Statistical Inferences Based on Two Samples

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Statistical Inferences Based on Two Samples

• 9.2 Comparing Two Population Means Using Small
  Independent Samples and Assuming Sigmas are
  Unknown
• 9.3 Paired Difference Experiments

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Sampling Distribution of
Normal, if each of the sampled populations is
normal and approximately normal if the sample
sizes n1 and n2 are large
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Sampling Distribution of
(Continued)
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Large Sample Confidence Interval, Difference in
Mean
If two independent samples are from populations
that are normal or each of the sample sizes is
large, 100(1 - a) confidence interval for m1 -
m2 is
If ?1 and ?2 are unknown estimate the sample
standard deviations by s1 and s2 and use the t
distribution
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9.2 Comparing Two Population Means Using
Independent Samples with Sigmas Unknown
If two independent samples are from populations
that are normal with equal variances, 100(1 - a)
confidence interval for m1 - m2 is
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Tests about Differences in Means When Variances
are Equal
If sampled populations are both normal with equal
variances, we can reject H0 ?1 - ?2 D0 at the
? level of significance if and only if the
appropriate rejection point condition holds or,
equivalently, if the p-value is less than ?.
Reject H0 if
p-Value
Alternative
t?, t?/2 and p-values are based on (n1 n2 2)
df
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Hypothesis Test and Confidence Interval Example
Exercise 9.21, pg. 369

• What are we given? n1 22 s1 225 xbar1
  1500 n2 22 s2 251 xbar2 1300 ? .05
• First assume equal population variances
• Step 1, establish hypotheses
• H0 ?1 - ?2 0 vs. Ha ?1 - ?2 gt 0
• Step 2, set significance level. a .05 (given)
• Step 3, compute the test statistic, but first the
  pooled variance

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Hypothesis Test and Confidence Interval
Example Exercise 9.21, pg. 369

• Step 4a, determine the rejection point, t.05,42
  1.684
• Step 4b, estimate the p-value. Using df 42,
  t-table gives P(T gt 3.307) .001 and P(T gt
  2.704) .005 Since 2.704 lt (t 2.78) lt 3.307,
  p-value is between 0.001 and 0.005
• Step 5, decision reject Ho since (a) test
  statistic, t (2.78) gt rejection point (1.684) or
  (b) p-value (between .001 .005) lt ? .05

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Using df 42, t-table gives P(T gt 2.704) .005
and P(T gt 3.307) .001. With t 2.78,
p-value is between .001 and .005
0.005
0.001
t 2.704 2.78
3.307
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Hypothesis Test Example

• Step 6, conclusion within context there is very
  strong evidence that type A training results in
  higher mean weekly sales than does type training.
  

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MegaStat Output for Example
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100(1 - a) confidence interval for m1 - m2 is
given by
We are 95 confident that mean weekly sales with
type A train-ing exceeds that with type B
training by between 42 and 358
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9.3 Paired Difference Experiments

• Before, we drew random samples from two different
  populations
• Now, have two different processes (or methods)
• Draw one random sample of units and use those
  units to obtain the results of each process
• For instance, use the same individuals for the
  results from one process vs. the results from the
  other process
• E.g., use the same individuals to compare
  before and after treatments
• By using the same individuals, we eliminate any
  differences in the individuals themselves and
  just compare the results from the two processes

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Paired Difference ExperimentsContinued

• Let md be the mean of population of paired
  differences
• md m1 m2 , where m1 is the mean of
  population 1 and m2 is the mean of population 2
• Let and sd be the mean and standard deviation
  of a sample of paired differences that has been
  randomly selected from the population
• is the mean of the differences between pairs
  of values from both samples

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t-Based Confidence Interval forPaired
Differences in Means
If the sampled population of differences is
normally distributed with mean ?d, then a
(1-a)100 confidence interval for md m1 - m2 is
where for a sample of size n, ta/2 is based on n
1 degrees of freedom
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Test Statistic for Paired Differences

• The test statistic is
• D0 m1 m2 is the claimed or actual difference
  between the population means
• D0 varies depending on the situation
• Often D0 0, and the null means that there is no
  difference between the population means
• The sampling distribution of this statistic is a
  t distribution with (n 1) degrees of freedom

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Paired Differences Testing Rules
where ta, ta/2, and p-values are based on (n 1)
degrees of freedom.
either t gt ta/2 or t lt ta/2
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Example on Inferences with Paired Samples
Exercise 9.32, pg. 377
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Example on Inferences with Paired Samples
Exercise 9.32, pg. 377

• Key sample information n 10 s 3.02 xbar
  4.0 ? .10 - .001
• Step 1, establish hypotheses
• H0 ?d ?pst - ?pre 0 vs. Ha ?d ?pst -
  ?pre gt 0
• Step 2, set significance level. a .05
  (mid-range)
• Step 3, compute the test statistic

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Hypothesis Test and Confidence Interval
Example Exercise 9.21, pg. 369

• Step 4a, determine the rejection point, t.05,9
  1.833
• Step 4b, estimate the p-value. Using df 9,
  t-table gives P(T gt 3.25) .005 and P(T gt
  4.297) .001 Since 3.25 lt (t 4.19) lt 4.297,
  p-value is between 0.001 and 0.005
• Step 5, decision reject Ho since (a) test
  statistic, t (4.19) gt rejection point (1.833) or
  (b) p-value (between .001 .005) lt ? .05.
  Note, we would F.T.R. Ho only at ? .001 of
  given range

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Using df 9, t-table gives P(T gt 3.25) .005 and
P(T gt 4.297) .001. With t 4.19, p-value is
between .001 and .005
0.005
0.001
t 3.250 4.19
4.297
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Hypothesis Test Example

• Step 6, conclusion within context there is very
  strong evidence that post-exposure attitude
  scores are higher on average than pre-exposure
  attitude scores. In other words, advertisement
  appears to increase mean attitude scores.

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100(1 - a) confidence interval for m1 - m2 is
given by
We are 95 confident that mean post-exposure
score exceeds the mean pre-exposure score by an
amount between 1.84 and 6.16. So 1.84 is a
reasonable estimate of the minimum difference
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MegaStat Output for Paired Diff. Example