Question-29
Suppose \(X \sim Normal (\mu, 4)\). For \(n=20\) i.i.d. samples of \(X\), the observed sample mean is 5.2. What conclusion would a z-test reach if the null hypothesis assumes \(\mu = 5\) (against an alternative hypothesis \(\mu \neq 5\)) at a significance level of \(\alpha = 0.05 ?\)
Use \(F_Z(-1.9599) = 0.025\)
If \(X \sim \text{Normal}(\mu, \sigma^2)\), then \(\overline{X} \sim \text{Normal}\left(\mu,\dfrac{\sigma^2}{n}\right)\). Thus, If \(X \sim \text{Normal}(\mu, 4)\), then \(\overline{X} \sim \text{Normal}\left(\mu,\dfrac{4}{20}\right)\)
Test will be : Reject \(H_0\) if \(|\overline{X} - 5| > c \implies \overline{X} > 5+c ~~\text{or}~~ \overline{X} < 5-c\)
Use \(\alpha = P\left(\text{Reject}~~ H_0 ~|~ H_0 \right) = P\left( |\overline{X} - 5| > c ~|~ \mu = 5 \right)\). Find the value of \(c\) by solving the above part and then make conclusion.
If \(X \sim \text{Normal}(\mu, 4)\), then \(\overline{X} \sim \text{Normal}\left(\mu,\dfrac{4}{20}\right)\)
Test will be : Reject \(H_0\) if \(|\overline{X} - 5| > c \implies \overline{X} > 5+c ~~\text{or}~~ \overline{X} < 5-c\)
Use \(\alpha = P\left(\text{Reject}~~ H_0 ~|~ H_0 \right) = P\left( |\overline{X} - 5| > c ~|~ \mu = 5 \right)\). \(\implies 0.05 = P\left( |\overline{X} - 5| > c ~|~ \mu = 5 \right)\) \(\implies 0.05 = P\left( \left|\dfrac{\overline{X} - 5 }{2/\sqrt{20}} \right| > \dfrac{c}{2/\sqrt{20}}\right)\)
\(\implies 0.05 = P\left( \left| Z \right| > \dfrac{c}{2/\sqrt{20}}\right)\)
\(\implies 0.05 = 2 F_Z \left(\dfrac{-c}{2/\sqrt{20}}\right) \implies F_Z \left(\dfrac{-c}{2/\sqrt{20}}\right) = 0.025\)
\(\implies c = -\dfrac{2}{\sqrt{20}} F_Z^{-1}(0.025) = -\dfrac{2}{\sqrt{20}} \times (-1.9599) = 0.8765\)
Since \(5.2-5 < c\), therefore we will accept \(H_0\).