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Page 342 The pvalue approach to hypothesis testing: Twotailed tests

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... from the sample data, given that the null hypothesis Ho is true. ... Null Hypothesis = 368. Level of Significance = .05. Population Standard Deviation = 15 ... – PowerPoint PPT presentation

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Title: Page 342 The pvalue approach to hypothesis testing: Twotailed tests


1
Page 342 - The p-value approach to hypothesis
testing (Two-tailed tests)
  • In recent years? widely available statistical
    software? alternative approach to hypothesis
    testing.
  • The p-value is the probability of obtaining a
    test statistic equal to or more extreme than the
    result obtained from the sample data, given that
    the null hypothesis Ho is true.
  • Also called the observed level of significance
  • It is the smallest level at which Ho can be
    rejected for a given set of data.
  • p-value ? ?, fail to reject Ho
  • p-value lt ? , reject Ho
  • Easy for the normal distribution ? but the
    computation of the p-value can be very difficult
    for other distributions.
  • Excel routinely present the p-value as part of
    the output for many hypothesis-testing
    procedures.
  • P343 Phstat One-Sample Tests Z test for the
    mean, ? known

2
  • Cereal box filling example Ho 368, ? 15.
  • n 25, X-bar 372.5, ? .05, two tailed error
  • Traditional hypothesis testing
  • critical value ? ? 1.96
  • Z (Xbar - ?)/(?/n½) 1.50
  • -1.96 lt 1.50 lt 1.96.
  • Fail to reject Ho. There is no evidence that the
    mean is different from the hypothesized value.
  • p-value approach to hypothesis testing p342
  • Compute the probability of obtaining a Z value
    greater than 1.50 along with the probability of
    obtaining a Z value less than -1.50.
  • Above 1.50 is .5000 -.4332.0668.
  • Standard normal distribution is symmetric, p
    .0668.0668.1336.
  • The probability of obtaining a result equal to or
    more extreme than the one observed is .1336.
  • (p-value .1336) ? (? .05)
  • p-value ? ?, Barely fail to reject Ho.
  • P342 343 10 steps - p-value approach
  • 11 Draw a picture.

3
Page 345 Confidence intervals estimation verse
hypothesis testing
  • Two major components of statistical inference
  • Same set of concepts, use them for different
    purposes.
  • In many situations we can use confidence
    intervals to do a test of a null hypothesis.
  • If the hypothesized value of ? falls into the
    confidence interval, the null hypothesis would
    not be rejected.
  • If the hypothesized value of ? does not fall into
    the confidence interval, the null hypothesis
    would be rejected, because it would then be
    considered an unusual value.
  • Cereal box filling example Ho 368, ? 15.
  • n 25, X-bar 372.5, ? .05, two tailed error
  • confidence interval estimate of ?. ? Xbar?Z(?/n½)
  • 372.5 ? 5.88
  • 95 CI of ? ? 366.62 ?? ? 378.38
  • 95 CI includes Ho ? fail to reject

4
E7.3 p343 Hypothesis test steps
  • 1. State the null hypothesis, Ho. (stated in
    statistical terms)
  • 2. State the alternative hypothesis, H1.
  • 3. Choose the level of significance, ?.
  • (This along with the sample size, determines ß)
  • 4. Choose the sample size, n.
  • Sample size is determined after taking into
    account the specified risks of committing Type I
    and Type II errors i.e., selected levels of ? and
    ß
  • 5. Determine the appropriate statistical
    technique and corresponding test statistic to
    use.
  • (? known ? Z test) (? unknown ? t test)
  • 6. Collect the data and compute the sample value
    of the appropriate test statistic.
  • 7. Calculate the p-value based on the test
    statistic.
  • 8. Compare the p-value to ?
  • 9. Make the statistical decision
  • reject Ho if p-value lt ?
  • fail to reject Ho if p-value gt ?
  • 10. Express the statistical decision in terms of
    the problem.
  • 11. Put it all in a picture.

5
  • Book example p343. Is the population mean weight
    of cereal per box different from at 368 grams?
  • 1) Ho ? 368 grams
  • 2) H1 ? ? 368 grams
  • 3) ? .05
  • 4) n 25
  • 5) ? 15 grams, so ? is known, use Z-test,
    normal distribution.
  • 6) Collect data and calculate test statistic
  • X-bar 372.5 grams. e7.1, p340 Z (Xbar -
    ?)/(?/n½) 1.50
  • 7) Calculate the p-value
  • Phstat One-Sample Tests Z test for the
    mean, ? known
  • Null Hypothesis ? 368
  • Level of Significance .05
  • Population Standard Deviation 15
  • Sample Size 25
  • Sample Mean 372.5
  • Two Tailed Test
  • Output results p-Value .1336
  • 8) p-Value .1336 gt ? .05
  • 9) Barely fail to reject Ho. Borderline
    situation. There is no evidence that the mean is
    significantly different from the hypothesized
    value.

