Business Statistics for Managerial Decision

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Business Statistics for Managerial Decision

• Inference for the Population Mean

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Test for a Population Mean

• There are four steps in carrying out a
  significance test
• State the hypothesis.
• Calculate the test statistic.
• Find the p-value.
• State your conclusion by comparing the p-value to
  the significance level ?.

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Example Blood pressures of executives

• The medical director of a large company is
  concerned about the effects of stress on the
  companys younger executives. According to the
  National Center for health Statistics, the mean
  systolic blood pressure for males 35 to 44 years
  of age is 128 and the standard deviation in this
  population is 15. The medical director examines
  the records of 72 executives in this age group
  and finds that their mean systolic blood pressure
  is . Is this evidence that the
  mean blood pressure for all the companys young
  male executives is higher than the national
  average? Use ? 5.

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Example Blood pressures of executives

• Hypotheses
• H0 ? 128
• Ha ? gt 128
• Test statistic
• P-value

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Example Blood pressures of executives

• Conclusion
• About 14 of the time, a SRS of size 72 from the
  general male population would have a mean blood
  pressure as high as that of executive sample. The
  observed is not significantly
  higher than the national average.

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Example A company-wide health promotion campaign

• The company medical director institutes a health
  promotion campaign to encourage employees to
  exercise more and eat a healthier diet. One
  measure of the effectiveness of such a program is
  a drop in blood pressure. Choose a random sample
  of 50 employees, and compare their blood
  pressures from annual physical examination given
  before the campaign and again a year later. The
  mean change in blood pressure for these n 50
  employees is . We take the population
  standard deviation to be ? 20. Use ? 5.

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Example A company-wide health promotion campaign

• Hypotheses
• H0 ? 0
• Ha ? lt 0
• Test statistic
• P-value

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Example A company-wide health promotion campaign

• Conclusion
• A mean change in blood pressure of 6 or better
  would occur only 17 times in 1000 samples if the
  campaign had no effect on the blood pressures of
  the employees. This is convincing evidence that
  the mean blood pressure in the population of all
  employees has decreased.

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ExampleTesting Pharmaceutical products

• The Deely Laboratory analyzes pharmaceutical
  products to verify the concentration of active
  ingredients. Such chemical analyses are not
  perfectly precise. Repeated measurements on the
  same specimen will give slightly different
  results. The results of repeated measurements
  follow a Normal distribution quite closely, the
  analysis procedure has no bias, so that the mean
  ? of the population of all measurements is the
  true concentration in the specimen. The standard
  deviation of this distribution is a property of
  the analytical procedure and is known to be ?
  0.0068 gram per liter. The laboratory analyzes
  each specimen three times and reports the mean
  results.

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ExampleTesting Pharmaceutical products

• A client sends a specimen for which the
  concentration of active ingredient is supposed to
  be 0.86. Deelys three analyses give
  concentrations
• 0.8403 0.8363 0.8447
• Is there significant evidence at the 1 level
  that the true concentration is not 0.86?

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ExampleTesting Pharmaceutical products

• Hypotheses
• H0 ? 0.86
• Ha ? ? 0.86
• Test Statistic The mean of the three analyses is
  . The one sample z test statistic
  is therefore

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ExampleTesting Pharmaceutical products

• We do not need to find the exact P-value to
  assess significance at the ? 0.01 level. Look
  in the table A under tail area 0.005 because the
  alternative is two-sided. The z-values that are
  significant at the 1 level are z gt 2.575 and
  z lt -2.575.
• Our observed z -4.99 is significant

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P-value versus fixed ?

• In our example , we concluded that the test
  statistic z -4.99 is significant at the 1
  level.
• The observed z is far beyond the critical value
  for ? 0.01, and the evidence against H0 is far
  stronger than 1 significance suggests.
• The P-value P .0000006 (from a statistical
  software) gives a better sense of how strong the
  evidence is.
• The P-value is the smallest level ? at which the
  data are significant.

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Inference for the Mean of a Population

• Both confidence intervals and tests of
  significance for the mean ? of a Normal
  population are based on the sample mean ,
  which estimates the unknown ?.
• The sampling distribution of depends on
  ?.
• There is no difficulty when ? is known.
• When ? is unknown, we must estimate it.
• The sample standard deviation s is used to
  estimate the population standard deviation ?.

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The t-distribution

• Suppose we have a simple random sample of size n
  from a Normally distributed population with mean
  ? and standard deviation ?.
• The standardized sample mean, or one-sample z
  statistic
• has the standard Normal distribution N(0, 1).
• When we substitute the standard deviation of the
  mean (standard error) s /?n for the ?/?n, the
  statistic does not have a Normal distribution.

