Download Example: You wish to estimate the average number of housing starts

Example: You wish to estimate the
average number of housing starts in all
large cities in the United States. You have
a random sample of 25 cities and obtain
the number of housing starts in each. The
sample mean is 525 with a sample
standard deviation of 40.
Given values:
n=25,
sample standard deviation =40,
Sample mean = 525
Confidence = 95% (default)
Identifier: “What is (or estimate) the
population mean?”
First calculate the typical error in a
sample mean. This is value is 40 divided
by the square root of 25 = 8. Therefore
when using this sample mean the typical
error you would expect is estimated to be
eight.
Next determine how far you have to go
either side of the sample mean for the
specified confidence. With 95%
confidence and 24 degrees of freedom,
you have to go 2.064 standard errors
(2.064*8 = 16.512) or 16.512 either side
of the sample mean to have 95%
confidence that the population mean is
within the interval.
With 95% confidence we can say that
average number of housing starts in all
cities of interest is 525 with a maximum
possible error of  16.512
Requirements to use a t table:
Original population of “number of
housing starts” values must be normal or a
very large sample or cities
Simple random sample of cities
Exercise: You are measuring the size of
houses in a city in thousands of square feet.
From a random sample of 225 houses, you
find a sample mean and standard deviation
of 2.0 and .2 respectively. With 90%
confidence, what is the estimate of the
average house size of all houses in the city
and what is the estimate’s margin of error?
Given values:
n = 225
sample standard deviation = .2
sample mean = 2.0
confidence = 90%
What is the estimate of the average house
size
Typical error = .2 / sqrt(225) = 0.01333
With 90% confidence and 224 degrees of
freedom, you have to go 1.6577 standard
errors or 0.02204 either side of the sample
mean to have a 90% confidence that the
population mean is within the interval.
1.677 * 0.01333 = 0.022
With 90% confidence we can say that the
average house size is 2,000 square feet with
a maximum possible error of  22 square
feet
The average paper length in a
manufacturing process has been 11 inches
in the past. You think the process is
producing paper that is too short. You take
a random sample of 36 sheets and
determine that the sample average is 10.98
and the sample standard deviation is 0.06
inches. At the five percent level of
significance can you say the process is out
of control.
Given values:
n=36,
sample mean = 10.98
S = 0.06,
 = 0.05
Hypothesized mean = 11
Identifier: Does the population mean take
on a specific set of values? In this case is
the average paper less than 11?
Determine the null and alternative. (What
you wish to show goes in the alternative
and the equal sign goes in the null)
The null is that the average paper length is
11 inches
The alternative is that the average paper
length is less than 11 inches.
Determine the values of the sample mean
that would be unlikely if the average paper
length was 11 and would support the
alternative. In this case small sample means
would cause you support a small
population mean and lead you to reject the
null and support the alternative. Using the
t-table with 35 degrees of freedom, any
sample mean more than 1.690 sample
standard errors or more below the mean
would occur only five percent of the time.
(Only one side is considered so the alpha is
not divided by two.)
Rejection Region: Reject Ho if t < -1.690
(1.690 or more below)
A real estate agent claims that the average
house size in a city is 2,500 square feet.
You take a random sample of 225 houses
in that city measuring the size of the houses
in thousands of square feet. You find a
sample mean and standard deviation of
2400 and 200 respectively. At a 5% level
of significance, can you conclude that the
real estate agent is incorrect? Give the
conclusion as if you were talking to
someone who has not had statistics.
Given values:
n=
sample mean =
S=
=
Hypothesized mean =
Determine how far your data is below the
hypothesized value. In this case the sample
standard error is 0.01 (0.06 divided by the
square root of 36). Your sample mean then
falls two standard errors below 11 inches.
Test Statistic: t = (10.98 – 11) / 0.01 = -2
( two sample standard errors below)
This is an unlikely number and would
cause you to say that the null is false and
support the alternative
Conclusion: We can say the
process is out of control and
producing paper that is too
short on average.