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I'm having trouble finding the right answer to this question:

If $H_0: \mu= 240$ is tested against $H_1: \mu < 240$ at the $α= 0.01$ level of significance with a random sample of twenty-five normally distributed observations, what pro- portion of the time will the procedure fail to recognize that $\mu$ has dropped to $220$? Assume that $\sigma = 50$.

This is what I did, but I got the wrong answer

$=p(y-240/50/\sqrt{25} \leq 220-240/50/\sqrt{25})$

$=p(Z \leq -2.0)$

$=0.9772$

Any help would be appreciated.

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  • $\begingroup$ For starters, $P(Z \leq -2.0) = P(Z > 2.0) = 1 - P(Z < 2.0)$. I've not actually looked at your other working out properly yet so can't say if $P( Z \leq -2)$ is correct or not, but your value for $P(Z \leq -2)$ is wrong. $\endgroup$ – Kaish Apr 13 '13 at 20:35
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Your test statistic would be $\frac{\bar{Y}-240}{\sigma/\sqrt{n}}$. If the null hypothesis were true, this would be N(0,1) distributed. So, you would first need to establish your rejection rule so that your $\alpha$ level would be 0.01. It will be of the form "Reject the null hypothesis when $\frac{\bar{Y}-240}{\sigma/\sqrt{n}} < ***$". What you then need to find is $Pr(\frac{\bar{Y}-240}{\sigma/\sqrt{n}}<***$ when in reality $\mu$ is 220).

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