistics for Business and Economics

    Chapter 7Inferences Based on a Single Sample:

    Tests of Hypotheses

    2011 Pearson Education, Inc 1

    Content

    1. The Elements of a Test of Hypothesis

    2. Formulating Hypotheses and Setting Up the Rejection Region

    3. Observed Significance Levels: p-Values4. Test of Hypothesis about a Population Mean:

    Normal (z) Statistic5. Test of Hypothesis about a Population Mean:

    Students t-Statistic

    2011 Pearson Education, Inc 2

    Content

    6. Large-Sample Test of Hypothesis about a Population Proportion

    7. Test of Hypothesis about a Population Variance

    8. Calculating Type II Error Probabilities: More about b*

    2011 Pearson Education, Inc 3

    Learning Objectives

    1. Test a specific value of a population parameter (mean or proportion), called a test of hypothesis

    2. Provide a measure of reliability for the hypothesis test, called the significance level of the test

    2011 Pearson Education, Inc 4

  • Learning Objectives

    3. Test a specific value of a population parameter (mean, proportion or variance) called a test of hypothesis

    4. Show how to estimate the reliability of a test

    2011 Pearson Education, Inc 5

    7.1

    The Elements ofa Test of Hypothesis

    2011 Pearson Education, Inc 6

    Hypothesis Testing

    2011 Pearson Education, Inc 7

    Population

    JJ

    J

    J

    J

    JJ

    I believe the population mean age is 50 (hypothesis).

    Mean`X = 20

    Random sample

    J

    J J

    Reject hypothesis! Not close.

    J

    Whats a Hypothesis?

    A statistical hypothesis is a statement about the numerical value of a population parameter.

    2011 Pearson Education, Inc 8

    I believe the mean GPA of this class is 3.5!

    1984-1994 T/Maker Co.

  • Null Hypothesis

    The null hypothesis, denoted H0, represents the hypothesis that will be accepted unless the data provide convincing evidence that it is false. This usually represents the status quo or some claim about the population parameter that the researcher wants to test.

    2011 Pearson Education, Inc 9

    Alternative Hypothesis

    The alternative (research) hypothesis, denoted Ha, represents the hypothesis that will be accepted only if the data provide convincing evidence of its truth. This usually represents the values of a population parameter for which the researcher wants to gather evidence to support.

    2011 Pearson Education, Inc 10

    Alternative Hypothesis1. Opposite of null hypothesis2. The hypothesis that will be accepted only if

    the data provide convincing evidence of its truth

    3. Designated Ha4. Stated in one of the following forms

    Ha: m (some value)Ha: m < (some value)Ha: m > (some value)

    2011 Pearson Education, Inc 11

    Identifying Hypotheses

    Example problem: Test that the population mean is not 3Steps:

    State the question statistically (m 3) State the opposite statistically (m = 3)

    Must be mutually exclusive & exhaustive Select the alternative hypothesis (m 3)

    Has the , sign State the null hypothesis (m = 3)

    2011 Pearson Education, Inc 12

  • What Are the Hypotheses?

    State the question statistically: m = 12

    State the opposite statistically: m 12

    Select the alternative hypothesis: Ha: m 12

    State the null hypothesis: H0: m = 12

    2011 Pearson Education, Inc 13

    Is the population average amount of TV viewing 12 hours?

    What Are the Hypotheses?

    State the question statistically: m 12

    State the opposite statistically: m = 12

    Select the alternative hypothesis: Ha: m 12

    State the null hypothesis: H0: m = 12

    2011 Pearson Education, Inc 14

    Is the population average amount of TV viewing different from 12 hours?

    What Are the Hypotheses?

    State the question statistically: m 20

    State the opposite statistically: m > 20

    Select the alternative hypothesis: Ha: m > 20

    State the null hypothesis: H0: m = 20

    2011 Pearson Education, Inc 15

    Is the average cost per hat less than or equal to $20?

    What Are the Hypotheses?

    State the question statistically: m > 25

    State the opposite statistically: m 25

    Select the alternative hypothesis: Ha: m > 25

    State the null hypothesis: H0: m 25

    2011 Pearson Education, Inc 16

    Is the average amount spent in the bookstore greater than $25?

