#### Transcript Stat Chapter 14

```Chapter 14 - Confidence Intervals: The Basics
• Confidence Interval = estimate + margin of error
• Purpose is to estimate an unknown population parameter
and give some indication of how accurate the estimate is.
ex: Test Scores: (0-500), n=840 people
x  272
  60
What is the mean for this test in the
population?
60
s
 2.1 • According to the 68-95-99.7 Rule, 95%
of all samples should have a mean that
840
falls within 2 std. devs. (s) of the
population mean
• 272 is our estimate of the population
mean…
• 2s = 2(2.1) = 4.2
• So 95% of all samples should have a mean somewhere
between 272 + 4.2
 267.8, 276.2 We call this a “95% Confidence Interval”
272 + 4.2
estimate
margin of error
“Level C” Confidence Interval
• C = decimal form of confidence percentage
•ex: 95% conf. int. --> C = .95
• Definition: An interval computed from sample data by a
method that has probability “C” of producing an interval
containing the true value of the parameter.”
• z* = unknown number of standard deviations to produce a
given confidence interval.
• ex: Find z* for an 80% confidence interval.
90%
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Need a z-value
with .9 to its
left…
80%
10%
10%
Calculator Steps
Select STAT--TESTS--#7
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Set to the following values:
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Interval Formula
  
x  z *  
 n 
ex: Blood sample drug concentration
x  .8404
  .0068
n3
Find a 99% confidence interval…
Z* = 2.576

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.0068 

Interval = .8404  2.576
 3 
 .8404  .0101
 .8303, .8505

Confidence Interval Behavior
Margin of Error (m) =
  
z *  
 n 
• Changing any of the values in the equation will change the
margin of error…

• The margin of error will decrease when:
• z* gets smaller
•  gets smaller
• n gets larger (4n to reduce m by half…)
• ex 5.5, pg 339
Choosing the sample size
The confidence interval for a population mean will have a
specified margin of error (m) when the sample size is:
z *  2
n  

 m 
ex: Blood sample drug concentration
m = .005
z* = 1.96 for C=.95
=.0068
Produce results accurate to within
+0.005 with 95% confidence… how
many measurements will we need to
average?
1.96.0068
n  
 = 7.1
.005


2
** n generally will need to be whole numbers - so we must take
8 measurements…

```