Let \{a_n \}_{n \geq 1} be a sequence of real numbers such that a1+5m=2 , a2+5m=3 , a3+5m=4 , a4+5m=5, a5+5m=6 , m=0,1,2, … . Then
\limsup_{n \rightarrow \infty} a_n + \liminf_{n \rightarrow \infty} a_n equals ____________________________ .
It is evident that an is defined for all values of n=1,2,3,…. . And so
sup \quad a_n = 6 ,\quad inf \quad a_n = 2 \\
So, \limsup_{n \rightarrow \infty} a_n + \liminf_{n \rightarrow \infty} a_n = 8[Q No. 51-60]
Let X_1,X_2,\dots,X_9 be a random sample from a N(\mu,\sigma^2) distribution, where \mu \in \mathbb{R} \text{ and } \sigma>0 are unknown. Let the observed values of \bar{X} = \frac{1}{9} \sum_{i=1}^9 \text{and} S^2 = \frac{1}{8} \sum_{i=1}^{9} ( X_i -\bar{X} )^2 be 9.8 and 1.44, respectively. If the likelihood ratio test is used to test the hypothesis H_0: \mu=8.8 \text{ against } H_1:\mu>8.8, then the p-value of the test equals __________________ (round off to 3 decimals)
We know that,
Y=\frac{(\bar{X}-\mu)}{\sigma} \sim N(0,1) \\
and \\
Z^2=\frac{8S^2}{\sigma^2} \sim \chi^2_{(8)} \text{independently}Thus, the test statistic that should to be used for testing the null vs alternative would be:
T = \frac{Y}{Z/\sqrt{8}} \sim t_{(8)} Judging from the alternative the rule of rejection of Null Hypothesis is: Reject H_0 if the tabulated value is larger than the tabulated value, t1 (say).
Now, the p-value of the test is given as:
Pr [T > T_{tab} / H_0 \text{ is true} ] = Pr \left[ T > \frac{ (9.8 - 8.8)}{\sqrt{8 (1.44)} } \right] \\= Pr \left[T > \frac{1}{1.2 (2\sqrt{2})} \right] = 0.0185