  • Test Statistic

    The test statistic is a sample statistic, computed from information provided in the sample, that the researcher uses to decide between the null and alternative hypotheses.

    2011 Pearson Education, Inc 17

    Test Statistic - ExampleThe sampling distribution of assuming = 2,400. The chance of observing more than 1.645 standard deviations above 2,400 is only .05 if in fact the true mean is 2,400.

    2011 Pearson Education, Inc 18

    Type I Error

    A Type I error occurs if the researcher rejects the null hypothesis in favor of the alternative hypothesis when, in fact, H0 is true.

    The probability of committing a Type I error is denoted by a.

    2011 Pearson Education, Inc 19

    Rejection Region

    The rejection region of a statistical test is the set of possible values of the test statistic for which the researcher will reject H0 in favor of Ha.

    2011 Pearson Education, Inc 20

  • Type II Error

    A Type II error occurs if the researcher accepts the null hypothesis when, in fact, H0 is false.

    The probability of committing a Type II error is denoted by b.

    2011 Pearson Education, Inc 21

    Conclusions and Consequences for a Test of Hypothesis

    2011 Pearson Education, Inc 22

    True State of NatureConclusion H0 True Ha TrueAccept H0(Assume H0 True)

    Correct decision Type II error (probability b)

    Reject H0(Assume Ha True)

    Type I error (probability a)

    Correct decision

    Elements of a Test of Hypothesis

    1. Null hypothesis (H0): A theory about the specific values of one or more

    population parameters. The theory generally represents the status quo, which

    we adopt until it is proven false.

    2011 Pearson Education, Inc 23

    2. Alternative (research) hypothesis (Ha): - A theory that contradicts the null hypothesis. - The theory generally represents that which we will

    adopt only when sufficient evidence exists to establish its truth.

    Elements of a Test of Hypothesis

    3. Test statistic: A sample statistic used to decide whether to reject the null hypothesis.

    2011 Pearson Education, Inc 24

    4. Rejection region: The numerical values of the test statistic for which the null

    hypothesis will be rejected. The rejection region is chosen so that the probability is a

    that it will contain the test statistic when the null hypothesis is true, thereby leading to a Type I error.

    The value of a is usually chosen to be small (e.g., .01, .05, or .10) and is referred to as the level of significance of the test.

  • Elements of a Test of Hypothesis

    5. Assumptions: Clear statement (s) of any assumptions made about the population (s) being sampled.

    2011 Pearson Education, Inc 25

    6. Experiment and calculation of test statistic: Performance of the sampling experiment and determination of the numerical value of the test statistic.

    Elements of a Test of Hypothesis

    7. Conclusion:

    a. If the numerical value of the test statistic falls in the rejection region, we reject the null hypothesis and conclude that the alternative hypothesis is true. We know that the hypothesis-testing process will lead to this conclusion incorrectly (Type I error) only 100a% of the time when H0 is true.

    2011 Pearson Education, Inc 26

    Elements of a Test of Hypothesis

    7. Conclusion:

    b. If the test statistic does not fall in the rejection region, we do not reject H0. Thus, we reserve judgment about which hypothesis is true. We do not conclude that the null hypothesis is true because we do not (in general) know the probability b that our test procedure will lead to an incorrect acceptance of H0 (Type II error).

    2011 Pearson Education, Inc 27

    Determining theTarget Parameter

    2011 Pearson Education, Inc 28

    Parameter Key Words or Phrases Type of Data Mean; average Quantitativep Proportion; percentage;

    fraction; rateQualitative

    s2 Variance; variability; spread

    Quantitative

  • 7.2

    Formulating Hypotheses and Setting Up the Rejection Region

    2011 Pearson Education, Inc 29

    Steps for Selecting the Null and Alternative Hypotheses

    1. Select the alternative hypothesis as that which the sampling experiment is intended to establish. The alternative hypothesis will assume one of three forms:a. One-tailed, upper-tailed (e.g., Ha: > 2,400) b. One-tailed, lower-tailed (e.g., Ha: < 2,400) c. Two-tailed (e.g., Ha: 2,400)

    2011 Pearson Education, Inc 30

    Steps for Selecting the Null and Alternative Hypotheses

    2. Select the null hypothesis as the status quo, that which will be presumed true unless the sampling experiment conclusively establishes the alternative hypothesis. The null hypothesis will be specified as that parameter value closest to the alternative in one-tailed tests and as the complementary (or only unspecified) value in two-tailed tests.

