Answer

We will solve the problem using \(p-value\) as well as critical regions. Let \(\bar{X}\) denote the mean of the samples. First, lets find the critical region to reject \(H_{0}\).

\[ \begin{aligned} P(\lvert \bar{X} \rvert > c \lvert H_{0} \quad\text{is true}) &= \alpha = 0.05\newline P \bigg(\bigg\lvert \frac{\bar{X} - 30}{20/\sqrt{10}} \bigg\rvert > c \bigg) &= 0.05 \end{aligned} \]

Since this is a double sided test, \(c = z_{0.05/2} = z_{0.025} = 1.96\). We reject the hypothesis if either the test statistic \(> 1.96\) or \(< -1.96\). The test statistic in this case is \((27-30) \sqrt{10} / 20 = -2.12\) rejecting \(H_{0}\).

For solution using p-value, we calculate the following

\[ \begin{aligned} P \bigg(Z > \bigg\lvert \frac{\bar{X} - 30}{20/\sqrt{10}} \bigg\rvert \bigg) &= P(Z > 2.12 \cup Z < -2.12)\newline &= 2P(Z > 2.12) = 2(1-\Phi(2.12)) = 2(1-0.98300)\newline &= 0.034 < 0.05 \end{aligned} \]

Since the p-value is smaller than the significance level, we accept \(H_{a}\) which states that the mean is different than \(30\).