6
7.3 One-tailed tests
  • Ho , H1 ? , ? contained two possibilities
  • In some situations the alternative hypothesis
    focuses in a particular direction
  • also called directional test ? entire rejection
    region is contained in one tail of the sampling
    distribution
  • Upper tail - reject Ho when the sample mean is
    significantly above(greater than) ?
  • Lower tail - reject Ho when the sample mean is
    significantly below (less than)?
  • Chief financial officer (CFO) would be
    interested in filling above 368 grams
  • Ho ? ? 368, H1 ?gt368
  • ? 15, n 25, X-bar 372.5, ? .05,
  • one-tailed error, so CV1.645, Z still 1.50
  • We would conclude that there is no evidence that
    the average fill per box is above 368 grams.
  • The result from the sample is deemed due to
    chance or sampling error, it is not statistically
    significant.
  • One tail with ? .01 ? CV2.33
  • One tail with ? .10 ? CV1.28

7
p-value for one tailed test
  • Depending on the direction of the alternative
    hypothesis
  • Compute the probability of obtaining a value
    greater than the computed test statistic, or
  • Compute the probability of obtaining a value less
    than the computed test statistic
  • Ho ? ? 368, H1 ?gt368
  • X-bar 372.5 Z 1.50
  • 1.50 to the normal distribution table ? .4332
  • p .5000 - .4332 .0668
  • So... the probability of obtaining a value
    greater than 372.5 is .0668 (or 6.68 percent of
    the time).
  • In Class GMAT example
  • Homework 7.40 and others on web page

8
  • GMAT example. The Director of Admissions at MSU
    believes that their MBA students are above the
    national average. The population average GMAT
    score is 500 with a population standard deviation
    of 100. A sample of 12 MSU MBA students is
    selected at random. The sample mean is 537. Use
    a level of significance of .01.
  • 1) Ho ? lt 500 points
  • 2) H1 ? gt 500 points
  • 3) ? .01
  • 4) n 12
  • 5) ? 100 points, so ? is known, use Z-test,
    normal distribution.
  • 6) Collect data and calculate test statistic
  • X-bar 537 grams. e7.1, p340 Z (Xbar -
    ?)/(?/n½)
  • Z (537 - 500) / (100 / 12½) 1.28
  • 7) Calculate the p-value
  • Phstat One-Sample Tests Z test for the
    mean, ? known
  • Null Hypothesis ? 500
  • Level of Significance .01
  • Population Standard Deviation 100
  • Sample Size 12
  • Sample Mean 537
  • Upper Tailed Test
  • Output results p-Value .09997
  • 8) p-Value .09997 gt ? .01

9
Review
  • Critical Values - Normal Distribution
  • t distribution, each one is different
  • p-value analysis - very subjective
  • p-value gt.2 ? Not close to rejecting. Sample
    statistic is close to the hypothesized mean. A
    strong fail to reject.
  • p-value lt .009 ? Little chance of failing to
    reject. Sample statistic is far from the
    hypothesized mean. Sample statistic is way out
    in the tail. A strong rejection.
  • .009 lt p-value lt .2 ? Boarder line situation.
    Barely rejecting or barely failing to reject.
    When p value and ? are similar.

10
7.4 t-test of hypothesis for the mean (?
Unknown)
  • In most hypothesis-testing situations, the
    standard deviation of the population (?) is
    unknown. It is estimated by S, the standard
    deviation of the sample.
  • If the population is assumed to be normally
    distributed ? the sampling distribution of the
    mean will follow a t distribution with n-1
    degrees of freedom.
  • In practice ? As long as the sample size is not
    very small and the population is not very skewed.
  • E7.2 p350 ? t (Xbar - ?)/(S/n½) ? is the test
    statistic for determining the difference between
    the sample mean X-bar and the population mean ?.

11
Assumptions of the one-sample t Test (review)
  • Each of the rows in T.E3 corresponds to a
    particular t distribution.
  • The one-sample t test is considered a classical
    parametric procedure. Assumptions required ? a
    random sample from a population that is normally
    distributed
  • Need to determine how closely the actual data
    match the normal distributions theoretical
    properties.
  • We can use a stem-and-leaf display, a
    box-and-whisker plot, and a normal probability
    plot (p357)
  • If the sample size is small (n less than 30)
    and/or we can not meet the normality assumption ?
    Distribution-free testing procedures are likely
    to be more powerful
  • Two excel files on web page to calculate p-values
    for the t-distribution
  • p-value_t_given_info.xls
  • p-value_t_rawdata.xls
  • In class 7.50 p356 Cost of Books, Coin-Op
    Laundry Homework 7.54 and others on web page.