    (e.g., H0: = 2,400)

    2011 Pearson Education, Inc 31

    One-Tailed Test

    A one-tailed test of hypothesis is one in which the alternative hypothesis is directional and includes the symbol < or >.

    2011 Pearson Education, Inc 32

  • Two-Tailed Test

    A two-tailed test of hypothesis is one in which the alternative hypothesis does not specify departure from H0 in a particular direction and is written with the symbol .

    2011 Pearson Education, Inc 33

    Basic Idea

    2011 Pearson Education, Inc 34

    Sample Meansm = 50H0

    Sampling DistributionIt is unlikely that we would get a sample mean of this value ...

    20

    ... if in fact this werethe population mean

    ... therefore, we reject the hypothesis that m = 50.

    H0: = 50

    Ha: < 50

    Rejection Region (One-Tail Test)

    2011 Pearson Education, Inc 35

    HoValueCritical

    Value

    a

    Sample Statistic

    RejectionRegion

    Fail to RejectRegion

    Sampling Distribution

    1 a

    Level of Confidence

    H0: = 0Ha: < 0

    Rejection Regions (Two-Tailed Test)

    2011 Pearson Education, Inc 36

    HoValue Critical

    ValueCriticalValue

    1/2 a1/2 a

    Sample Statistic

    RejectionRegion

    RejectionRegion

    Fail to RejectRegion

    Sampling Distribution

    1 a

    Level of ConfidenceH0: = 0Ha: 0

  • Rejection Regions

    2011 Pearson Education, Inc 37

    Alternative Hypotheses

    Lower-Tailed

    Upper-Tailed

    Two-Tailed

    a = .10 z < 1.28 z > 1.28 z < 1.645 or z > 1.645

    a = .05 z < 1.645 z > 1.645 z < 1.96 or z > 1.96

    a = .01 z < 2.33 z > 2.33 z < 2.575 or z > 2.575

    7.3

    Observed Significance Levels:p-Values

    2011 Pearson Education, Inc 38

    p-ValueThe observed significance level, or p-value, for a specific statistical test is the probability (assuming H0 is true) of observing a value of the test statistic that is at least as contradictory to the null hypothesis, and supportive of the alternative hypothesis, as the actual one computed from the sample data.

    2011 Pearson Education, Inc 39

    p-Value Probability of obtaining a test statistic more

    extreme ( or ) than actual sample value, given H0 is true

    Called observed level of significance Smallest value of a for which H0 can be

    rejected

    Used to make rejection decision If p-value a, do not reject H0 If p-value < a, reject H0

    2011 Pearson Education, Inc 40

  • Steps for Calculating the p-Value for a Test of Hypothesis

    1. Determine the value of the test statistic zcorresponding to the result of the sampling experiment.

    2011 Pearson Education, Inc 41

    Steps for Calculating the p-Value for a Test of Hypothesis

    2a. If the test is one-tailed, the p-value is equal to the tail area beyond z in the same direction as the alternative hypothesis. Thus, if the alternative hypothesis is of the form > , the p-value is the area to the right of, or above, the observed z-value. Conversely, if the alternative is of the form < , the p-value is the area to the left of, or below, the observed z-value.

    2011 Pearson Education, Inc 42

    Steps for Calculating the p-Value for a Test of Hypothesis

    2b. If the test is two-tailed, the p-value is equal to twice the tail area beyond the observed z-value in the direction of the sign of z that is, if z is positive, the p-value is twice the area to the right of, or above, the observed z-value. Conversely, if z is negative, the p-value is twice the area to the left of, or below, the observed z-value.

    2011 Pearson Education, Inc 43

    Reporting Test Results asp-Values: How to Decide Whether

    to Reject H0

    1. Choose the maximum value of a that you are willing to tolerate.

    2. If the observed significance level (p-value) of the test is less than the chosen value of a, reject the null hypothesis. Otherwise, do not reject the null hypothesis.

    2011 Pearson Education, Inc 44

  • Two-Tailed z Test p-Value Example

    Does an average box of cereal contain 368 grams of cereal? A random sample of 64 boxes showed x = 372.5. The company has specified s to be 25 grams. Find the p-Value.

