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Math Help - not sure which on to choose

  1. #1
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    not sure which on to choose

    ) A tyre manufacturer claims that his new steel-belted tire has a mean life expectancy of 40,000 miles. A consumer association decides to test these claims against the alternatives that the mean life expectancy is less than 40,000 miles. 100 tyres are selected at random and the sample mean is found to be 38,500 miles with the sample standard deviation of 2,000 miles.


    Perform the hypothesis test at the 5% (or 0.05) level of significance.



    because its a sample standard deviation i should use xbar- meu = a and then SD/ n square root= b
    so is it 38500- 40000 = 1500 and the 2000 / sq of 100 = 10 = 2000/ 10 = 200 so
    1500/ 200 = 7.5



    so
    h0 meu= 40000
    h1 meu doesnt=40000
    5%
    Last edited by crashuk; June 6th 2008 at 06:58 AM.
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  2. #2
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    xbar = 30 s squared = 100 and n = 64
    Test the hypothesis
    q H0 : μX = 25 and
    H1 : μX ≠ 25:

    would it be 30-25/ s= 10 n= 10=
    Last edited by crashuk; June 6th 2008 at 06:30 AM.
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  3. #3
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    Quote Originally Posted by crashuk View Post
    ) A tyre manufacturer claims that his new steel-belted tire has a mean life expectancy of 40,000 miles. A consumer association decides to test these claims against the alternatives that the mean life expectancy is less than 40,000 miles. 100 tyres are selected at random and the sample mean is found to be 38,500 miles with the sample standard deviation of 2,000 miles.


    Perform the hypothesis test at the 5% (or 0.05) level of significance.



    because its a sample standard deviation i should use xbar- meu = a and then SD/ n square root= b
    so is it 38500- 40000 = 1500 and the 2000 / sq of 100 = 10 = 2000/ 10 = 200 so
    1500/ 200 = 7.5



    so
    h0 meu= 40000
    h1 meu doesnt=40000
    5%
    You need to say what test you're using. I assume you're using a z-test since, although the population sd is unknown, n is large and the assumption of normality is reasonable.

    So you have z = -7.5. What's the critical value of z for a one-sided test at the 0.05 level of significance? Is -7.5 larger or smaller than this value? What conclusion do you draw?
    Last edited by mr fantastic; June 9th 2008 at 03:56 AM. Reason: Added the negative .... I should've checked the arithmetic more closely
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    i thought it was a two sided test?
    = and not= ?
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  5. #5
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    Quote Originally Posted by crashuk View Post
    i thought it was a two sided test?
    = and not= ?
    " test these claims against the alternatives that the mean life expectancy is less than 40,000 miles."

    is the only alternative stated in the question. This implies a one sided test.
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    Quote Originally Posted by mr fantastic View Post
    You need to say what test you're using. I assume you're using a z-test since, although the population sd is unknown, n is large and the assumption of normality is reasonable.

    So you have z = 7.5. What's the critical value of z for a one-sided test at the 0.05 level of significance? Is 7.5 larger or smaller than this value? What conclusion do you draw?
    how do you work that out?
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  7. #7
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    Quote Originally Posted by crashuk View Post
    how do you work that out?
    Usually you would use a table of critical values to get the critical value (it can of course be got from the usual four figure tables):

    The critical value of z is -1.64.

    -7.5 < -1.64 therefore the result is significant at the 0.05 level.

    Therefore you reject the null hypothesis.

    The manufacturers claim appears to be bogus.
    Last edited by mr fantastic; June 9th 2008 at 03:57 AM. Reason: Added the negatives.
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  8. #8
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    Unbiased estimate?

    Shouldn't you use an unbiased estimate of the sample variance, because it's not normally distributed? like:

    \frac{n}{n-1}(2000^2) = 4040404.04

    Then if z<-1.6448, reject H_{o}: \mu=40000

    and z= \frac{38500-40000}{\sqrt{\frac{4040404.04}{100}}}

    =-7.462405778<-1.6448, so reject H_{o}
    Last edited by Nyoxis; June 9th 2008 at 09:35 AM.
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  9. #9
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    Quote Originally Posted by Nyoxis View Post
    Shouldn't you use an unbiased estimate of the sample variance, because it's not normally distributed? like:

    \frac{n}{n-1}(2000^2) = 2020.20202

    Then if z<-1.6448, reject H_{o}: \mu=40000

    and z= \frac{38500-40000}{\sqrt{\frac{4040404.04}{100}}}

    =-7.462405778<-1.6448, so reject H_{o}
    I think n = 100 is large enough for the correction factor to be ignored. Also, I think an asusmption of normality is reasonable (the lifetime of a tyre might just follow a Weibull distribution, say. But the value of the parameters would I think make a normal approximation reasonable)
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