12
  • Problem 7.50 page 356, The director of admissions
    at a large university advises parents of incoming
    students about the cost of textbooks during a
    typical semester. A sample of 100 students
    enrolled in the university indicates a sample
    mean cost of 315.40, with a sample standard
    deviation of 43.20. Use a level of significance
    of 0.01. Is there evidence that the population
    average is above 300.
  • 1) Ho ? lt 300
  • 2) H1 ? gt 300
  • 3) ? .01
  • 4) n 100
  • 5) ? unknown, only sample information, t-test
  • 6) Collect data and calculate test statistic
  • X-bar 315.40. e7.2, p350 t (Xbar -
    ?)/(S/n½)
  • t (315.4 - 300) / ( 43.2 / 100½) 3.56
  • 7) Calculate the p-value
  • Phstat One-Sample Tests t test for the
    mean, ? unknown
  • Null Hypothesis ? 300
  • Level of Significance .10
  • Sample Standard Deviation 43.2
  • Sample Size 100
  • Sample Mean 315.40
  • Upper Tail Test
  • Output results p-Value .00028
  • 8) p-Value .00028 lt ? .10

13
  • You are considering the purchase of a coin
    operated laundry facility. Cash flows of 675
    per day are needed to cover the cost of capital,
    or the bank will not loan you the money. The
    current owner has presented accounting
    information stating that the average daily cash
    intake is 675 per day. A sample of 30 days is
    collected. The sample mean is 625 per day, with
    a sample standard deviation of 75. Use a level
    of significance of 0.01. Should you buy this
    laundry facility, are the cash flows at least
    675 per day?
  • 1) Ho ? gt 675
  • 2) H1 ? lt 675
  • 3) ? .01
  • 4) n 30
  • 5) ? unknown, only sample information, t-test
  • 6) Collect data and calculate test statistic
  • X-bar 625. e7.2, p350 t (Xbar -
    ?)/(S/n½)
  • t (625 - 675) / ( 75 / 30½) - 3.65
  • 7) Calculate the p-value
  • Phstat One-Sample Tests t test for the
    mean, ? unknown
  • Null Hypothesis ? 675
  • Level of Significance .01
  • Sample Standard Deviation 75
  • Sample Size 30
  • Sample Mean 625
  • Lower Tail Test
  • Output results p-Value .0005
  • 8) p-Value .0005 lt ? .01

14
7.5 One sample Z test for the proportion
  • The sample proportion ( psX/n ) is compared to
    the hypothesized value of the parameter, p
  • Evaluate the magnitude of the difference between
    ps and p ? e7.3 p357 ? Z test statistic
  • Evaluate the magnitude of the difference between
    the of successes in the sample and the
    hypothesized of successes in the population ?
    e7.4 p358 ? Z test statistic
  • e7.3 and e7.4 will provide the same result
  • Z test statistic will be approximately normally
    distributed if the sample size is large enough ?
    np?5 and n(1-p)?5
  • We will not work any problems of this type now,
    but we will work some in chapter 10, two sample
    with categorical data.

15
7.6 Pitfalls and Ethical Issues
  • Fundamental hypothesis testing methodology. It
    is used for analyzing differences between sample
    estimates (i.e., statistics) of hypothesized
    population characteristics (i.e., parameters).
    It is used to make decisions.
  • Ensure proper methodology
  • 1. What is the goal?
  • 2. Two-tailed or one-tailed?
  • 3. Can a random sample be drawn?
  • 4. Measurements - numerical or categorical?
  • 5. Significance level - ?
  • 6. Sample size vs. power given ?
  • 7. What statistical test - normal or t
  • 8. Conclusions and interpretations
  • A person with substantial statistical training
    should be consulted and involved early in the
    process. Not after the data have been collected.

16
Ethical Issues
  • Data collection method - many issues,
    self-selecting subjects
  • Human subjects - consent
  • Type of test - Prior information leads to a
    specifically directed one-tailed test.
    Interested only in differences then use a
    two-tailed test. In the overwhelming majority of
    research studies, a two-tailed test should be
    employed.
  • Level of significance - ? is selected in advance
    of data collection. Solution ? always report the
    p-value
  • Data snooping - steps must be done first before
    the data are collected.
  • Cleansing and discarding of data - flagging
    outlier observations. Look at the displays.
    Should an observation be removed from a study?
  • Reporting of findings - document both good and
    bad results
  • Next upExam 2, same format, 20 MC, problems
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