    2011 Pearson Education, Inc 45

    368 gm.

    Two-Tailed z Test p-Value Solution

    2011 Pearson Education, Inc 46

    z0 1.44z value of sample statistic (observed)

    Two-Tailed Z Test p-Value Solution

    2011 Pearson Education, Inc 47

    1/2 p-Value1/2 p-Value

    z value of sample statistic (observed)

    p-value is P(z 1.44 or z 1.44)

    z0 1.441.44

    From z table: lookup 1.44

    .4251

    .5000 .4251

    .0749

    Two-Tailed z Test p-Value Solution

    2011 Pearson Education, Inc 48

    1/2 p-Value.0749

    1/2 p-Value.0749

    p-value is P(z 1.50 or z 1.50) = .1498

    z0 1.441.44

  • Two-Tailed z Test p-Value Solution

    2011 Pearson Education, Inc 49

    0 1.441.44 z

    Reject H0Reject H0

    1/2 p-Value = .07491/2 p-Value = .0749

    1/2 a = .0251/2 a = .025

    p-Value = .1498 a = .05 Do not reject H0.

    Test statistic is in Do not reject region

    One-Tailed z Test p-Value Example

    Does an average box of cereal contain more than 368 grams of cereal? A random sample of 64 boxes showed x = 372.5. The company has specified sto be 25 grams. Find the p-Value. How does it compare to a = .05?

    2011 Pearson Education, Inc 50

    368 gm.

    One-Tailed z Test p-Value Solution

    2011 Pearson Education, Inc 51

    z0 1.44

    z value of sample statistic

    One-Tailed z Test p-Value Solution

    2011 Pearson Education, Inc 52

    Use alternative hypothesis to find direction

    p-Value is P(z 1.44)

    z value of sample statistic

    p-Value

    z0 1.44From z table: lookup 1.44

    .4251

    .5000 .4251

    .0749

  • One-Tailed z Test p-Value Solution

    2011 Pearson Education, Inc 53

    p-Value.0749

    z value of sample statistic

    From z table: lookup 1.44

    Use alternative hypothesis to find direction

    .5000 .4251

    .0749

    p-Value is P(z 1.44) = .0749

    z0 1.44

    .4251

    One-Tailed z Test p-Value Solution

    2011 Pearson Education, Inc 54

    a = .05

    0 1.44 z

    Reject H0

    p-Value = .0749

    (p-Value = .0749) (a = .05). Do not reject H0.

    Test statistic is in Do not reject region

    p-Value Thinking Challenge

    Youre an analyst for Ford. You want to find out if the average miles per gallon of Escorts is less than 32 mpg. Similar models have a standard deviation of 3.8mpg. You take a sample of 60Escorts & compute a sample mean of 30.7 mpg. What is the value of the observed level of significance (p-Value)?

    2011 Pearson Education, Inc 55

    p-Value Solution*

    2011 Pearson Education, Inc 56

    Use alternative hypothesis to find direction

    z02.65

    z value of sample statistic

    From z table: lookup 2.65

    .4960

    p-Value.004

    .5000 .4960

    .0040

    p-Value is P(z -2.65) = .004.p-Value < (a = .01). Reject H0.

  • Converting a Two-Tailedp-Value from a Printout to a One-

    Tailed p-Value

    if Ha is of the form > and z is positiveor Ha is of the form < and z is negative

    2011 Pearson Education, Inc 57

    if Ha is of the form > and z is negativeHa is of the form < and z is positive

    7.4

    Test of Hypotheses about a Population Mean:Normal (z) Statistic

    2011 Pearson Education, Inc 58

    Large-Sample Test of Hypothesis about

    One-Tailed Test Two-Tailed TestH0: = 0 H0: = 0Ha: < 0 Ha: 0

    (or Ha: > 0)

    Test Statistic: Test Statistic:

    2011 Pearson Education, Inc 59

    Large-Sample Test of Hypothesis about

    One-Tailed TestRejection region:

    z < za(or z > za when Ha: > 0)

    where za is chosen so thatP(z > za) = a

    2011 Pearson Education, Inc 60

  • Large-Sample Test of Hypothesis about

    Two-Tailed TestRejection region:

    |z| > za/2where za/2 is chosen so that

    P(|z| > za/2) = a/2

    Note: 0 is the symbol for the numerical value assigned to under the null hypothesis.

