### 3.5 Normal Distributions and z

```2.6 Confidence Intervals and
Margins of Error
What you often see in reports
These results are accurate to within +/- 3.7%,
19 times out of 20.
Margin of error
Confidence level
95% probability that x is
somewhere in the range
(x – 3.7, x+3.7)
Confidence interval
Confidence Intervals
• x is sample mean
• We don’t usually know population mean, 
(mu)
• We can find confidence intervals
– Ranges of values  likely to be in
– E.g., a 95% confidence level has 0.95
probability of containing 
Notation
•  = allowed error, or probability of error
• (1 – ) = confidence level
• z   = z-score for that confidence interval
1 
 2
– E.g. z0.975 is the z-score for a 95% confidence
interval
Note: the proper notation is actually z
2
Confidence Intervals
• A (1 – ) or (1 – ) x 100% confidence
interval for , given population standard
deviation , sample size n, and sample
mean x , represents the range of values
x  z


1 
 2
n
   x  z

1



 2
x  z

n
What’s
this?
Population mean/Sample Means
• So far, mean of sample = mean of
population
• Means from different samples of the same
population are different
• Sample means have normal distribution
 2 
X  N  ,

n 

Common confidence levels and
their associated z-scores
Confidence
Level
Tail size,

2
z-score,
z
90%
0.05
1.645
95%
0.025
1.960
99%
0.005
2.576
 
1 
 2
Example 1: Drying Times
• A paint manufacturer knows from experience that
drying times for latex paints have a standard
deviation of 10.5 min. The manufacturer wants to
use the slogan “Dries in T min.” on its advertising.
Twenty test areas of equal size are painted and the
mean drying time is found to be 75.4 min.
• A) Find a 95% confidence interval for the actual
mean drying time of the paint.
• B) What would be a reasonable value for T?
Example 1
• A) For a 95% confidence level, the acceptable
probability error is  = 5% = 0.05
z

1 
 2
 z(0.975)
 1.960
x  z


1 
 2
n
   x  z

1 
 2

n
10.5
10.5
75.4  (1.960)
   75.4  (1.960)
20
20
70.8    80.0
The manufacturer can be 95% confident that the actual
mean drying time is between 70.8 min and 80.0 min.
• B) It would be reasonable to advertise “Dries in 80
min.”
Margin of Error
and Sample Sizes
• Consider the confidence interval width, w
– E.g. 70.8 <  < 80.0,
–
w = 9.2
• Margin of Error = half the confidence interval width
–the maximum difference between the observed sample mean x
and the true value of the population mean 
E  z


1 
 2
w = 2E
n
Sample Size
• We can use this to calculate the minimum sample size
necessary for a given confidence level
– Often used in opinion polls and other surveys
• If sample size too large, waste of resources/time/money
• If sample size too small, inaccurate results
x  z


1 
 2
Solving for sample size, n:
n
w  2 z

1 
 2

n
 2z  
 1 2 
n
w







2
Notes
• Need to know  in advance
• Estimate it by doing a pre-survey/study
• Margin of error decreases as sample size
increases, but only to a point
Example 2: ISPs
• We would like to start an Internet Service
Provider (ISP) and need to estimate the
average Internet usage of households in one
week for our business plan and model. How
many households must we randomly select
to be 90% sure that the sample mean is
within 1 minute of the population mean .
Assume that a previous survey of household
usage has shown  = 6.95 minutes.
Example 2
• For a 90% confidence level, the acceptable
probability error is  = 10% = 0.10
 2z   
z    z(0.995)
2
1 
 2
 1.645
E = 1 min
w = 2E
= 2 min

n




1 
 2
w




 2(1.645)(6.95) 


2


2
 130.7
You would need a sample of about 131 households.
```