Introduction to Inference

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Chapter 6

• Introduction to Inference

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Motivational Scenario

• A market research agency has been given the task
  to estimate the average number of hours per week
  that young adults spend surfing the web.
• The agency surveys a random sample of 100 young
  adults and obtains a mean of 20 hours and a
  standard deviation of 5 hours
• Can the agency conclude that the true mean number
  of hours per week spent by all young adults
  surfing the web is exactly 20 hours?

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Motivational Scenario contd

• Because the market research agency recognizes
  that the 20 hours was obtained from just one of
  many possible samples of the population they are
  unwilling to say the population mean is exactly
  equal to 20 hours.
• To allow for the variation in the sample estimate
  they may cautiously conjecture that the true mean
  is somewhere between 18 and 22 hours, between 15
  and 25 hours, etc.

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Establishing an Interval for Estimation

• How wide should they make the interval?
• How confident should they be that the named
  interval does indeed contain the true mean?
• On what basis should the choice be made?
• They can use an established fact about how sample
  means vary when random samples are repeatedly
  drawn from any population the central limit
  theorem

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Confidence Intervals for the Mean - Rationale

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Based on this relative frequency idea, if only
one random sample of size n is drawn we can
express 95 confidence that the interval x 1.96
sx will contain m. This interval is called a
95 confidence interval for m.
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Confidence level C refers to probability the
interval will contain the true mean before the
sample data are collected
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Margin of error given by z times std. error z
is determined from the Normal curve based on C,
the confidence level
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For a given sample size, the interval width is
narrower for lower levels of confidence.
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For a given level of confidence the interval
width is narrower for larger samples.
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Inference on a Single Population Mean
Suppose you have collected a sample of
20 observations, your sample mean is 5.5 and the
assumed population standard deviation is 1.7 The
95 CI for m is 5.5 .7451 yielding 4.755 to
6.245 Suppose you originally thought the mean
was actually 5.0 Do your data support your
belief?
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In the previous case with the 95 C.I. for m
being 4.755 to 6.245, suppose, instead, you
originally thought the mean was 6.5. Do your
data support this belief?
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Hypothesis Testing

• Formal way to determine whether or not the data
  support a belief or hypothesis.

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Hypothesis Testing in the Judicial System

• In our judicial system we have the following
  hypotheses
• The accused is innocent The accused is guilty
• We can make two errors
• Convicting the innocent Letting the guilty go
  free

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Hypothesis Testing in the Judicial System

• It is desirable to minimize the chance of
  committing either error. But guarding against
  one usually results in increasing the chance of
  committing the other.
• Society favors guarding against convicting the
  innocent.

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Procedure for Guarding Against Convicting Innocent

• Assume accused is innocent.
• Gather evidence to prove guilt
• Convict only if evidence is strong enough

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Equivalent Procedure in Statistics

• To test the belief m gt 20 (alternative
  hypothesis, Ha)
• Assume m not gt 20 (Null hypothesis, Ho m ? 20)
• Gather random sample from population compute
  sample mean, x
• Conclude m gt 20 (Ha) only if evidence is strong
  enough, i.e. if x is so many standard deviations
  away from 20, the probability of this occurring
  by pure random chance is very small

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Hypothesis Testing Definitions

• Type I error concluding Ha when Ho is true
  (convicting the innocent)
• Type II error concluding Ho when in fact it is
  false (letting the guilty go free)
• ? Prob (Type I error) significance level
• b Prob (Type II error)

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Formal Hypothesis Testing

• Establish the Null Hypothesis and the Alternative
  Hypothesis
• H0 ? 20.00 Ha ? ? 20.00 (two tailed)
• OR
• H0 ? 20.00 Ha ? gt 20.00 (right tailed)
• OR
• H0 ? 20.00 Ha ? lt 20.00 (left tailed)
• Ho must always have an equal sign and Ha must be
  what you want to prove

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Formal Hypothesis Testing for m

• Select ?, the probability of a Type I error.
• For example, ?.05.
• You set your standard for how extreme the
  sample results must be (in support of the
  alternative hypothesis in order for you to reject
  the null.
• Here, the sample results must be strong
  enough in favor of Ha that you would falsely
  reject the null only 5 of the time.

