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Math Help - Confidence Intervals

  1. #1
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    Confidence Intervals

    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?
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  2. #2
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    Quote Originally Posted by sebchase0625 View Post
    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?
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  3. #3
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    Quote Originally Posted by TKHunny View Post
    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.
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  4. #4
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    That's what I get for being up that late. Thanks for the redraft.
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