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Transcript
Section 8.2 ~
Estimating Population Means
Introduction to Probability and Statistics
Ms. Young
Sec. 8.2
Objective

After this section you will learn how to
estimate population means and compute the
associated margins of error and confidence
intervals.
Sec. 8.2
Confidence Interval

Recall that a confidence interval is a range of values that is likely to
contain the population parameter

Ex. ~ The population for a survey is all adults in the U.S. A sample of adults
in the U.S. was taken and the people were asked how many hours per day
they spend watching TV. The sample mean was found to be x  2.3 hours. If
the margin of error for this was found to be .8 hours, what is the
confidence interval?
2.3  .8  1.5  3.1



So based on this sample and the margin of error, we are 95% confident that
adults in the U.S. watch between 1.5 hours and 3.1 hours of TV per day
In other words, this range of values is 95% likely to contain the true population
mean (which we would not know unless we took a census)
There are other levels of confidence (90% and 99%), but we will be
working with the 95% confidence level and how to calculate the margin
of error using the sample mean
Sec. 8.2
Margin of Error

The margin of error for a 95% confidence interval is found using
the following formula where s = the standard deviation from the
sample and n = the sample size
Margin of error (for 95% confidence) = E 

2s
n
Once the margin of error is found, the confidence interval is
written as follows:
x E   x E
or
xE

Ex. ~ For the confidence interval found earlier (between 1.5 hours
and 3.1 hours) it could be written as:
1.5 hours    3.1 hours
or
2.3  .8
Sec. 8.2
Example 1

Compute the margin of error and find the 95% confidence interval for
the protein intake sample of n = 267 men, which has a sample mean of
x  77 grams and a sample standard deviation of s = 58.6 grams.
2s
Margin of error (for 95% confidence) = E 
n
2(58.6)
E
 7.2
267
The sample mean is x  77 grams, so the 95% confidence interval extends
approximately from 77.0 – 7.2 = 69.8 grams to 77.0 + 7.2 = 84.2 grams.
69.8 grams    84.2 grams
We can conclude with 95% confidence, that the average protein intake of
men is between 69.8 grams and 84.2 grams.
Sec. 8.2
Choosing Sample Size

Often times in statistical surveys and experiments, we know in advance
the margin of error we would like to achieve


Ex. ~ We might want to estimate the mean cost of a new car within $200
If you know the margin of error that you would like to achieve, you can
manipulate the formula to solve for the appropriate sample size needed
to obtain that margin of error
2s
E
n

 E n  2s

n
2s
E
 2s 
 n 
E
2
To find the exact sample size needed, you would use the population
standard deviation, but often times you will not know this since you just
study a sample

When using the population standard deviation to find the sample size, you
always round the result up
 2 
n

 E 
2
Sec. 8.2
Example 2
You want to study housing costs in the country by sampling recent house sales in
various regions. Your goal is to provide a 95% confidence interval estimate of the
housing cost. Previous studies suggest that the population standard deviation is
about $7,200. What sample size should be used to ensure that the sample mean is at
least
a. $500 of the true population mean?
 2 
n

 E 
2
 2  7200 
n

 500 
2
 n  28.22
 n  829.4
The sample should include at least 830 prices
b. $100 of the true population mean?
 2 
n

 E 
2
 2  7200 
n

100


2
 n  1442
The sample should include at least 20,736 prices
 n  20, 736