Introduction to Inference Chapter 6

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Introduction to InferenceChapter 6
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Statistical Inference

• Want to draw conclusions based on sample data
• i.e.-- say something about an entire population
  based on information in a sample.
• Conclusions are subject to sampling error
• We want to quantify the margin of error we are
  likely to encounter.
• Examples
• 1. Im interested in estimating the mean
  income of loggers in the interior of BC.
• 2. Gallup poll.

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Confidence intervals

• Our goal is to obtain an estimate of some
  parameter of a population.
• We want our estimate to be of the general form
• Best guess /- error of estimation
• To get our estimate, we take a random sample from
  the population and proceed on the basis of the
  information we obtain from the sample.
• What is our best guess??
• How do we obtain our error of estimation ?

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Confidence intervals for population mean

• Best Guess
• Our best guess is xbar, the sample mean.
• How to determine the estimation error?

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Determining the estimation error

• To determine estimation error, well use our
  knowledge of sampling distribution of xbar.
• We know that xbar has mean µ, with standard
  deviation, ?/sqrt(n), where n is our sample
  size.
• If our original population is normal, we also
  know that xbar is normally distributed.
• Even if the original population isnt normal,
  xbar approximately follows a normal distribution
  if the sample size, n , is large (by CLT).

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Determining the error of estimation -- continued

• For example, we are about 95 sure that xbar lies
  in the range
• µ /- 2 ?/sqrt(n). ()
• E.g., if ? 45 and n 100, we are 95 sure
  that xbar lies in the range µ /- 9. (verify).
• How to use () to get our confidence interval??

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A confidence interval for population mean

• It turns out that our 95 c.i. for µ is just
• xbar /- 2 ?/sqrt(n).
• Why does this work?
• From our picture ( to be added in class), we see
  that our interval will trap µ approximately
  95 percent of the time.

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A numerical example

• I collect SRS of n 100 loggers. I find
  average income in sample is 17,000.
• Obtain a 95 c.i. for the mean income of all
  loggers. Assume that std. dev. of loggers
  incomes is known to be ? 2500.
• (In Ch 7 we will drop the assumption that ? is
  known -- ? will be estimated from the sample data)

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Level C confidence intervals
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Tradeoffs

• For a given sample size, higher level of
  confidence leads to wider confidence interval.

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Designing a confidence interval

• Suppose I want a level C confidence interval of
  the form
• Xbar /- m,
• where m is a desired margin of error that I
  supply in advance. I also specify C in advance.
• Required sample size is
• n (z?/m)2
• where z is the z value required for level C
  confidence
• Where does this formula come from?

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Example--Designing a confidence interval

• Recall the logger example. I have ? 2500.
  Suppose I want a 99 c.i. of the form xbar /-
  300. How big a sample do I need?
• What if I want a 99 confidence interval of form
  xbar /- 150?

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Example 6.12 (Text)

• A study of career paths of hotel general managers
  sent questionnaires to a SRS of 160 hotels
  belonging to major US hotel chains. There were
  114 responses. The average time these 114
  general managers had spent with their current
  company was 11.78 years. Give a 99 c.i. for the
  mean number of years general managers of
  major-chain hotels have spent with their current
  company. (Take it as known that the std dev of
  time with the company for general managers is 3.2
  years).
• A margin of error of /- 1 year is considered
  acceptable. What is the minimum sample size that
  would be required to achieve this level of
  accuracy with a confidence level of 99

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Confidence interval summary

• Assume I have a random sample and that the
  population std dev, ?, is known.
• A level C c.i. for the population mean is
• xbar /- z?/sqrt(n).
• (value of z will depend on C)
• The c.i, is exact when the underlying population
  is normal.
• If the pop. is not normal, the c.i. is
  approximately correct for large samples (by CLT).
• We can find the sample size, n, required to
  obtain a c.i. with a specified margin error, m,
  by using the formula
• n (z?/m)2

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Tests of Significance

• Confidence interval
• Goal is to estimate some parameter of a
  population.
• Test of Significance
• Goal is to assess the evidence provided by the
  data in favor of some claim about the population.

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Motivational Example

• Four randomly selected students do a 20 hour SAT
  prep course at a special school. After the
  course, they write the SAT. Their scores turn
  out to be 560, 600, 590, 490. We find xbar
  560.
• Scores of students who take the course are known
  to be normally distributed with a standard
  deviation of 50.
• National SAT test scores are normally distibuted
  with a mean of 500 and a standard deviation of
  50.
• Does the sample data provide strong support for
  the schools claim that the prep course is
  effective in increasing SAT scores?

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Motivational Example (cont.)

• Lets use what we know about the behavior of xbar
  to assess this claim.
• In particular, we ask
• What is the probability of observing a sample
  mean of 560 or larger if the population mean
  score for those who took the course is 500?
  (i.e., course doesnt help on average)

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Motivational Example (cont.)

• What if the sample mean had been xbar 700?
• What if the sample mean were xbar 520?