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The t-distribution

• It has a distribution called t-distribution.
• The t-distribution
• Suppose that a SRS of size n is drawn from a N(?,
  ?) population. Then the one sample t statistic
• has the t-distribution with n-1 degrees of
  freedom.
• There is a different t distribution for each
  sample size.
• A particular t distribution is specified by
  giving the degrees of freedom.

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The t-distribution

• We use t(k) to stand for t distribution with k
  degrees of freedom.
• The density curves of the t-distributions are
  symmetric about 0 and are bell shaped.
• The spread of t distribution is a bit greater
  than that of standard Normal distribution.
• As degrees of freedom k increase, t(k) density
  curve approaches the N(0, 1) curve.

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The one Sample t Confidence Interval

• Suppose that an SRS of size n is drawn from a
  population having unknown mean ?. A level C
  confidence interval for ? is
• Where t is the value for the t (n-1) density
  curve with area C between t and t. The margin
  of error is
• This interval is exact when the population
  distribution is Normal and is approximately
  correct for large n in other cases.

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Example Estimating the level of Vitamin C

• The following data are the amount of vitamin C,
  measured in milligram per 100 grams (mg/100g) of
  the corn soy blend (dry basis), for a random
  sample of size 8 from a production run
• 26 31 23 22 11 22 14 31
• We want to find a 95 confidence interval for
  ?, the mean vitamin C content of the corn soy
  blend (CSB) produced during this run.

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Example Estimating the level of Vitamin C

• The sample mean and the standard
  deviation s 7.19 with degrees of freedom n-1
  8-1 7. The standard error of is
• From table we find t 2.365. The 95 confidence
  interval is

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The one-sample t testSummary
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Example Is the Vitamin C level correct?

• The specifications for the CSB state that the
  mixture should contain 2 pounds of vitamin premix
  fro every 2000 pounds of product. These
  specifications are designed to produce a mean (?)
  vitamin C content in the final product of
  40mg/100 g. We test the null hypothesis that the
  mean vitamin C content of the production run in
  the previous example conforms to these
  specifications. Use ? 5.

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Example Is the Vitamin C level correct?

• Hypotheses
• H0 ? 40
• Ha ? ? 40
• Test statistic
• P-value
• Because the degrees of freedom are n-1 7, this t
  statistic has t(7) distribution.

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Example Is the Vitamin C level correct?

• From the largest entry in the df 7 line of the
  table we see that
• We conclude that the P-value is less than
  2?0.0005, or P lt .001.
• We reject H0 and conclude that the vitamin C
  content for this run is below the specification.

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Matched Pairs t procedures

• Comparative studies are usually preferred to
  single-sample investigations because of the
  protection they offer against confounding.
• In a matched pairs study, subjects are matched in
  pairs and the outcomes are compared within each
  matched pair.
• One situation calling for matched pairs is
  before-and-after observations on the same
  subjects.

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Matched Pairs t procedures

• A matched pair analysis is needed when there are
  two measurements or observations on each
  individual and we want to examine the change from
  the first to the second.
• For each individual, subtract the before
  measure from the after measure.
• Analyze the difference using the one-sample
  confidence interval and significance testing
  procedures.

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Example The effect of language instruction

• A company contracts with a language institute to
  provide individualized instruction in foreign
  languages for its executives who will be posted
  overseas. Is the instruction effective? Last year
  20 executives studied French. All had some
  knowledge of French, so they were given the
  Modern Language Associations listening test of
  understanding of spoken French before the
  instruction began. After several weeks of
  immersion in French, the executives took the
  listening test again. The following table gives
  the pretest and posttest scores.

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Example The effect of language instruction
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Example The effect of language instruction

• To analyze these data
• Subtract pretest score from the posttest score.
• These differences appear in the gain column in
  previous table.
• These 20 differences form a single sample.
• To assess whether the institute significantly
  improved the executives comprehension of spoken
  French we test
• H0 ? 0
• Ha ? gt 0

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Example The effect of language instruction

• Here ? is the mean improvement that would be
  achieved if the entire population of executives
  received similar instruction.
• The null hypothesis says that no improvement
  occurs, and the alternative hypothesis says that
  posttest scores are higher on the average.
• The 20 differences have

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Example The effect of language instruction

• The one sample t statistic is therefore
• P-value is found from the t(19) distribution.
• T-table shows that 3.86 lies between the upper
  .001 and .0005 critical values of the t(19)
  distribution. The P-value therefore lies
  between these values.

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Example The effect of language instruction

• Conclusion
• The improvement in scores was significant. We
  have strong evidence that the posttest scores are
  systematically higher.
• A statistically significant but very small
  improvement in language ability would not justify
  the expense of the individualized instruction. A
  confidence interval allows us to estimate the
  amount of improvement.

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Example The effect of language instruction

• Find a 90 confidence interval for the mean
  improvement in the entire population.
• The critical value t 1.729 from t-table for
  90 confidence.
• The confidence interval is