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So lution:-(1) given,\begin{array}{l}H_{0}: \mu=80 \quad H_{a}: \mu \neq 80 \text { (Twotailed test) } \\\text { Sample mean }=\bar{x}=83\end{array}\begin{array}{l}\text { Sample size }=n=39 \\\text { Standard deviation }=\sigma=5\end{array}level of significance =\alpha=0.05\rightarrow Decision or interpretation 'ull:-If Z<Z_{\alpha / 2} \Rightarrow Accept H_{0}If z>z_{\alpha / 2} \Rightarrow Reject H_{0}\rightarrow Test statistic:- z=\frac{\bar{x}-\mu}{\sigma / \sqrt{n}}z=\frac{83-80}{5 / \sqrt{39}}=\frac{3}{0.8006}=3.747\rightarrow Critical value Z_{\alpha 12}=1.96 (from z critical at \alpha / 2=0.05 & two tailed test\rightarrow Interpretation és conclusion As Z>Z_{\alpha / 2} (we reject H_{0} ) i.e, we accept H_{a} \Rightarrow \mu \neq 80(2) given.H_{0}: \mu=7.5 \quad H_{a}: \mu>7.5 (right tailed test)Sample mean =\bar{x}=8.3Sample size =n=52standard deviation =\sigma=3.17The population follows normal distributionx \sim N(\mu, \sigma)significance level =\alpha=0.01\rightarrow Test statestic z=\frac{\bar{x}-\mu}{\sigma / \sqrt{n}}z=\frac{8.3-7.5}{3.17 / \sqrt{52}}=\frac{0.8}{0.4396}=1.819\rightarrow Critical value z_{\alpha}=2.33 (from zcritical at \alpha_{1}=0.01 \& right tailed test value table)\rightarrow Interpretation & Conclusion As z<Z_{\alpha} (we accept Ho) i.e., we fail to reject H_{0} Hence we conclude that \Rightarrow \mu=7.5 ...