    2011 Pearson Education, Inc 61

    Conditions Required for a Valid Large-Sample Hypothesis Test for

    1. A random sample is selected from the target population.

    2. The sample size n is large (i.e., n 30). (Due to the Central Limit Theorem, this condition guarantees that the test statistic will be approximately normal regardless of the shape of the underlying probability distribution of the population.)

    2011 Pearson Education, Inc 62

    Possible Conclusions for a Test of Hypothesis

    1. If the calculated test statistic falls in the rejection region, reject H0 and conclude that the alternative hypothesis Ha is true. State that you are rejecting H0 at the a level of significance. Remember that the confidence is in the testing process, not the particular result of a single test.

    2011 Pearson Education, Inc 63

    Possible Conclusions for a Test of Hypothesis

    2. If the test statistic does not fall in the rejection region, conclude that the sampling experiment does not provide sufficient evidence to reject H0 at the a level of significance. [Generally, we will not acceptthe null hypothesis unless the probability b of a Type II error has been calculated.]

    2011 Pearson Education, Inc 64

  • Two-Tailed z Test Example

    Does an average box of cereal contain 368 grams of cereal? A random sample of 25 boxes had x = 372.5. The company has specified s to be 25 grams. Test at the .05level of significance.

    2011 Pearson Education, Inc 65

    368 gm.

    Two-Tailed z Test Solution

    H0: Ha: a = n = Critical Value(s):

    2011 Pearson Education, Inc 66

    Test Statistic:

    Decision:

    Conclusion:

    m = 368m 368.0525

    z0 1.961.96

    .025

    Reject H 0 Reject H0.025

    Do not reject at a = .05

    No evidence average is not 368

    Two-Tailed z Test Thinking Challenge

    Youre a Q/C inspector. You want to find out if a new machine is making electrical cords to customer specification: average breaking strength of 70 lb. with s = 3.5 lb. You take a sample of 36 cords & compute a sample mean of 69.7 lb. At the .05 level of significance, is there evidence that the machine is notmeeting the average breaking strength?

    2011 Pearson Education, Inc 67

    Two-Tailed z Test Solution*

    H0: Ha: a = n = Critical Value(s):

    2011 Pearson Education, Inc 68

    Test Statistic:

    Decision:

    Conclusion:

    m = 70m 70.05

    36

    z0 1.961.96

    .025Reject H 0 Reject H0

    .025Since -1.96 < z = -0.51

  • One-Tailed z TestExample

    Does an average box of cereal contain more than 368 grams of cereal? A random sample of 25 boxes showed x = 372.5. The company has specified s to be 25 grams. Test at the .05level of significance.

    2011 Pearson Education, Inc 69

    368 gm.

    One-Tailed z Test Solution

    H0: Ha: a = n = Critical Value(s):

    2011 Pearson Education, Inc 70

    Test Statistic:

    Decision:

    Conclusion:

    m = 368m > 368.0525

    z0 1.645

    .05

    Reject Do not reject at a = .05

    No evidence average is more than 368

    One-Tailed z Test Thinking Challenge

    Youre an analyst for Ford. You want to find out if the average miles per gallon of Escorts is at least 32 mpg. Similar models have a standard deviation of 3.8 mpg. You take a sample of 60 Escorts & compute a sample mean of 30.7mpg. At the .01 level of significance, is there evidence that the miles per gallon is less than 32?

    2011 Pearson Education, Inc 71

    One-Tailed z Test Solution*

    H0: Ha: a = n = Critical Value(s):

    2011 Pearson Education, Inc 72

    Test Statistic:

    Decision:

    Conclusion:

    m = 32m < 32.0160

    z0-2.33

    .01

    Reject Reject at a = .01

    There is evidence average is less than 32

  • 7.5

    Test of Hypothesis about a Population Mean:

    Students t-Statistic

    2011 Pearson Education, Inc 73

    Small-Sample Test of Hypothesis about

    One-Tailed TestH0: = 0Ha: < 0 (or Ha: > 0)

    Test statistic:

