HAWKES LEARNING SYSTEMS
math courseware specialists
Copyright © 2008 by Hawkes Learning
Systems/Quant Systems, Inc.
All rights reserved.
Section 8.3
Estimating Population Means
(Small Samples)
σ unknown, and a small sample (n < 30)
AND population is normally distributed
Large sample:
Use z, the normal
distribution.
Small sample:
Use the Student
t-distribution.
This is where we
were in the
previous lesson.
TODAY
z
t
(added content by D.R.S.)
HAWKES LEARNING SYSTEMS
Confidence Intervals
math courseware specialists
8.3 Estimating Population Means
(Small Samples)
Criteria for estimating the population mean for small samples:
• All possible samples of a given size have an equal
probability of being chosen.
• The size of the sample is less than 30 (n < 30).
• The distribution of the population is approximately
normal.
• 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.
HAWKES LEARNING SYSTEMS
Confidence Intervals
math courseware specialists
8.3 Estimating Population Means
(Small Samples)
Margin of Error, E, for Small Samples:
ta/2 = the critical t-value
a=1–c
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
math courseware specialists
8.3 Estimating Population Means
(Small Samples)
Find the margin of error:
Assuming the population is approximately normal and all
samples have an equal probability of being chosen, find the
margin of error for a sample size of 10 given that s = 15.5
and the level of confidence is 95%. Printed table lookup:
Solution:
C=95% or
α=.05 or α/2=0.025
d.f. = n – 1 = 9
n = 10, s = 15.5, c = 0.95, a = 1 – 0.95 = 0.05
t0.05/2 = t0.025 = 2.262
HAWKES LEARNING SYSTEMS
Confidence Intervals
math courseware specialists
8.3 Estimating Population Means
(Small Samples)
Construct a confidence interval:
A student records the repair cost for 20 randomly selected
computers. A sample mean of $216.53 and a standard deviation
of $15.86 are subsequently computed. Determine the 98%
confidence interval for the mean repair cost for all computers.
Assume the criteria for this section are met.Printed table lookup: C=98%
or α=.02 or α/2=0.01
Solution:
d.f. = n – 1 = 19
n = 20,
= 216.53, s = 15.86, c = 0.98, a = 1 – 0.98 = 0.02
t0.02/2 = t0.01 = 2.539
216.53 – 9.00 <  < 216.53+ 9.00
$207.53 <  < $225.53
($207.53, $225.53)
Using TI-84 to find critical value of t
•
•
•
•
•
Similar to invNorm (area to left ) = critical z
But t involves degrees of freedom, too
invT(area to left, df) = critical t value
invT is only available on TI-84
TI-83/Plus owners:
– Use printed tables if asked for critical value
– If asked only for the interval, you do have TInterval
(added content by D.R.S.)
Using TI-84 to find critical value of t
• From the two examples they demonstrated
with primitive formulas, we use invT
• C = 95%, α = 0.05, α/2 = 0.025, n = 10, df = 9
• C = 98%, α = 0.02, α/2 = 0.01, n = 20, df = 19
• Their table values
were 2.262 and 2.539
(added content by D.R.S.)
HAWKES LEARNING SYSTEMS
Confidence Intervals
math courseware specialists
8.3 Estimating Population Means
(Small Samples)
Construct a confidence interval: Using TI-84 STAT, TESTS, TInterval
A student records the repair cost for 20 randomly selected
computers. A sample mean of $216.53 and a standard deviation
of $15.86 are subsequently computed. Determine the 98%
confidence interval for the mean repair cost for all computers.
Assume the criteria for this section are met.
Solution:
Inputs
Outputs
Note that these
results differ from
theirs by a penny.
This is because
the calculator’s
T critical value is
more precise.
(added content by D.R.S.)