ISS (Statistical Services) Statistics Paper II (Old Subjective Pattern): Questions 34 - 38 of 39

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Question 34

Minimum Variance Bound Estimators

Appeared in Year: 2015

Describe in Detail

Essay▾

Let X1 , X2 , … , Xn be a random sample from the probability distribution with density

= 0; otherwise

where 0 < θ < ∞ . Show that is a minimum variance bound estimator and has variance .

Explanation

By using Cramer-Rao lower bound we find the minimum variance

The Fisher information is

Here E (X) = θ. The unbiased estimator of exponential distribution is .

The Lower bound is

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Question 35

Complete Statistics

Appeared in Year: 2015

Describe in Detail

Essay▾

Let X1 , X2 , … Xn be a random sample from the binomial distribution with probability mass function

Examine whether the statistic is complete for this distribution.

Explanation

We know that is sufficient statistic for this distribution. Then,

The definition of completeness is

Assume is free from and

This equation is true iff

This is also true if and only if this is a polynomial of n th degree in u and at = 0, t = 0,1, … , n

So, the statistic is complete sufficient statistic for this distribution.

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Question 36

Appeared in Year: 2015

Describe in Detail

Essay▾

Let X1 , X2 , … , Xn be a random sample from an population with , a positive definite matrix. Derive 100 (1-α) % simultaneous confidence interval for for all .

Explanation

Let X1 , X2 , … , Xn be a random sample from an and assume the linear combination of the random sample is

From, the theorem of linear combinations of multivariate normal distribution is that every linear combination of X follows an univariate normal distribution.

The sample mean and variance of the observed values are

where and S are the sample mean&#8230;

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Question 37

Appeared in Year: 2015

Describe in Detail

Essay▾

Let

Determine

(i) The principal components y1 , y2 and y3 .

(ii) The proportion of variance explained each one of them.

(iii) Correlation between the first principal components y1 and the third original random variable.

Explanation

First find the eigen value and eigen vector pairs

To solve this equation, we get the eigen values is

The corresponding eigen vector for each eigen value is by using normalize the eigen vector by the equation

For λ1 = 6,

To solve these equations, we get the eigen vector is

For λ2 = 3,

To solve this equation, we get the eigen vector is

For λ3 = 2,

To solve&#8230;

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Question 38

Appeared in Year: 2015

Describe in Detail

Essay▾

Fins a Most powerful test for testing the simple hypothesis against the simple alternative hypothesis based on n random observations from N (µ, σ 2) where µ is know. Show that this MP test is UMP (Uniformly most powerful) .

Explanation

Let Xi ′ s are n random observations from N (µ, σ 2) and the pdf is

The hypothesis is

To find the most powerful test, we use following step

The N-P test is

Given α, we can find k by solving if

iff

Iff

where c is constant contained k.

We know that

Hence

The most powerful test for null hypothesis is

Power of the test

We see that under null hypothesis the po&#8230;

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