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A sample mean, sample size and population standard deviation are provided below. Use the one-mean z-test to perform the requi
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\\ $ The hypotheses are $ \\ $ Null hypothesis $ H_{o} : \mu = 25 \\ $ Alternate hypothesis $ H_{a} : \mu \neq 25 \\\\ $ Population standard deviation $ (\sigma) = 8 \\\\ $ The sample data is $ \\ $ Sample size $ (n) = 25 \\ $ Sample mean $(\bar{x}) = 27 \\\\ $ Test statistic $ : z = \dfrac{\bar{x} - \mu} { \frac{\sigma}{\sqrt{n}}} = \dfrac{27 - 25}{\frac{8}{\sqrt{25}}} = \textbf{1.25} \\ \\\\ $ As $ H_{a} $ has $ "\neq" $ sign $ , $ so test is 2-tailed $ \\ $ Thus, there are 2 critical values $ \\\\ $ Significance level $ (\alpha) = 10\% = 0.10 \\ $ So, critical values $ : z_{c} = \pm z_{\alpha/2} \\ \hspace*{15 mm}= \pm z_{0.10/2} = \pm z_{0.05} = \mathbf{\pm 1.645}\textbf{ Option A)} \\ ( P( z < -1.64 ) = 0.0505 $ and $ P( z < -1.65) = 0.0495 \\ $ from Z table , so $ P( z < -1.645) $ is closest to 0.05 \\ by extrapolation $)

\\ | z | = | 1.25 | = 1.25 \\ | z_{c} | = | \pm 1.645 | = 1.645 \\\\ $ As $ | z | \ngtr | z_{c} | , $ so $ \textbf{Fail to reject} $ the null hypothesis $. $ The data$ \\ \textbf{ does not have} $ sufficient evidence to conclude that the $ \\ $ mean is $ \textbf{different from 25}

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