    Rejection region: t < ta(or t > ta when Ha: > 0)where ta and ta/2 are based on (n 1) degrees of

    freedom 2011 Pearson Education, Inc 74

    Small-Sample Test of Hypothesis about

    Two-Tailed TestH0: = 0Ha: 0

    Test statistic:

    Rejection region: |t| > ta/2

    2011 Pearson Education, Inc 75

    Conditions Required for a Valid Small-Sample Hypothesis Test for

    1. A random sample is selected from the target population.

    2. The population from which the sample is selected has a distribution that is approximately normal.

    2011 Pearson Education, Inc 76

  • Two-Tailed t TestExample

    Does an average box of cereal contain 368 grams of cereal? A random sample of 36 boxes had a mean of 372.5 and a standard deviation of 12 grams. Test at the .05 level of significance.

    2011 Pearson Education, Inc 77

    368 gm.

    Two-Tailed t Test Solution

    H0: Ha: a = df = Critical Value(s):

    2011 Pearson Education, Inc 78

    Test Statistic:

    Decision:

    Conclusion:

    m = 368m 368.0536 1 = 35

    t0 2.030-2.030

    .025Reject H0 Reject H0

    .025Reject at a = .05

    There is evidence population average is not 368

    Two-Tailed t TestThinking Challenge

    You work for the FTC. A manufacturer of detergent claims that the mean weight of detergent is 3.25 lb. You take a random sample of 64containers. You calculate the sample average to be 3.238 lb. with a standard deviation of .117 lb. At the .01 level of significance, is the manufacturer correct?

    2011 Pearson Education, Inc 79

    3.25 lb.

    Two-Tailed t Test Solution*

    H0: Ha: a = df = Critical Value(s):

    2011 Pearson Education, Inc 80

    Test Statistic:

    Decision:

    Conclusion:

    m = 3.25m 3.25.0164 1 = 63

    t0 2.656-2.656

    .005Reject H0 Reject H0

    .005Do not reject at a = .01

    There is no evidence average is not 3.25

  • One-Tailed t TestExample

    Is the average capacity of batteries less than 140ampere-hours? A random sample of 20 batteries had a mean of 138.47 and a standard deviation of 2.66. Assume a normal distribution. Test at the .05 level of significance.

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    One-Tailed t Test Solution

    H0: Ha: a = df = Critical Value(s):

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    Test Statistic:

    Decision:

    Conclusion:

    m = 140m < 140.0520 1 = 19

    t0-1.729

    .05

    Reject H0 Since t = -2.57 < -1.729, reject H0 at a = .05

    There is evidence population average is less than 140

    One-Tailed t TestThinking Challenge

    Youre a marketing analyst for Wal-Mart. Wal-Mart had teddy bears on sale last week. The weekly sales ($ 00) of bears sold in 10 stores was:8 11 0 4 7 8 10 5 8 3At the .05 level of significance, is there evidence that the average bear sales per store is more than 5 ($ 00)?

    2011 Pearson Education, Inc 83

    One-Tailed t Test Solution*

    H0: Ha: a = df = Critical Value(s):

    2011 Pearson Education, Inc 84

    Test Statistic:

    Decision:

    Conclusion:

    m = 5m > 5.0510 1 = 9

    t0 1.833

    .05

    Reject H0 Do not reject at a = .05

    There is no evidence average is more than 5

  • 7.6

    Large-Sample Test of Hypothesis about a Population Proportion

    2011 Pearson Education, Inc 85

    Large-Sample Test of Hypothesis about p

    One-Tailed TestH0: p = p0Ha: p < p0 (or Ha: p > p0)

    Test statistic:

    Rejection region:z < za (or z > za when Ha: p > p0)

    Note: p0 is the symbol for the numerical value of passigned in the null hypothesis

    2011 Pearson Education, Inc 86

    Large-Sample Test of Hypothesis about p

    Two-Tailed TestH0: p = p0Ha: p p0

    Test statistic:

    Rejection region: |z| > za /2

    Note: p0 is the symbol for the numerical value of passigned in the null hypothesis

    2011 Pearson Education, Inc 87

    Conditions Required for a Valid Large-Sample Hypothesis Test for

    p

    1. A random sample is selected from a binomial population.

    2. The sample size n is large. (This condition will be satisfied if both np0 15 and nq0 15.)

    2011 Pearson Education, Inc 88

  • One-Proportion z Test Example

    The present packaging system produces 10% defective cereal boxes. Using a new system, a random sample of 200 boxes had 11 defects. Does the new system produce fewer defects? Test at the .05level of significance.

