Discussion
A simple random sample of size 10 from \(N(\mu, \sigma^{2})\) gives 98% confidence interval (20.49, 23.51). Then the null hypothesis \(H_{0} : \mu = 20.5\) against \(H_{A} : \mu \neq 20.5\)
Answer: C
The hypothesis can be rejected at 10% level of significance.
The interval is centred at \(\bar{x} = 22\) with half-width 1.51. With \(n = 10\) and \(\sigma\) unknown, the 98% interval uses \(t_{0.01, 9} = 2.821\), so the standard error is:
\[ \begin{aligned} s/\sqrt{n} &= 1.51/2.821 = 0.535 \\ t &= (22 - 20.5)/0.535 = 2.80 \end{aligned} \]
Since 20.5 lies just inside the 98% interval, \(H_{0}\) cannot be rejected at 2%, so the first option is wrong. But \(2.80\) exceeds \(t_{0.025, 9} = 2.262\) and \(t_{0.05, 9} = 1.833\), so \(H_{0}\) is rejected at 5% and at 10%. That makes 'cannot be rejected at 5%' and 'cannot be rejected at any level' wrong. The same conclusion follows if \(\sigma\) is treated as known (z = 2.31 against 1.96 and 1.645).
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