ISS (Statistical Services) Statistics Paper I (Old Subjective Pattern): Questions 85 - 89 of 164

Given that there are four outcomes. Define A = (O₁ , O₄) , B = (O₂ , O₄) , C = (O₃ , O₄)

Their intersection is and Union is AUBUC = (O₁ , O₂ , O₃ , O₄)

P (A) = P (B) = P (C) =

P (AUBUC) = 1

The events A, B and C are not pairwise independent because

But the events A and B are pairwise independent because

Similarly the events B and C or A and C are pairw…

Let X₁ , X₂ , … , X_n and Y₁ , Y₂ , … Y_m are the samples taken from independent N (µ₁ , σ₁ ²) and N (µ₂ , σ₂ ²) .

In this question the hypothesis for testing is

Let s₁ ² and s₂ ² be the estimates variances of σ₁ ² and σ₂ ² based on sample sizes n and m.

and

Then

This two chi-square random variable are independent. So, H₀ can be test by F-test

where s₁ ²…

Let X be a random variable with mean µ and variance σ ² . Then any k > 0, the Chebyshev՚s inequality is

or ‎

σ ² = E [X ²] - (E [X] ) ² = 4

Then, a lower bound for the probability

Using Chebyshev՚s inequality

Given that P (A) = , P (B) = , P (C) =

then probability that at least one event of these three events happens is

The events are independent because only one event happens

Let X₁ , X₂ , … , X_m i. i. d. random variables with common p. m. f. is P (X = k) which is a binomail random variables with common parameters n and p respectively. Then, the p. m. f. of S_m = X₁ + X₂ + … . + X_m , sum of random variables are found by moment generating function. The moment generating function of binomial distribution is

So, the moment g…