•

Module 11: Confidence Intervals Part 1
B. Creating confidence intervals to estimate the population mean with small samples (n 30).
Example 6
:
The auditor at an equipment rental firm wants to estimate the mean number of days a piece of ditch digging
equipment is rented out per person per time. She takes a random sample of 14 rental invoices, the results of which are listed
in the table below. Use this data to construct a 99% confidence interval for the average number of days the ditch digging
machine is rented out per person per time.
First we use Excel to calculate the sample mean ()and sample standard deviation(s).
= 2.1429, s = 1.2924, n = 14, and n - 1 = 13
With a 99% level of confidence, α/2 = .005
Next we use Excel to calculate
t
/2, n-1
=
t
.005, 13
•
2.1429
=AVERAGE(A1:A14
)
1.2924
=STDEV(A1:A14)
t
α/2, n-1
t
.005, 13
3.0123
=TINV(2*0.005
,13)
3
1
3
2
5
1
2
1
4
2
1
3
1
1

Module 11: Confidence Intervals Part 1
B. Creating confidence intervals to estimate the population mean with small samples (n 30).
Example 6
:
From the previous slide, we have:
= 2.1429, s = 1.2924, n = 14, and n - 1 = 13, and =
t
.005, 13
=
3.0123
To construct our confidence, we just plug the data into our formula:
±
t
/2, n-1
= 2.1429 ± 3.0123
= 2.1429 ± 1.0405
P(1.1024
3.1834) = 99%
Interpretation:
We're 99% confident that the mean length of time the ditch digging machine is rented out is between 1.1024
and 3.1834 hours.
•

Module 11: Confidence Intervals Part 1
B. Creating confidence intervals to estimate the population mean with small samples (n 30).
Example 7
:
A researcher with way too much time oh his hands wants to estimate the average amount of money Cuyahoga
County CPAs spend on lunch. Toward that end he randomly samples 20 Cuyahoga County CPAs and finds a sample mean of
$11.24 with a sample standard deviation of $1.79. Assuming there are 330 CPAs in Cuyahoga County, construct a 95%
confidence interval for the amount of money Cuyahoga County CPAs spend on lunch.
Since we're given a population size, we must use the finite correction factor:
±
t
/2, n-1
From the text of the question we know the following:
= $11.24, s = $1.79, n = 20, n - 1 = 19 N = 330, and α/2 = .025 (95% Confidence)
All we need is
calculate
t
/2, n-1
=
t
.025, 19 = 2.0930
±
t
/2, n-1
=
11.24
± 2.0930
= 11.24 ± .8132
P(10.4268
12.0532) = 95%
Interpretation:
We're 95% confident that the mean amount of money a Cuyahoga County CPA spends on lunch is between
$10.43 and $12.05.
•
t
α/2, n-1
t
.025, 19
2.0930
=TINV(2*0.025,19)

Module 11: Confidence Intervals Part 1
B. Creating confidence intervals to estimate the population mean with small samples (n 30).
Example 8
:
A random sample of 22 is taken from a population of 700. This produces a sample mean of 111.7 and a sample
standard deviation of 13. Construct a 98% confidence interval for the population mean.
Since we're given a population size, we must use the finite correction factor:
±
t
/2, n-1
From the text of the question we know the following:
= 111.7, s = 13, n = 22, n - 1 = 21, N = 700 and α/2 = .01 (98% Confidence)
All we need is
calculate
t
/2, n-1
=
t
.01, 21 =
2.5176
±
t
/2, n-1
=
111.7
± 2.5176
= 111.7 ± 6.8723
P(104.8277
118.5723) = 98%
Interpretation:
We're 98% confident that the mean for the population from which the sample was drawn is between 104.8277
and 118.5723.

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- Spring '14
- DebraACasto
- Statistics, Normal Distribution, Standard Deviation