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Section 8.2
Estimating Population Means (Large Samples)
HAWKES LEARNING SYSTEMS
math courseware specialists
Copyright © 2008 by Hawkes Learning
Systems/Quant Systems, Inc.
All rights reserved.
σ unknown, but a large sample (n ≥ 30) or population is normally distributed
Large sample:Use z, the normal distrib’n.
Small sample:Use the Student t-distribution.
z t
TODAY
HAWKES LEARNING SYSTEMS
math courseware specialists
Criteria for estimating the population mean for large samples:
Confidence Intervals
8.2 Estimating Population Means
(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.
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Find the critical value:
Confidence Intervals
8.2 Estimating Population Means
(Large Samples)
Find the critical value for a 95% confidence interval.
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.
Solution:
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Critical Value, zc:
Confidence Intervals
8.2 Estimating Population Means
(Large Samples)
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
HAWKES LEARNING SYSTEMS
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Critical Value, zc:
Confidence Intervals
8.2 Estimating Population Means
(Large Samples)
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
TI-84 confirmation of these values using invNorm(area to left)= z
c = area in the middleSo area in each tail = (1 – c) / 2(1 – 0.90) / 2 = .05(1 – 0.95) / 2 = .025(1 – 0.99) / 2 = .005
Take absolute value of –z.Round Responsibly.
(added content by D.R.S.)
HAWKES LEARNING SYSTEMS
math courseware specialists
Margin of Error, E, for Large Samples:
Confidence Intervals
8.2 Estimating Population Means
(Large Samples)
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.
zc = the critical z-values = the sample standard deviationn = the sample size
This E is the value thatwe will + or – from the sample mean to build a Confidence Interval
Find the margin of error for a 99% confidence
interval, given a sample of size 100 with a sample
standard deviation of 15.50.
Find the margin of error:
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n = 100, s = 15.50, c = 0.99
z0.99 =
Solution:
Confidence Intervals
8.2 Estimating Population Means
(Large Samples)
2.575
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.
Construct a confidence interval:
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c = 0.95, n = 85, s = 14, = 67
z0.95 =
Solution:
Confidence Intervals
8.2 Estimating Population Means
(Large Samples)
1.96
$64.02 < < $69.98
($64.02, $69.98)
67 – 2.98 < < 67 + 2.98
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.
Construct a confidence interval:
HAWKES LEARNING SYSTEMS
math courseware specialists
Confidence Intervals
8.2 Estimating Population Means
(Large Samples)
Large sample, so we choose z.STAT, TESTS, ZintervalUse “Data” if data is in a List.We have “Stats” already given.We don’t know σ so use s.Sample mean and sample size.Confidence Level as a decimal.Highlight Calculate.Press ENTER.
TI-84 easy way
(added content by D.R.S)
TI-84 Solution, continued
Inputs (as described on the previous slide.)
Outputs: “We are 95% confident that the mean monthly maintenance cost is $64.02 < μ < $69.98”
What’s the smallest sample size I need to use?
• Confidence Interval scenario:– I want to estimate the mean of a population– It’s going to take time and cost money to collect a
sample.– If sample is too small, my result is not accurate
enough.,– If sample is too big, I’ve worked too hard and
spent too much for precision I didn’t need.• There’s a way to find it out in advance, for z.
(added content by D.R.S.)
HAWKES LEARNING SYSTEMS
math courseware specialists
Finding the Minimum Sample Size for Means:
Confidence Intervals
8.2 Estimating Population Means
(Large Samples)
When calculating the sample size, round to up to the next whole
number.
zc = the critical z-values = the population standard deviationE = the margin of error
To find the minimum sample size necessary to estimate an average,
use the following formula:
This applies when you qualify to use z, not t.
always bump it upExample: 80.01 bumps up to n = 81 sample size needed
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.
Find the minimum sample size:
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c = 0.99, = 6.5, E = 2
z0.99 =
You will need a minimum sample size of 71.
Solution:
Confidence Intervals
8.2 Estimating Population Means
(Large Samples)
2.575
You can find zc from printed tables or by
using TI-84 invNorm(, as described earlier.
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?
Find the minimum sample size:
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c = 0.99, = 1.8, E = 0.12, z0.99 = 2.575
You will need a minimum sample size of 1492 families.
Solution:
Confidence Intervals
8.2 Estimating Population Means
(Large Samples)