Transcript Section 8-2 Estimating Population Means

8.2 Estimating Population Means
LEARNING GOAL
Learn to estimate population means and compute the
associated margins of error and confidence intervals.
Copyright © 2009 Pearson Education, Inc.
Estimating a Population Mean: The Basics
When we have only a single sample, the sample
mean is the best estimate of the population mean, μ.
However, we do not expect the sample mean to be
equal to the population mean, because there is likely
to be some sampling error. Therefore, in order to
make an inference about the population mean, we
need some way to describe how well we expect it to
be represented by the sample mean.
The most common method for doing this is by way
of confidence intervals.
Copyright © 2009 Pearson Education, Inc.
Slide 8.2- 2
A precise calculation shows that if the distribution of
sample means is normal with a mean of μ, then 95%
of all sample means lie within 1.96 standard
deviations of the population mean; for our purposes
in this book, we will approximate this as 2 standard
deviations.
A confidence interval is a range of values likely to
contain the true value of the population mean.
Copyright © 2009 Pearson Education, Inc.
Slide 8.2- 3
95% Confidence Interval for a Population Mean
The margin of error for the 95% confidence interval is
2s
margin of error = E ≈
n
where s is the standard deviation of the sample.
We find the 95% confidence interval by adding and
subtracting the margin of error from the sample mean.
That is, the 95% confidence interval ranges
from (xx̄ – margin of error)
to
x̄ + margin of error)
(x
We can write this confidence interval more formally as
x̄x – E < μ < x̄x + E
or more briefly as
x̄x ± E
Copyright © 2009 Pearson Education, Inc.
Slide 8.2- 4
Copyright © 2009 Pearson Education, Inc.
Slide 8.2- 5
EXAMPLE 1 Computing the Margin of Error
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 xx̄ = 77.0 grams and a sample standard
deviation of s = 58.6 grams.
Solution: The sample size is n = 267 and the standard
deviation for the sample is s = 58.6, so the margin of error is
E ≈ 2s = 2 × 58.6 = 7.2
n
267
Copyright © 2009 Pearson Education, Inc.
Slide 8.2- 6
EXAMPLE 1 Computing the Margin of Error
Solution: (Cont.)
The sample mean is x̄ = 77.0 grams, so the 95% confidence
interval extends approximately from 77.0 – 7.2 = 69.8 grams to
77.0 + 7.2 = 84.2 grams. We write this result more formally as
69.8 grams < μ < 84.2 grams
or more simply as 77.0 ± 7.2 grams.
We are 95% confident that the population mean for protein intake
of all American men is between 69.8 and 84.2 grams.
It is interesting to note that even the lower number in this
confidence interval (69.8 grams) is greater than the recommended
daily protein allowance for men of 55–60 grams, suggesting that
actual protein consumption is significantly greater than
recommended.
Copyright © 2009 Pearson Education, Inc.
Slide 8.2- 7
Interpreting the Confidence Interval
Figure 8.10 This figure
illustrates the idea behind
confidence intervals. The
central vertical line
represents the true
population mean, μ. Each of
the 20 horizontal lines
represents the 95%
confidence interval for a
particular sample, with the sample mean marked by the dot in the
center of the confidence interval. With a 95% confidence interval, we
expect that 95% of all samples will give a confidence interval that
contains the population mean, as is the case in this figure, for 19 of the
20 confidence intervals do indeed contain the population mean. We
expect that the population mean will not be within the confidence
interval in 5% of the cases; here, 1 of the 20 confidence intervals (the
sixth from the top) does not contain the population mean.
Copyright © 2009 Pearson Education, Inc.
Slide 8.2- 8
EXAMPLE 2 Constructing a Confidence Interval
A study finds that the average time spent by eighth-graders
watching television is 6.7 hours per week, with a margin of error
of 0.4 hour (for 95% confidence). Construct and interpret the
95% confidence interval.
Solution: The best estimate of the population mean is the sample
mean, x̄ = 6.7 hours.
We find the confidence interval by adding and subtracting the
margin of error from the sample mean, so the interval extends from
6.7 – 0.4 = 6.3 hours to 6.7 + 0.4 = 7.1 hours.
Copyright © 2009 Pearson Education, Inc.
Slide 8.2- 9
EXAMPLE 2 Constructing a Confidence Interval
Solution: (cont.)
We can therefore claim with 95% confidence that the average time
spent watching television for the entire population of eighthgraders is between 6.3 and 7.1 hours, or
6.3 grams < μ < 7.1 grams
If 100 random samples of the same size were taken, we would
expect the confidence intervals of 95 of those samples to contain
the population mean.
Copyright © 2009 Pearson Education, Inc.
Slide 8.2- 10
Choosing Sample Size
Solve the margin of error formula E ≈ 2s /
 2s 
n   
E 
n for n.
2
Choosing the Correct Sample Size
In order to estimate the population mean with a specified
margin of error of at most E, the size of the sample should
2
be at least
 2 

n  
E 
where σ is the population standard deviation (often
estimated by the sample standard deviation s).
Copyright © 2009 Pearson Education, Inc.
Slide 8.2- 11
EXAMPLE 6 Constructing a Confidence Interval
You want to study housing costs in the country by sampling
recent house sales in various (representative) 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 (at a minimum)
should be used to ensure that the sample mean is within
a. $500 of the true population mean?
b. $100 of the true population mean?
Solution:
a. With E = $500 and σ estimated as $7,200, the minimum sample
size that meets the requirements is
 2
n  
E
2
  2  7,200 
2
  
  28.8  829.4
  500 
2
Copyright © 2009 Pearson Education, Inc.
Slide 8.2- 12
EXAMPLE 6 Constructing a Confidence Interval
Solution:
a. (cont.) Because the sample size must be a whole number, we
conclude that the sample should include at least 830 prices.
b. With E = $100 and σ = $7,200, the minimum sample size that
meets the requirements is
 2
n  
E
2
  2  7,200 
2
  
  144  20,736
  100 
2
Notice that to decrease the margin of error by a factor of 5
(from $500 to $100), we must increase the sample size by a
factor of 25. That is why achieving greater accuracy
generally comes with a high cost.
Copyright © 2009 Pearson Education, Inc.
Slide 8.2- 13
TIME OUT TO THINK
If you decide you want a smaller margin of error for a
confidence interval, should you increase or decrease the
sample size? Explain.
Copyright © 2009 Pearson Education, Inc.
Slide 8.2- 14
The End
Copyright © 2009 Pearson Education, Inc.
Slide 8.2- 15