12.2 Inference for a population mean when the stdev is unknown; one more example

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Lecture 5

• 12.2 Inference for a population mean when the
  stdev is unknown one more example
• 12.3 Testing a population variance
• 12.4 Testing a population proportion

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Announcements

• Answer to Problem 12.77 Sample size of 752.
• Extra office hour this week Wednesday, 9-10
• Homework due Thursday, see web page for
  correction on last problem
• Type II error calculation from last lecture
  solution on web page

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Hypothesis Testing Basic Steps

1. Set up alternative and null hypotheses
2. Choose appropriate test statistic and values of
   test statistic that will be considered evidence
   in favor of H1, e.g., for testing
   , reject for large values of z-score
3. Find critical values and compare the observed
   test statistic to critical value (rejection
   region method) or find p-value (p-value method)
4. Make substantive conclusions.

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Estimating m when s is unknown

• Example 12.2
• An investor is trying to estimate the return on
  investment in companies that won quality awards
  last year.
• A random sample of 83 such companies is selected,
  and the return on investment is calculated had he
  invested in them.
• Construct a 95 confidence interval for the mean
  return.
• Is there evidence that the returns are greater
  than 10?

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Estimating m when s is unknown

• Solution
• The problem objective is to describe the
  population of annual returns from buying shares
  of quality award-winners.
• Given x-bar15.02, s8.31, n83
• Data Xm12-02
• There is evidence that the returns are gt10 at
  the 2.5 significance level. (Why?)

t.025,82_at_ t.025,80
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12.3 Inference About a Population Variance

• Sometimes we are interested in making inference
  about the variability of processes.
• Examples
• The consistency of a production process for
  quality control purposes.
• Investors use variance as a measure of risk.
• To draw inference about variability, the
  parameter of interest is s2.

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12.3 Inference About a Population Variance

• The sample variance s2 is an unbiased, consistent
  and efficient point estimator for s2.
• The statistic has a
  distribution called Chi-squared, if the
  population is normally distributed.

d.f. 5
d.f. 10
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Confidence Interval for Population Variance

• From the following probability statement P(c21-
  a/2 lt c2 lt c2a/2) 1-awe have (by substituting
  c2 (n - 1)s2/s2.)

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Testing the Population Variance

• Example 12.3 (operation management application)
• A container-filling machine is believed to fill 1
  liter containers so consistently, that the
  variance of the filling will be less than 1 cc
  (.001 liter).
• To test this belief a random sample of 25 1-liter
  fills was taken, and the results recorded
  (Xm12-03). s20.8659.
• Do these data support the belief that the
  variance is less than 1cc at 5 significance
  level?
• Find a 99 confidence interval for the variance
  of fills.

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JMP implementation of two-sided test
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12.4 Inference About a Population Proportion

• When the population consists of nominal data
  (e.g., does the customer prefer Pepsi or Coke),
  the only inference we can make is about the
  proportion of occurrence of a certain value.
• When there are two categories (success and
  failure), the parameter p describes the
  proportion of successes in the population. The
  probability of obtaining X successes in a random
  sample of size n from a large population can be
  calculated using the binomial distribution.

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12.4 Inference About a Population Proportion

• Statistic and sampling distribution
• the statistic used when making inference about p
  is

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Testing and Estimating a Proportion

• Test statistic for p

• Interval estimator for p (1-a confidence level)

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Why are Proportions Different?

• The true variance of a proportion is determined
  by the true proportion
• The CI of a proportion is NOT derived from the
  z-test
• The denominator of the z-statistic is NOT
  estimated, but the width of the CI is estimated.
  
• gt CI test and z-test can differ sometimes.

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Testing the Proportion

• Example 12.5 (Predicting the winner in election
  day)
• Voters are asked by a certain network to
  participate in an exit poll in order to predict
  the winner on election day.
• The exit poll consists of 765 voters. 407 say
  that they voted for the Republican candidate.
• The polls close at 800. Should the network
  announce at 801 that the Republican candidate
  will win?

