Mar. 22 Statistic for the day: Percent of Americans 18 or older who believe Martha Stewart

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Mar. 22 Statistic for the dayPercent of
Americans 18 or older who believe Martha
Stewarts sentence should include jail time 53
Source gallup.com
Assignment Read Chapter 18 Exercises p329
1, 2, 5, 6, 8, 10
These slides were created by Tom Hettmansperger
and in some cases modified by David Hunter
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Sample percentages categorical variable
Do you believe Martha Stewart got a fair
trial? Do you believe Martha Stewarts sentence
should include jail time?
Gallup Poll
Yes No No opinion
Fair trial 66 27 7
Jail time 53 40 7
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The Gallup Poll was based on 1005 telephone
interviews. Based on the sample of 1005 we
estimate that 53 of the population of millions
believes that Martha Stewarts sentence should
include jail time. If we take a new sample of
1005 we will get a new sample percentage. It
will generally not be exactly 53. If we take
lots of samples of 1005 we will get lots of
sample percentages. Next we look at the
histogram for the percentages.
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200 percentages based on 200 samples of 1005
each. Mean 53 (or .53) Standard deviation
(57 - 50) / 4 1.75 (or .0175)
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How do we measure and assess the uncertainty
in the sample percentage?
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So in our example, if the sample size is
1600, then the MARGIN OF ERROR is
Or 3.2 And we report 53 3.2
We defined the margin of error to be 2 standard
deviations. We estimated the standard deviation
from the histogram to be .0175. This nearly
agrees since 2x.0175 .035. Pretty close!
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Summary Gallup Poll

• We have a simple random sample from
• the population of telephone owners.
• The sample size used was 1005.
• We find the percentage from our sample.
• The MARGIN OF ERROR is 1 divided by
• the square root of the sample size.
• For 1005 the MARGIN OF ERROR is .032.
• Hence we report PERCENTAGE .032
• The margin of error does not depend on the
• population size, only on the sample size!

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Goals

• To refine the idea of standard deviation (for
  later
• use in a refined margin of error).
• We also want to relate this to the normal curve.
• In the past we
• used a sample to get a sample proportion
• used a formula to get the margin of error
• reported the sample proportion the margin of
  error
• Now we want a formula for the standard deviation.
  
• Then we will use the new standard deviation
  formula
• to calculate a new margin of error.

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Formula for estimating the standard deviation of
a sample proportion (dont need histogram)
If we happen to know the true population
proportion we use it instead of the sample
proportion.
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(No Transcript)
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• Summary
• We take a sample of 1005 phone interviews
• We estimate the percent of the American public
• that thinks that Martha Stewart should go to
  jail 53
• To assess the uncertainty in the 53 sample
• figure, we think of a normal curve of percentages
• centered at .530 with standard deviation of
• .016.
• So the normal curve has 95 of its distribution
• between .530 2x.016 and .530 2x.016 or
• Estimate 53 (.53) with 50 to 56 (.50 to .56)
  the reasonable interval of values.

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What to expect from sample proportions
Facts fingerprints may be influenced by
prenatal hormones. Most people have more ridges
on right hand than left. People who have more on
the left hand are said to have leftward
asymmetry. Women are more likely to have this
trait than men. The proportion of all men who
have this trait is about 15
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In a study of 186 heterosexual and 66 homosexual
men 26 (14) heterosexual men showed the trait
and 20 (30) homosexual men showed the
trait (Reference Hall, J. A. Y. and Kimura, D.
"Dermatoglyphic Asymmetry and Sexual Orientation
in Men", Behavioral Neuroscience, Vol. 108, No.
6, 1203-1206, Dec 94. ) Is it unusual to observe
a sample of 66 men and observe a sample
proportion of 30? We now know what the
distribution of sample proportions based on a
sample of 66 should look like. We will
suppose that the true proportion in the
population of men is 15.
Standard deviation
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The sample proportion for homosexual men (30) is
too large to come from the expected distribution
of sample proportions.
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Sample means measurement variables
Suppose we want to estimate the mean weight at PSU
Data from stat 100 survey. Sample size 237. Mean
value is 152.5 pounds. Standard deviation is
about (240 100)/4 35
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What is the uncertainty in the mean? We need a
margin of error for the mean. Suppose we take
another sample of 237. What will the mean
be? Will it be 152.5 again? Probably
not. Consider what happens if we take 1000
samples each of size 237 and compute 1000 means.
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Standard deviation is about (157 148)/4
9/4 2.25
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Formula for estimating the standard deviation of
the sample mean (dont need histogram)
Just like in the case of proportions, we would
like to have a simple formula to find the
standard deviation of the mean without having to
resample a lot of times. Suppose we have the
standard deviation of the original sample. Then
the standard deviation of the sample mean is
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So in our example of weights The standard
deviation of the sample is about 35. Hence by
our formula Standard deviation of the mean is
35 divided by the square root of 237
35/15.4 2.3 (Recall we estimated it to
be 2.25) So the margin of error of the sample
mean is 2x2.3 4.6 Report
152.5 4.6 or 147.9 to 157.1
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Example SAT scores
Suppose nationally we know that the SAT has
a mean of 425 points and a standard deviation of
120 points. Draw by hand a picture of what you
expect the distribution of sample means based on
samples of size 100 to look like.
Sample means have a normal distribution mean
425 standard deviation 120/10 12 So draw a
bell shaped curve, centered at 425, with 95 of
the bell between 425 24 401 and 425 24 449
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A sample of 100 SATs with a mean of 460 would be
very unusual. A sample of 100 with a mean of 440
would not be unusual.