Download Chapter 7.4 Estimating a Population Mean: σ not known

Chapter 7.4 Estimating a Population Mean: σ not known
We can construct a confidence interval for the population mean using the normal
distribution if the following requirements are met:
•
•
•
Simple Random Sample
Normal Distribution or n > 30
σ known
In most “real life” situations σ is not known. For if it is known then most likely µ
would also be known and there would be no need to estimate it.
If σ is not known we can still estimate µ by using the t–distribution. The t–distribution is
a bell–shaped curve similar to the Normal distribution. Use table A–3.
The number of degrees of freedom is the number of sample values that can vary after
certain restrictions have been imposed on all the data values. For this section the
degrees of freedom: df = n – 1
Requirements:
• Sinple Random Sample
• Population is Normal or n > 30
Point Estimate for the population mean, µ: x .
Margin of Error: E = t α / 2 s
n
Confidence Interval: x – E < µ < x + E
Using table A–3
Note that the area (probability) is found on the top rows and the t–score (like the z–score)
is found in the center. Also the degrees of freedom are on the left column.
Example:
For a 98% confidence level with n = 12, find tα/2
df – 12–1 = 11
α = .02 (second row)
α/2 = .01 (first row)
tα/2 = 2.718
Notice the top row is α/2 and the bottom row is α.
<–– α/2
<–– α
Example:
A department store wishes to estimate the number of customers who daily pass a
particular location in their store. A sampling of the traffic on 18 randomly selected
days at this location showed a mean of 1246 customers and a standard deviation of
145. Assume the distribution is approximately normal.
a.
Find a 90% confidence interval for the mean number of customers who daily
pass this location.
x = 1246
s = 145
n = 18
df = 17
CL = .90
α = .10
α/2 = .05
tα/2 = 1.740
b.
E = (1.74) 145 = 59.4677
18
1246 – 59.4677 < µ < 1246 + 59.4677
1186.5323 < µ < 1305.4677
Based on the confidence interval, does it appear that more than 1000 customers
pass this location? Why or why not?
Answer: Yes, because the confidence interval values are greater than 1000.
c.
Based on the confidence interval, does it appear that 1200 customers pass this
location? Why or why not?
Answer: Yes, because 1200 is contained in the confidence interval.
d.
Based on the confidence interval, does it appear that 2000 customers pass this
location? Why or why not?
Answer: No, because the confidence interval values do not contain 2000.
e.
Based on the confidence interval, does it appear that fewer than 1000 customers
pass this location? Why or why not?
Answer: No, because the confidence interval values are greater than 1000.