Confidence Intervals

**sebchase0625** · July 30th 2008, 08:26 PM

I have read and re-read the chapter on confidence intervals but I can't seem to grasp the concept. Please help!!

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?

**TKHunny** · July 30th 2008, 09:12 PM

Originally Posted by sebchase0625

I have read and re-read the chapter on confidence intervals but I can't seem to grasp the concept. Please help!!

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

$\sum(x) = 965$

$\sum(x^2) = 79275$

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

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

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

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

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

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

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

**mr fantastic** · July 30th 2008, 10:05 PM

Originally Posted by TKHunny

n = 12

$\sum(x) = 965$

$\sum(x^2) = 79275$

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

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

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

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

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

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

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

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

To the OP: This is because the population sd is not known and so the sample sd is used.

**TKHunny** · July 31st 2008, 05:13 AM

That's what I get for being up that late. Thanks for the redraft.

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