11. Relative to a normal population, test at the 97 percent level of significance the null hypothesis $H_{0}: \mu=460$ versus the alternative hypothesis $H_{1}: \mu<460$, where $\mu$ is the population mean, $n=13$ is the sample size, $\bar{x}=472.11$ is the sample mean, and $s=11.96$ the sample standard deviation. Let $Q_{1}$ be the degrees of freedom for the $t$-distribution, $Q_{2}$ the $t$-statistic, and $Q_{3}=1$ if we reject the null hypothesis $H_{0}$, and $Q_{3}=0$ otherwise. Let $Q=\ln \left(3+\left|Q_{1}\right|+2\left|Q_{2}\right|+3\left|Q_{3}\right|\right)$. Then $T=5 \sin ^{2}(100 Q)$ satisfies:- (A) $0 \leq T<1$. - (B) $1 \leq T<2$. (C) $2 \leq T<3$. (D) $3 \leq T<4$. (E) $4 \leq T \leq 5$.

Step 1 ：Given that the sample size \(n = 13\), the sample mean \(\bar{x} = 472.11\), the sample standard deviation \(s = 11.96\), and the population mean under the null hypothesis \(\mu = 460\). We are testing the null hypothesis \(H_{0}: \mu = 460\) against the alternative hypothesis \(H_{1}: \mu < 460\) at the 97 percent level of significance.

Step 2 ：First, we calculate the degrees of freedom for the t-distribution, which is \(n - 1\). So, \(Q_{1} = 12\).

Step 3 ：Next, we calculate the t-statistic using the formula \(t = (\bar{x} - \mu) / (s / \sqrt{n})\). So, \(Q_{2} = 3.65\).

Step 4 ：Then, we find the critical value for the t-distribution at the 97 percent level of significance, which is \(t_{critical} = 2.08\).

Step 5 ：Since the calculated t-statistic is greater than the critical value, we reject the null hypothesis. So, \(Q_{3} = 1\).

Step 6 ：Now, we calculate \(Q = \ln (3 + |Q_{1}| + 2|Q_{2}| + 3|Q_{3}|)\), which gives \(Q = 3.10\).

Step 7 ：Finally, we calculate \(T = 5 \sin^{2}(100Q)\), which gives \(T = 1.38\).

Step 8 ：From the given options, the value of \(T\) falls in the range \(1 \leq T < 2\).

Step 9 ：\(\boxed{\text{Final Answer: (B) } 1 \leq T < 2}\)