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Selecting the Sample Size to Estimate the
Proportion

• Recall The confidence interval for the
  proportion is
• Thus, to estimate the proportion to within W, we
  can write
• The required sample size is

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Sample Size to Estimate the Proportion

• Example
• Suppose we want to estimate the proportion of
  customers who prefer our companys brand to
  within .03 with 95 confidence.
• Find the sample size needed
• Solution
• W .03 1 - a .95,
• therefore a/2 .025,
• so z.025 1.96

Since the sample has not yet been taken, the
sample proportion is still unknown.
We proceed using either one of the following two
methods
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Sample Size to Estimate the Proportion

• Method 1
• There is no knowledge about the value of
• Let . This results in the largest
  possible n needed for a 1-a
  confidence interval of the form .
• If the sample proportion does not equal .5, the
  actual W will be narrower than .03 with the n
  obtained by the formula below.
• Method 2
• There is some idea about the value of
• Use the value of to calculate the sample size

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Chapter 12 Introduction

• Variety of techniques are presented whose
  objective is to compare two populations.
• We are interested in
• The difference between two means.
• The ratio of two variances.
• The difference between two proportions.

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Inference about the Difference between Two Means

• Example 13.1
• Do people who eat high-fiber cereal for breakfast
  consume, on average, fewer calories for lunch
  than people who do not eat high-fiber cereal for
  breakfast?
• A sample of 150 people was randomly drawn. Each
  person was identified as an eater or non-eater of
  high fiber cereal.
• For each person the number of calories consumed
  at lunch was recorded. There were 43 high-fiber
  eaters who had a mean of 604.02 calories for
  lunch with s64.05. There were 107 non-eaters
  who had a mean of 633.23 calories for lunch with
  s103.29.

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13.2 Inference about the Difference between Two
Means Independent Samples

• Two random samples are drawn from the two
  populations of interest.
• Because we compare two population means, we use
  the statistic .

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The Sampling Distribution of

1. is normally distributed if the
   (original) population distributions are normal .
2. is approximately normally
   distributed if the (original) population is not
   normal, but the samples size is sufficiently
   large (greater than 30).
3. The expected value of is m1 -
   m2
4. The variance of is s12/n1
   s22/n2

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Making an inference about m1 m2

• If the sampling distribution of is
  normal or approximately normal we can write
• Z can be used to build a test statistic or a
  confidence interval for m1 - m2

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Making an inference about m1 m2

• Practically, the Z statistic is hardly used,
  because the population variances are not known.

?
?

• Instead, we construct a t statistic using the
• sample variances (s12 and s22) to estimate

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Making an inference about m1 m2

• Two cases are considered when producing the
  t-statistic
• The two unknown population variances are equal.
• The two unknown population variances are not
  equal.

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Inference about m1 m2 Equal variances

• Calculate the pooled variance estimate by

The pooled variance estimator
n2 107
n1 43
Example 1 s12 4103.02 s22 10669.77 n1
43 n2 107.
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Inference about m1 m2 Equal variances

• Construct the t-statistic as follows

• Perform a hypothesis test
• H0 m1 - m2 0
• H1 m1 - m2 gt 0

or lt 0
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Example 13.1

• Assuming that the variances are equal, test the
  scientists claim that people who eat high-fiber
  cereal for breakfast consume, on average, fewer
  calories for lunch than people who do not eat
  high-fiber cereal for breakfast at the 5
  significance level.
• There were 43 high-fiber eaters who had a mean of
  604.02 calories for lunch with s64.05. There
  were 107 non-eaters who had a mean of 633.23
  calories for lunch with s103.29.

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Inference about m1 m2 Unequal variances
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Inference about m1 m2 Unequal variances
Conduct a hypothesis test as needed, or, build
a confidence interval
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Which case to useEqual variance or unequal
variance?

• Whenever there is insufficient evidence that the
  variances are unequal, it is preferable to
  perform the equal variances t-test.
• This is so, because for any two given samples

The number of degrees of freedom for the equal
variances case
The number of degrees of freedom for the unequal
variances case
³
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Example 13.1 continued

• Test the scientists claim about high-fiber
  cereal eaters consuming less calories than
  non-high fiber cereal eaters assuming unequal
  variances at the 5 significance level.
• There were 43 high-fiber eaters who had a mean of
  604.02 calories for lunch with s64.05. There
  were 107 non-eaters who had a mean of 633.23
  calories for lunch with s103.29.

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Practice Problems

• 12.58,12.77,12.98,13.34