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- #1

I'm assuming the hypothesis test is

$\displaystyle H_0 : \mu = 13 \quad \quad H_a : \mu < 13 $

We are given $\displaystyle \mu = 13, \quad \sigma = 2.47, \quad \bar{x} = 12.86 , \quad n = 20 $.

The test statistic is

$\displaystyle \begin{align*} z &= \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} \\

&= \frac{12.86 - 13}{\frac{2.47}{\sqrt{20}}} \\

&\approx -0.253\,481 \end{align*} $

Thus the p value is

$\displaystyle \begin{align*} p &= \textrm{Pr}\left( Z < -0.253\,481 \right) \\

&\approx \Phi \left( -0.25 \right) \textrm{ from the Z distribution table}\\

&= 0.401\,29 \end{align*} $

However, a CAS (or Linear Interpolation) could be used to get a more accurate value. Using technology, I find the p value to be $\displaystyle 0.399\,948 $, which our approximation is very close to.