• Properties of Likelihood Ratio Tests (LRT):
• Statement (I): Likelihood Ratio test is generally an UMP if an UMP test at all exists. $\implies$ Correct.
• By the Karlin-Rubin theorem and Neyman-Pearson lemma, for families with Monotone Likelihood Ratio (MLR), the Likelihood Ratio Test reduces to the Uniformly Most Powerful (UMP) test for one-sided composite hypotheses.
• Statement (II): In a Likelihood Ratio test, probability of type I error is controlled by effectively choosing a suitable cutoff point. $\implies$ Correct.
• The critical region is defined by $\lambda(x) \le k$, where the threshold $k$ is chosen such that $\sup_{\theta \in \Theta_0} P_\theta(\lambda(X) \le k) = \alpha$, strictly controlling the size (Type I error probability) of the test.
• Conclusion: Both Statement (I) and Statement (II) are correct (Option 1).

Correct option: (1) Both Statement (I) and Statement (I) are correct

• Expected Number of Binomial Outcomes:
• Number of tosses per set: $n = 10$
• Probability of head on a single toss: $p = 0.5$, tails: $q = 0.5$
• Probability of getting exactly 7 heads and 3 tails in a set of 10 tosses:

$$P(X = 7) = \binom{10}{7} p^7 q^3 = \binom{10}{3} (0.5)^{10} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} \times \frac{1}{1024} = \frac{120}{1024} = \frac{15}{128} \approx 0.1171875$$

• Total number of sets conducted: $N = 100$
• The expected number of sets showing 7 heads is:

$$E = N \times P(X = 7) = 100 \times 0.1171875 = 11.72 \approx 12$$

• Conclusion: 12 cases are expected (Option 1).

Correct option: (1) 12

• Relationship Between Dispersion Measures in Normal Distribution:
• For a perfectly normal distribution $X \sim \text{N}(\mu, \sigma^2)$:
• Standard Deviation (SD): $\text{SD} = \sigma$
• Mean Deviation (MD) about mean: $\text{MD} = \sqrt{\frac{2}{\pi}} \sigma \approx 0.7979 \sigma \approx \frac{4}{5}\sigma$
• Quartile Deviation (QD) (Semi-interquartile range): $\text{QD} = 0.6745 \sigma \approx \frac{2}{3}\sigma$
• Deriving the Proportional Ratio:

$$\text{QD} : \text{MD} : \text{SD} = \frac{2}{3}\sigma : \frac{4}{5}\sigma : \sigma$$

• Multiplying throughout by the common denominator 15:

$$\left(\frac{2}{3} \times 15\right) : \left(\frac{4}{5} \times 15\right) : (1 \times 15) = 10 : 12 : 15$$

• Conclusion: The relationship that holds good is 10 : 12 : 15 (Option 1).

Correct option: (1) 10:12:15