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A random sample of $20$ items is drawn from a population that is known to be normally distributed. The sample has a mean of $60$ and a standard deviation of $16$. Test the hypothesis that the true mean of the population from which the sample is taken is $68$.

My calculations:

Find std error of mean = $\frac{16}{\sqrt{20}}= 3.58$

Therefore, the population mean is very likely to be between 7.16-10.73 of $60$. Is my conclusion correct? Did I set this up correctly?

Mike Pierce
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David
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1 Answers1

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It isn't clear from what you have said what your conclusion is. You have also not stated the significance level you are working to.

A model answer will look something like this:

$H_0: $ population mean $\mu = 68$

$H_1: $ population mean $\mu \ne 68$

Assume that the standard deviation of the sample is a good estimate of the population standard deviation.

Conduct a two-tailed test at 5% significance level (which is customary).

z-statistic $z=\frac {\bar x-\mu}{\sigma / \sqrt{n}} = \frac {60-68}{16 / \sqrt{20}}=-2.236 $

This is greater in magnitude than the critical value for a 5% two-tailed test (which is 1.960), so there is evidence to suggest that the true mean is not equal to 68. Conclude that $H_1$ is true.

Alternatively you could calculate the $p$ value for the test statistic, which is 0.0254 or 2.54%

tomi
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  • If a sample size is smaller than 30, then one should use the t-dstribution (rule of thumb). I think this is worth to be mentioned. – callculus42 Jul 16 '15 at 00:15
  • Good point. That is why I made my simplifying assumption. The critical value for t-test is 2.086 so same conclusion. – tomi Jul 16 '15 at 00:32
  • But you the degrees of freedom are $df=n-1=19$. Therefore $t_{19;0.975}=2.093$ Your conclusion remains the same :) – callculus42 Jul 16 '15 at 00:48