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Formalization of our example

• First, state hypotheses
• H0 µ 500
• (course has no effect).
• Ha µ gt 500
• (course increases mean score).
• Terminology
• H0 is the null hypothesis.
• Ha is the alternative hypothesis.
• Usually, Ha is the claim we hope to establish.
  (Equivalently, H0 is the claim we want to
  falsify).
• In our example, the more the sample mean exceeds
  500, the more evidence we have in favor of Ha (
  i.e. against H0 ).

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Formalization of our example (cont.)

• Second We compute the test statistic
• z (xbar mu0)/(sigma/sqrt(n))
• (560 -500)/(50/sqrt(4)) 2.4
• Third Assess the strength of the evidence
  against the null hypothesis by computing a
  p-value.
• p-value Prob (z gt2.4) 0.0082
• (The p-value is obtained from the z value that
  results from the specific form of our
  hypotheses.)
• Fourth state a conclusion.
• strong evidence in favor of companys claim.

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Other possible forms of H0 and Ha in our example

• Suppose that I suspect that the school has a
  reverse effect and actually decreases average
  SAT performance.
• How would I set up hypotheses in such a way that
  a low average test score will support my
  suspicions?
• What if I suspect that the course has some effect
  on SAT scores, but Im not sure whether it
  increases or decreases the average score. How
  would I set up appropriate two sided hypotheses
  for this situation?
• Either a low or a high average test score will
  support my suspicions.

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General form of the z test
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Example

• SSHA is a psychological test that
  measures motivation, attitude towards school, and
  study habits of students. Scores range from 0 to
  200. The mean score for US college students is
  about 115 with a std. dev of about 30. A teacher
  who suspects that older students have better
  attitudes towards school gives the SSHA to 20
  students who are at least 30 years of age. Their
  mean score is 135.2
• State appropriate null and alternative
  hypotheses.
• Report the p-value of your test and state your
  conclusion clearly.
• Your test required 2 important assumptions in
  addition to the assumption that sigma 30. What
  are they? Which is more important?

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Another Example

• A study of the pay of corporate CEOs examined the
  increase in cash compensation for the CEOs of 104
  companies, adjusted for inflation, in a recent
  year. The average inflation adjusted increase
  in the sample was xbar 6.9 with a sample
  standard deviation of s 55. Is this good
  evidence that the mean real compensation
  increased in the past year?
• Because the sample size is large, s is close to
  the population sigma, so it is reasonable to
  assume that s 55.

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P-values and statistical significance
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Example statistical significance

• The mean yield of corn in the US is about 120
  bushels per acre. A survey of 40 farmers this
  year gives a sample mean of xbar 123.8 bushels
  per acre. We want to know if this provides good
  evidence that the national mean this year is not
  120 bushels per acre. Assume that the farmers
  surveyed constitute a SRS from the population of
  all commercial corn growers and that the
  population has a std dev of sigma 10 bushels
  per acre.
• (a) Set up the appropriate hypotheses. Give the
  p-value for the test. Is the result significant
  at the 5 level?
• (b) Are you convinced that the population mean is
  not 120 bushels per acre?
• (c) Is your conclusion correct if the
  distribution of corn yields is somewhat
  non-normal? Why?

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Another Example

• A computer has random number generator that
  generates random numbers uniformly distributed
  between 0 and 1. If this is true, the numbers
  generated come from a population with µ 0.5 and
  ? .2887. A command to generate 100 random
  numbers gives an average of 0.4365. Is the
  generator working properly?

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Another Example

• A union leader claims that the average school
  teacher makes less than 40,000 per year. A
  random sample of 400 school teachers finds a
  sample mean of xbar 39,650. The standard
  deviation in school teachers incomes is known to
  be 5,000. Assess the union leaders claim.

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Relationship between confidence intervals and two
sided tests

• A level ? 2-sided significance test rejects the
  hypothesis
• H0 µ µ0
• and accepts the alternative
• Ha µ ? µ0
• precisely when the value µ0 lies
• outside a level 1- ? confidence
• interval for µ.

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Example (confidence interval and 2 sided
hypothesis test)

• Diameters of a certain machine part are normally
  distributed with a standard deviation of 0.1 mm.
  A random sample of 25 parts yields an average
  diameter of 11.9 mm.
• Find a 99 c.i. for the true mean diameter.
• Based on your sample, can you conclude, at the 1
  level of significance, that the true mean
  diameter differs from 12 mm?

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Comments about hypothesis testing

• P-values tell us more than setting, in advance, a
  fixed level of significance.
• Statistical significance is not necessarily the
  same as practical significance.
• Statistical testing is not always valid e.g.
  faulty data bias in questionnaires etc.

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Two types of error

• Example If there are more than 32,000 trees on
  a plot of land it will be economical to log the
  plot. I sample the plot and get the following
  95 c.i. for the total number of trees
• 32,500 /- 3,500
• Should I log the lot?

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Type I and Type II errors