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GO Classes DPP | STATISTICS |HYPOTHESIS TESTING

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It is claimed that the lifetimes of light bulbs are normally distributed with a mean of $800$ hours and a standard deviation of $40$ hours. We wish to test the hypothesis that $\mu=800$ hours against the alternative that $\mu \neq 800$ hours with a sample size of $30$.

Find the power of the test against the alternative that the true mean life is $788$ hours. Enter the correct answer to two decimal places.

Note: Let $Z$ be a standard normal random variable and let $F_z(x)$ be its cumulative distribution function (i.e., $F_z(x)=P(Z \leq x)$). You may use the following values:

  • $F_z(4.38) \approx 0.99999$ (or $F_z(4.38) \approx 1$)
     
  • $F_z(-1.10) \approx 0.1357$

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$\begin{gathered}\text { Power }=1-\beta=P\left(\text { Reject } H_0 \mid H_A \text { is true }\right) \\ 1-\beta=P(\bar{X}>820 \text { or } \bar{X}<780 \mid \mu=788) \\ 1-\beta=P(\bar{X}>820 \mid \mu=788)+P(\bar{X}<780 \mid \mu=788) \\ 1-\beta=P\left(\frac{\bar{X}-788}{\sqrt{1600 / 30}}>\frac{820-788}{\sqrt{1600 / 30}}\right)+P\left(\frac{\bar{X}-788}{\sqrt{1600 / 30}}<\frac{780-788}{\sqrt{1600 / 30}}\right) \\ 1-\beta=P\left(z>\frac{32}{\sqrt{1600 / 30}}\right)+P\left(z<\frac{-8}{\sqrt{1600 / 30}}\right) \\ 1-\beta=1-F_z\left(\frac{32}{\sqrt{1600 / 30}}\right)+F_z\left(\frac{-8}{\sqrt{1600 / 30}}\right) \\ 1-\beta=0.1366\end{gathered}$
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