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Math Help - A random sample of grades of 50

  1. #1
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    A random sample of grades of 50

    A random sample of grades of 50 students in physics out of 200 students showed a mean of 75 and a standard deviation of 10. What are the 95% confidence limits for estimates of mean of the 200 grades? Since the population size is not very large compared with the sample size, make adjustments for this?

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  2. #2
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    Here we have an unknown population mean  \mu , and a known standard deviation  \sigma  based on a simple random sample of size n, which we are also given.

     \bar{X} = 75

     n = 50

     \sigma = 10

    Since our sample is large enough (usually n > 30 will suffice), we consider the sample mean  \bar{X} normally distributed, with the same expectation for the population mean  \mu , but with standard error  \frac{\sigma}{\sqrt{n}} . We standardize and get a random variable

      <br />
Z = \frac{\bar{X}-\mu}{\sigma / \sqrt{n}} = <br />
\frac{75-\mu}{1.4142}

    that depends on the population mean parameter \mu (which is to be estimated), but with a normal distribution independent of \mu. Thus, we can find the range endpoints -z and z, independent of the population mean parameter, which is the range that Z falls somewhere in between with 95% probability. So we set the following equation up:

     P(-z \leq Z \leq z) = 1 - \alpha = 0.95

     \implies P(Z \leq z) = 1 - \frac{\alpha}{2} =0.975


     \implies z = 1.96 (which you could have looked up easily enough (or had it memorized)

    So now we have

     0.95 = 1 - \alpha = P(-z \leq Z \leq z) = <br />
P(-1.96 \leq \frac{\bar{X}-\mu}{\sigma / \sqrt{n}} \leq 1.96)<br />

     <br />
= P\bigg{(}75 - 1.96(1.4142)<br />
\leq \mu \leq \bar{x} + 1.96 (1.4142 )\bigg{)}<br />

     <br />
= P(75 - 2.7719 \leq \mu \leq 75 + 2.7719)<br />


     CI =  (72.2281 ; 77.7719)
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