Confidence Intervals

• Jul 30th 2008, 08:26 PM
sebchase0625
Confidence Intervals

During the Rose Bowl, the length (in seconds) of 12 randomly chosen commercial breaks during timeouts (following touchdown, turnover, field goal, or punt) were
65 75 85 95 80 100 90 80 85 85 60 65
(a) Assuming a normal population, construct a 90 percent confidence interval for the mean length of a commercial break during the Rose Bowl.
(b) What are the limitations on your estimate?
How could they be overcome?
• Jul 30th 2008, 09:12 PM
TKHunny
Quote:

Originally Posted by sebchase0625

During the Rose Bowl, the length (in seconds) of 12 randomly chosen commercial breaks during timeouts (following touchdown, turnover, field goal, or punt) were
65 75 85 95 80 100 90 80 85 85 60 65
(a) Assuming a normal population, construct a 90 percent confidence interval for the mean length of a commercial break during the Rose Bowl.
(b) What are the limitations on your estimate?
How could they be overcome?

n = 12

$\displaystyle \sum(x) = 965$

$\displaystyle \sum(x^2) = 79275$

Sample Mean = $\displaystyle \frac{\sum(x)}{n}\;=\;70.41666...$

Sample Standard Deviation = $\displaystyle \sqrt{\frac{\sum(x^2) - \frac{\left(\sum(x)\right)^{2}}{n}}{n-1}}\;=\;12.33221$

Sample Standard Deviation of the Mean = Sample Standard Deviation / $\displaystyle \sqrt{n}\;=\;3.56002$

$\displaystyle \alpha\;=\;1 - 0.90\;=\;0.10$

$\displaystyle \frac{\alpha}{2}\;=\;0.05$

$\displaystyle z_{0.05}\;=\;1.645$ <== You just have to look this up.

Where does that leave us? Any closer? What's next?
• Jul 30th 2008, 10:05 PM
mr fantastic
Quote:

Originally Posted by TKHunny
n = 12

$\displaystyle \sum(x) = 965$

$\displaystyle \sum(x^2) = 79275$

Sample Mean = $\displaystyle \frac{\sum(x)}{n}\;=\;70.41666...$

Sample Standard Deviation = $\displaystyle \sqrt{\frac{\sum(x^2) - \frac{\left(\sum(x)\right)^{2}}{n}}{n-1}}\;=\;12.33221$

Sample Standard Deviation of the Mean = Sample Standard Deviation / $\displaystyle \sqrt{n}\;=\;3.56002$

$\displaystyle \alpha\;=\;1 - 0.90\;=\;0.10$

$\displaystyle \frac{\alpha}{2}\;=\;0.05$

$\displaystyle z_{0.05}\;=\;1.645$ <== You just have to look this up.

Where does that leave us? Any closer? What's next?

A small correction: The value of $\displaystyle t_{0.05}$ for 12 - 1 = 11 degrees of freedom is used, not $\displaystyle z_{0.05}$.

To the OP: This is because the population sd is not known and so the sample sd is used.
• Jul 31st 2008, 05:13 AM
TKHunny
That's what I get for being up that late. Thanks for the redraft.