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    One-Proportion z Test Solution

    H0: Ha: a = n = Critical Value(s):

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    Test Statistic:

    Decision:

    Conclusion:

    p = .10p < .10.05200

    z0-1.645

    .05Reject H0 Since z = -2.12 < -1.645

    Reject at a = .05

    There is evidence new system < 10% defective

    One-Proportion z Test Thinking Challenge

    Youre an accounting manager. A year-end audit showed 4% of transactions had errors. You implement new procedures. A random sample of 500 transactions had 25 errors. Has the proportion of incorrect transactions changed at the .05 level of significance?

    2011 Pearson Education, Inc 91

    One-Proportion z Test Solution*

    H0: Ha: a = n = Critical Value(s):

    2011 Pearson Education, Inc 92

    Test Statistic:

    Decision:

    Conclusion:

    p = .04p .04.05500

    z0 1.96-1.96

    .025Reject H0 Reject H0

    .025Do not reject at a = .05

    There is evidence proportion is not 4%

  • 7.7

    Test of Hypothesis about a Population Variance

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    Variance

    Although many practical problems involve inferences about a population mean (or proportion), it is sometimes of interest to make an inference about a population variance, s2.

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    Test of a Hypothesis about s 2

    One-Tailed TestH0: s 2 = s02Ha: s 2 < s02 (or Ha: s 2 > s02)

    Test statistic:

    Rejection region:(or c 2 > ca2 when Ha: s 2 > s02 )

    where s02 is the hypothesized variance and the distribution of c2 is a chi-square distribution with (n 1) degrees of freedom.

    2011 Pearson Education, Inc 95

    Test of a Hypothesis about s 2

    Two-Tailed TestH0: s 2 = s02Ha: s 2 s02

    Test statistic:

    Rejection region:

    where s02 is the hypothesized variance and the distribution of c2 is a chi-square distribution with (n 1) degrees of freedom.

    2011 Pearson Education, Inc 96

  • Conditions Required for a Valid Hypothesis Test for s 2

    1. A random sample is selected from the target population.

    2. The population from which the sample is selected has a distribution that is approximately normal.

    2011 Pearson Education, Inc 97

    Several c2 probability Distributions

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    Part of Table VI: Critical Values of Chi Square

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    Finding Critical Value Example

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    What is the critical c2 value given:Ha: s2 > 0.7n = 3a =.05?

    c20

    Upper Tail AreaDF .995 .95 .051 ... 0.004 3.8412 0.010 0.103 5.991

    c2 Table (Portion)

    df = n - 1 = 25.991

    Reject

    a = .05

  • Finding Critical Value Example

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    What is the critical c2 value given:Ha: s 2 < 0.7n = 3a =.05?

    What do you do if the rejection region is on the left?

    Finding Critical Value Example

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    What is the critical c2 value given:Ha: s 2 < 0.7n = 3a =.05?

    .103 c20

    Upper Tail AreaDF .995 .95 .051 ... 0.004 3.8412 0.010 0.103 5.991

    c2 Table (Portion)

    Upper Tail Area for Lower Critical Value = 1.05 = .95a = .05

    Reject H0

    df = n - 1 = 2

    Chi-Square (c2) Test Example

    Is the variation in boxes of cereal, measured by the variance, equal to 15 grams? A random sample of 25boxes had a variance of17.7 grams. Test at the .05level of significance.

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    Chi-Square (c2) Test Solution

    H0: Ha: a = df = Critical Value(s):

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    Test Statistic:

    Decision:

    Conclusion:

    s 2 = 15s 2 15.0525 1 = 24

    c20

    a/2 = .025

    39.36412.401

    Since 12.401

  • 7.8

    Calculating Type II Error Probabilities: More about b

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    Type II Error The Type II error probability b is calculated assuming

    that the null hypothesis is false because it is defined as the probability of accepting H0 when it is false.

    The situation corresponding to accepting the null hypothesis, and thereby risking a Type II error, is not generally as controllable.

