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QUESTION The times to carry out a certain job in an industrial
process is known to have mean 50 minutes and a standard deviation
of 4 minutes. It is claimed that a new method of doing the job
will save time.  A sample of 36 tests carried out using this new
method gave a mean time of 48 minutes.  Assuming the same standard
deviation for the new method, examine the claim at the $5\%$
level.  If the new mean was in fact 48 minutes find the
probability of type II error using the standard test.





ANSWER

$H_0:\mu=50\ \ H_1: \mu<50\ \ \alpha=5\%\\
  \sigma=4 $ given , test single mean 1a. \\
  $z=\frac{\overline{x}-\mu_0}{\frac{\sigma}{\sqrt{n}}}\sim
  N(0,1)$

  $\begin{array}{l}
  \overline{x}=48\\
  n=36\\
  z=\frac{48-50}{\frac{4}{6}}=-3
  \end{array}
  \ \ \
  \begin{array}{c}
  \epsfig{file=641-5-1.eps, width=60mm}
  \end{array}$

  Hence reject $H_0$, accept $H_1:\mu<50$.

  If $\mu_{\textrm{new}}=48\ \ \overline{x}\sim
  N(48,\frac{4^2}{36})$
  \begin{eqnarray*}
  P(\textrm{ Type II error})&=&P(\textrm{accept }h_0|H_1 \textrm{
  true})\\
  &=&P(Z>-1.6449)\\
  &=&P(\frac{\overline{x}-50}{6}>-1.6449)\\
  &=&P(\overline{x}>48.9034)\\
  &=&1-\Phi(\frac{48.9034-48}{\frac{4}{6}})\\
  &=&1-\Phi(1.3551)\approx 1-0.912 \approx 0.088
  \end{eqnarray*}

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