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Lesson 7:Confidence intervals
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Introduction
The difference between the present and the new commercial may be aresult of chance.
The example involves inferences about a population based on sampledata How can we analyse such problems?
Two equivalent methods:1. Confidence Intervals: estimating a value of a population parameter
2. Tests of significance: assessing evidence for a claim about a
population
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Remember the sample distribution of The sample mean is a random variable (if you pick a different sample
you will probably get a different sample mean).
The distribution of:
has mean and standard deviation
- If the population distribution is normal, then the distribution of is
normal.
- If the population distribution is not normal and n is large (in practice:
0), then is approximately normal (the Central Limit Theorem).
We assume that the population is known.
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Statistical confidence Example 6.5:
In a sample of 453 large firms the mean monthly premium paid
for the health insurance is 0 dollars. We assume that is
known to be 112.
What can we say about the population mean ?
In repeated sampling is approximately
,
,
11
, .
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Statistical confidence is approximately, .
The 69-95-99.7 rule says that the probability is
about 0.95 that is within 10.54 of .
2 3 + + 2 + 3
The 68-95-99.7 rule
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Statistical confidence To say that lies within 10.54 of is the same as saying that is
within 10.54 of ! (because distances are symmetrical)
So in 95% of all samples is within 10.54 of or put differently: in 95%
of all samples is in the interval from 10. to 10.
(with 0).
We say that we are 95% confident
that is between 394 and 416.
We call 0 10. , 1
a 95% confidence interval
for.
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Statistical confidence We are 95% confident that is between 394 and 416.
Be sure you understand that in reality could be outside this interval!
(Although this only happens in 5% of all samples)
We cannot know whether our sample is one of the 95% or one of the
unlucky 5%. The only thing we can say is: we arrived at the conclusion
that is between 394 and 416 by a process that gives correct results
95% of the time.
Implication: We dont need to take a lot of random samples to find the
sampling distribution and at its center. All we need is one sample and
relying on the properties of the sample means distribution to infer the
population mean .
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Confidence levels and critical values A confidence level C (e.g. 95%) gives the probability that the
interval will capture the true parameter in repeated samples.
The critical value z* for a confidencelevel C is the value so that the area
under the standard Normal curve
between z* and z* is C.
Common critical values:
o z*=2.576 (for C=99%)
o z*=1.960 (for C=95%)
o z*=1.645 (for C=90%)
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Confidence levels and critical values Exercise: Find the critical value for a 70% confidence level.
End solution:
z*=1.04
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Confidence intervals A confidence interval always has the form
where the margin of errorm shows the precision of the estimate.
The confidence interval for a population mean with a known andwith a confidence level Cis:
where
the margin of error.
To derive the confidence interval we had to assume that the distributionofis normal. So the interval is exact when the population distribution
is normal and is approximately correct when n is large in other cases
(because of the central limit theorem).
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Confidence intervalsA confidence interval is an
interval that contains the true
parameter in C percent of the
samples.
Experiment with the applet:http://bcs.whfreeman.com/psbe3e/#613741__630698__
Figure 6.5: Twenty-five samples from thesame population gave these 95%
confidence intervals.
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Application: quality controlA company makes 00 bottles of green tea. Assume we knowthe standard deviation is . In a sample of 10 bottles, themean content 01.. Is this convincing evidence that the
mean fill of all bottles differs from 00?
Two common mistakes: do not conclude that
- the mean is 01. therefore it is proven that the mean differs
from 00.- the difference of 1. is small compared to 00 so there isnothing unusual going on.
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Application: quality controlSo: 01. 10
Is this convincing evidence that the mean fill differs from 00?
We calculate a 95% confidence interval for :
C.I. for
01. 1.
00. , 0.
Since 500 is not included in the C.I., we are 95% confident that differs
from 500. So we can reject the hypothesis that the mean fill of all bottles
is 00.
Inferential analysis can be done either with confidence intervals or
significance tests (next lesson).
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Reducing the margin of errorConfidence interval for:
To reduce the margin of error (to make the confidence interval
smaller), there are only 3 ways:
1. Use a lower level of confidence (smaller C smaller )
2. Reduce
3. Increase the sample size
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Choosing the minimum sample sizeIt is possible to choose the sample size in a research so that inferences
about the population mean can be done with a predetermined level of
confidence and a predetermined margin of error:
Note: in practice is usually unknown, but if n is large then the sample standard
deviation s will be close to the unknown population standard deviation . Sosubstituting s for in the formula will result in an approximately correct minimum
sample size n.
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Exercise 1In a sample of 50 Belgian employees, the mean
gross monthly wage is 00 euro. The standard
deviation in the population is known to be 1000 euro.
a. Find the 80% confidence interval for.
b. Find the 95% confidence interval for.c. Find the 95% confidence interval for if the sample size
in this example 500 instead of 50.
d. What sample size would you need to limit the margin of
error to 50 euro with 95% confidence?
End solutions:
a. (2219,2581) b. (2123,2677) c. (2312,2488) d. 1536 cases
Confidence interval for:
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Exercise 2 An opinion poll asks:
Will you vote for Barack Obama in the elections: Yes or No?
Let be a variable taking the values 0 (No) and 1 (Yes), so that isthe percentage of Obama voters in the sample.
In a sample of 2400 respondents
o 0. (49% of the respondents in the sample say they will voteObama)
o Assume that standard deviation is 0.0.
Estimate the margin of error in this poll for a 95% confidence level.
End solution:
the margin of error in this poll is approximately 2 percentage points.
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