    For that reason, we adopted a policy of nonrejection of H0 when the test statistic does not fall in the rejection region, rather than risking an error of unknown magnitude.

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    Steps for Calculating b for a Large-Sample Test about

    1. Calculate the value(s) of x corresponding to the border(s) of the rejection region. There will be one border value for a one-tailed test and two for a two-tailed test. The formula is one of the following, corresponding to a test with level of significance a:

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    Steps for Calculating b for a Large-Sample Test about

    Upper-tailed test:

    Lower-tailed test:

    Two-tailed test:

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  • Steps for Calculating b for a Large-Sample Test about

    2. Specify the value of a in the alternative hypothesis for which the value of b is to be calculated. Then convert the border value(s) of to z-value(s) using the alternative distribution with mean a. The general formula for the z-value is

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    Steps for Calculating b for a Large-Sample Test about

    Sketch the alternative distribution (centered at a) and shade the area in the acceptance (nonrejection) region. Use the z-statistic(s) and Table IV in Appendix B to find the shaded area, which is b.

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    Power of Test

    Probability of rejecting false H0 Correct decision

    Equal to 1 b Used in determining test adequacy Affected by

    True value of population parameter Significance level a Standard deviation & sample size n

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    Two-Tailed z Test Example

    Does an average box of cereal contain less than 368grams of cereal? A random sample of 25 boxes had x = 372.5. The company has specified s to be 15 grams. Test at the .05 level of significance.

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    368 gm.

  • Finding PowerStep 1

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    `xm0 = 368

    Reject H0Do NotReject H0

    Hypothesis:H0: m 368Ha: m < 368 a = .05

    Draw

    Finding PowerSteps 2 & 3

    2011 Pearson Education, Inc 114`xm a = 360

    True Situation:ma = 360 (Ha)

    Draw

    Specify

    b1b

    `xm0 = 368

    Reject H0Do NotReject H0

    Hypothesis:H0: m 368Ha: m < 368 a = .05

    Draw

    Finding PowerStep 4

    2011 Pearson Education, Inc 115363.065

    `xm a = 360

    True Situation:ma = 360 (Ha)

    Draw

    Specify1b

    `xm0 = 368

    Reject H0Do NotReject H0

    Hypothesis:H0: m 368Ha: m < 368 a = .05

    Draw

    Finding PowerStep 5

    2011 Pearson Education, Inc 116363.065 `xma = 360

    True Situation:ma = 360 (Ha)

    Draw

    Specify

    = 368m `x0

    Reject H0Do NotReject H0

    Hypothesis:H0: m 368Ha: m < 368 a = .05

    Draw

    b = .1541b =.846z Table

  • Properties ofb and Power

    1. For fixed n and a, the value of bdecreases (the power increases) as the distance between the specified null value 0 and the specified alternative value aincreases.

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    Propertiesof b andPower

    2. For fixed n and values of 0 and a, the value of bincreases (the power decreases) as the value of ais decreased.

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    Properties of b and Power

    3. For fixed a and values of 0 and a, the value of bdecreases (the power increases) as the sample size n is increased.

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    Key Ideas

    Key Words for Identifying the Target Parameter

    m Mean, Averagep Proportion, Fraction, Percentage, Rate, Probabilitys 2 Variance, Variability, Spread

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  • Key Ideas

    Elements of a Hypothesis Test

    1. Null hypothesis (H0)

    2. Alternative hypothesis (Ha)

    3. Test statistic (z, t, or c2)

    4. Significance level (a)

    5. p-value

    6. Conclusion

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    Key Ideas

    Errors in Hypothesis Testing

    Type I Error = Reject H0 when H0 is true(occurs with probability a)

    Type II Error = Accept H0 when H0 is false(occurs with probability b )

    Power of Test = P(Reject H0 when H0 is false)= 1 b

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    Key Ideas

    Forms of Alternative Hypothesis

    Lower-tailed : Ha : m < 50

    Upper-tailed : Ha : m > 50

    Two-tailed : Ha : m 50

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    Key Ideas

    Using p-values to Decide

    1. Choose significance level (a )

    2. Obtain p-value of the test

    3. If a > p-value, reject H0

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  • Assignments

    #97,#121,#130, #132, #133, #142

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