Question

A copper smelting process is supposed to reduce the arsenic content of the copper to less than 1000 ppm. let μ denote the mean arsenic content for copper treated by this process, and assume that the standard deviation of arsenic content is σ=100 ppm. the sample mean arsenic content xˉ of 75 copper specimens will be computed, and the null hypothesis h0:μ≥1000 will be tested against the alternate h1:μ<1000.
a.a decision is made to reject h0 if xˉ≤980. find the level of this test.

Answer 1

The rejection region is give by 

$|z_{test}|\ \textgreater \ z_{\alpha/2}$

where the test statistics is given by

$\frac{\bar{x}-\mu}{\sigma/\sqrt{n}} = \frac{980-1000}{100/\sqrt{75}} \\ \\ = \frac{-20}{100/8.6603} = \frac{-20}{11.5470} =-1.73$

i.e. $|z_{test}|=|-1.73|=1.73$

Thus, $z_{\alpha/2}=1.73$

Using the statistical table, the level of the test is 0.04.