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Formal Hypothesis Testing for m

• Select ?, the probability of a Type I error.
• For example, ?.05
• Compute the test statistic (z if s known) from
  the sample. This tells you how many standard
  errors above or below the null value the sample
  mean is.

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Formal Hypothesis Testing for m

• Select ?, the probability of a Type I error.
• For example, ?.05
• Compute the test statistic z.
• Compute the probability of obtaining such an
  extreme test statistic z by pure chance, if the
  null hypothesis were true. This is called the
  p-value of the test.

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Formal Hypothesis Testing for m

• Select ?, the probability of a Type I error.
• For example, ?.05
• Compute the test statistic z.
• Compute the probability of obtaining such an
  extreme test statistic z by pure chance, if the
  Null hypothesis were true. This is called the
  p-value of the test.
• Reject or fail to reject the Null Hypothesis by
  determining whether the p-value is less than or
  greater than a.

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Interpreting p-values

• All statistical packages give p-values in the
  standard output.
• When we reject Ho we say the test is significant.
• If p-value lt .01, highly significant
  (overwhelming evidence in support of research
  hypothesis)
• If p-value between .01 and .05, significant
  (strong evidence)
• If p-value between .05 and .10, slightly
  significant (weak evidence)
• If p-value gt .10, not significant (no evidence)

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Two Tailed vs. One Tailed Tests
If the alternative is ? ? 20, the test is
two-tailed. Since a is shared between both tails
of the z-curve, the p-value twice the area cut
off at the tail by the computed z. This p-value
is then compared with a.
a/2 .025
?/2.025
½ p-value
z
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Two Tailed vs. One Tailed Tests
If the alternative hypothesis is ?gt20, this is a
right-tailed test with all the a .05 at the
right tail. The p-value (to compare with a) is
the area cut off at the right tail by the
calculated z.
? .05
p-value
z
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Two Tailed vs. One Tailed Tests

• If the alternative hypothesis is ? lt 20 and a
  .10,
• this is a left-tailed test with all the .10 at
  the left tail. The p-value (to compare with a) is
  the area cut off at the left tail by the
  calculated z.

?.1
p-value
z (usually negative)
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Hypothesis Test Example ex. 6.54

• What are we given? n 20 s 30 x 135.2 ?
  .01
• Step 1, establish hypotheses
• H0 ? 115 vs. Ha ? gt 115
• Step 2, set significance level. a .01 (given)
• Step 3, compute the test statistic
• z (135.2-115)/6.71 3.01
• Step 4, determine the p-value. Z-table gives P(Z
  lt 3.01) 0.9987. So, P(Z gt 3.01) 1- 0.9987
  .0013.
• Step 5, decision reject Ho since p-value (.0013)
  lt ? .01
• Step 6, conclusion within context Conclude older
  students appear to have better study attitude

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Hypothesis Test Example ex. 6.55

• What are we given? n 40 s 10 x 138.8 ?
  0.01
• Step 1, establish hypotheses
• H0 ? 135 vs. Ha ? ? 135
• Step 2, set significance level. a .01 (given)
• Step 3, compute the test statistic
• z (138.8-135)/1.58 2.40
• Step 4, determine the p-value. Z-table gives P(Z
  lt 2.40) 0.9918. So, P(Z gt 2.40) 1- 0.9918
  .0082. Since 2-tailed test, p-value 2.0082
  .0164
• Step 5, decision do not reject Ho since p-value
  (.0164) gt ? .01
• Step 6, conclusion within context insufficient
  evidence that national mean yield is not 135. But
  if we used a .05 we would reject Ho since
  p-value of .0164 lt .05. Conclusion would be
  there IS evidence mean yield is not 135.