Inferences about two proportions

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Inferences about two proportions

• Assumptions
• We have proportions from two simple random
  samples that are independent (not paired)
• For both samples, np 5 and nq 5
•  
• Possible alternative hypotheses
•  
• Null Hypothesis p1 p2

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Test Statistic
where p1 - p2 0 (assumed in null hypothesis)
(pooled estimate of p1 and p2)
Critical values and P-values come from the
z-score tables
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Example

• A drug company wants to determine that their new
  headache drug is effective. They give 500 people
  their new drug, and 400 people a placebo. In the
  experimental group, 350 said their headache went
  away in 15 minutes. In the placebo group, 235
  said their headache went away in 15 minutes. Is
  the success rate higher in the experimental
  group?

4
Define hypotheses
Lets use a 0.01 significance level Because were
working with a proportion, we use a normal
distribution.
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Value of the test statistic
Since we are working with a 0.01 significance
level, and this is a right-tailed test, the
critical value is 2.33. Since the test statistic
is larger than the critical value, we reject the
null hypothesis.   The sample data support the
claim that a larger proportion of people
recovered in 15 minutes using the new drug than
using the placebo.
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PValue P(z gt 3.516) 0.0001. Much smaller
than our critical value. This would again lead us
to reject the null hypothesis.
Confidence Intervals for two proportions
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Example The 99 confidence interval for our drug
test results
We are 99 confident that the difference between
the population proportions is between 0.03 and
0.195
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Inferences about two means

• Assumptions
• The two samples are simple random samples, and
  are independent
• Either both samples are large (gt30), or both
  samples come from normally distributed
  populations.
•  
• Test Statistic

Degrees of freedom is the smaller of n1-1 and
n2-1
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Example

• You want to test the theory that talking to
  plants makes a difference. You put 23 bean
  plants in one greenhouse and talk to them nicely
  each day. You put 21 bean plants in another
  greenhouse and ignore them. After 4 weeks, you
  find that the mean height of the talked-to plants
  is 38cm, with a standard deviation of 5cm. The
  mean height of the ignored plants is 34cm, with a
  standard deviation of 7cm. Test the claim that
  the results are different.

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Define hypotheses
Lets use a 0.05 significance level Because were
working with sample means, s unknown, we use a t
distribution.
Degrees of freedom 20
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Value of the test statistic
This is a two-tailed test with a 0.05
significance level. From our critical t table,
our critical values are -2.086 and 2.086 Since
the test statistic is larger than the critical
value, we reject the null hypothesis. The sample
data support the claim that plants that are
talked to grow differently than plants that are
ignored.
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P-Value Since it is a two-tailed test with the
test statistic to the right of center, we want to
find twice the area to the right of the test
statistic 2.P(t gt 2.163) 2.(0.020) 0.04
(using technology) Since this is less than our
significance level, we reject the null hypothesis

• Confidence Intervals for two means

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Example The 95 confidence interval for our
plant results above
We are 95 confident that the difference between
the populations is between 0.14cm and 7.86cm. In
other words, we are 95 confident that the plants
that are talked to grow between 0.14cm and 7.86cm
higher than plants that are ignored.
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Matched Pairs

• When comparing two populations where the samples
  are not independent, we must use the Matched
  Pairs test (8.4)
• Example of matched pairs
• People are given a test while listening to music,
  and another test in silence. Claim the mean
  test score while listening to music is higher
  than the mean test score in silence

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Homework

• 8.2 5, 7, 11, 21
• 8.3 1, 3, 5, 7, 15