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Thread: finding number of standard deviations between sample and population mean

  1. #1
    Member Jskid's Avatar
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    finding number of standard deviations between sample and population mean

    Suppose that the mean mileage of all pickup trucks of a particular make is unknown and will be estimated from a random sample of 36 trucks. The sample mean mileage was found to be 6 litres per 100 km. Assume that the standard deviation of the mileages of all trucks is .5 litres per 100 km. There is a 0.95 probability that the mean of the sample will be within z SDs of the unknown mileage. What is z?

    I believe I find the z-score associated with 0.95 but I'm not sure if I need to do something with the sample mean or SD.
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  2. #2
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    Quote Originally Posted by Jskid View Post
    Suppose that the mean mileage of all pickup trucks of a particular make is unknown and will be estimated from a random sample of 36 trucks. The sample mean mileage was found to be 6 litres per 100 km. Assume that the standard deviation of the mileages of all trucks is .5 litres per 100 km. There is a 0.95 probability that the mean of the sample will be within z SDs of the unknown mileage. What is z?

    I believe I find the z-score associated with 0.95 but I'm not sure if I need to do something with the sample mean or SD.
    A 0.95 probability that the sample mean is within z standard deviations
    means that you need to locate the two 2.5% regions at the tails.

    Hence, you need the z-score corresponding to 0.975 instead.
    Also, as you have the "population" SD

    $\displaystyle Z=\displaystyle\huge\frac{sample\ mean-\mu}{\frac{\sigma}{\sqrt{36}}}$

    where $\displaystyle \mu$ is the population mean.

    The above calculation is taken for sample mean greater than population mean.
    The graph is symmetrical however.
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  3. #3
    Member Jskid's Avatar
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    The area under the normal distribution curve for 0.25% is z=-1.96 and for 97.25% is z=1.96. The answer to my question is 1.96. So why do I need $\displaystyle Z=\displaystyle\huge\frac{sample\ mean-\mu}{\frac{\sigma}{\sqrt{36}}}$?
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    Quote Originally Posted by Jskid View Post
    The area under the normal distribution curve for 0.25% is z=-1.96 and for 97.25% is z=1.96. The answer to my question is 1.96. So why do I need $\displaystyle Z=\displaystyle\huge\frac{sample\ mean-\mu}{\frac{\sigma}{\sqrt{36}}}$?
    You don't need it for Z.
    You could use that Z to get a reading for population mean if needed.
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