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Section 8.2
Estimating Population Means
(Large Samples)
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Confidence Intervals
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8.2 Estimating Population Means
(Large Samples)
Criteria for estimating the population mean for large samples:
• All possible samples of a given size have an equal
probability of being chosen.
• The size of the sample is at least 30 (n ≥ 30).
• The population’s standard deviation is unknown.
When all of the above conditions are met, then the
distribution used to calculate the margin of error for the
population mean is the Student t-distribution.
However, when n ≥ 30, the critical values for the t-distribution
are almost identical to the critical values for the normal
distribution at corresponding levels of confidence.
Therefore, we can use the normal distribution to approximate
the t-distribution.
HAWKES LEARNING SYSTEMS
Confidence Intervals
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8.2 Estimating Population Means
(Large Samples)
Find the critical value:
Find the critical value for a 95% confidence interval.
Solution:
To find the critical value, we first need to find the values for
–z0.95 and z0.95.
Since 0.95 is the area between –z0.95 and z0.95, there will be
0.05 in the tails, or 0.025 in one tail.
HAWKES LEARNING SYSTEMS
Confidence Intervals
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8.2 Estimating Population Means
(Large Samples)
Critical Value, zc:
Critical z-Values for Confidence Intervals
Level of Confidence, c
zc
0.80
1.28
0.85
1.44
0.90
1.645
0.95
1.96
0.98
2.33
0.99
2.575
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Confidence Intervals
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8.2 Estimating Population Means
(Large Samples)
Margin of Error, E, for Large Samples:
zc = the critical z-value
s = the sample standard deviation
n = the sample size
When calculating the margin of error, round to one more decimal place
than the original data, or the same number of places as the standard
deviation.
HAWKES LEARNING SYSTEMS
Confidence Intervals
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8.2 Estimating Population Means
(Large Samples)
Find the margin of error:
Find the margin of error for a 99% confidence
interval, given a sample of size 100 with a sample
standard deviation of 15.50.
Solution:
n = 100, s = 15.50, c = 0.99
z0.99 = 2.575
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Confidence Intervals
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8.2 Estimating Population Means
(Large Samples)
Construct a confidence interval:
A survey of 85 homeowners finds that they spend on average $67
a month on home maintenance with a standard deviation of $14.
Find the 95% confidence interval for the mean amount spent on
home maintenance by all homeowners.
Solution:
c = 0.95, n = 85, s = 14,
= 67
z0.95 = 1.96
67 – 2.98 <  < 67 + 2.98
$64.02 <  < $69.98
($64.02, $69.98)
HAWKES LEARNING SYSTEMS
Confidence Intervals
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8.2 Estimating Population Means
(Large Samples)
Finding the Minimum Sample Size for Means:
To find the minimum sample size necessary to estimate an average,
use the following formula:
zc = the critical z-value
 = the population standard deviation
E = the margin of error
When calculating the sample size, round to up to the next whole
number.
HAWKES LEARNING SYSTEMS
Confidence Intervals
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8.2 Estimating Population Means
(Large Samples)
Find the minimum sample size:
Determine the minimum sample size needed if you wish to
be 99% confident that the sample mean is within two units
of the population mean, given that  = 6.5. Assume that the
population is normally distributed.
Solution:
c = 0.99,  = 6.5, E = 2
z0.99 = 2.575
You will need a minimum sample size of 71.
HAWKES LEARNING SYSTEMS
Confidence Intervals
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8.2 Estimating Population Means
(Large Samples)
Find the minimum sample size:
The electric cooperative wishes to know the average household
usage of electricity by its non-commercial customers. They
believe that the mean is 15.7 kWh per day for each family with a
variance of 3.24 kWh.
How large of a sample would be required in order to estimate the
average number of kWh of electricity used daily per family at the
99% confidence level with an error of at most 0.12 kWh?
Solution:
c = 0.99,  = 1.8, E = 0.12, z0.99 = 2.575
You will need a minimum sample size of 1492 families.