# ISS (Statistical Services) Statistics Paper II (Old Subjective Pattern): Questions 13 - 18 of 39

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## Question 13

Appeared in Year: *2014*

### Describe in Detail

Essay▾Find 3-sigma control limits for a

(i) C-chart with process average equal to 4 non-conformities

(ii) U chart with process average c = 4, and n = 4.

### Explanation

the 3-sigma control limits is

(i) Given that

Control limits for the C chart with a known non-conformities is

(ii) Given that c = 4 and n = 4

Control limits for the U chart is

## Question 14

Appeared in Year: *2014*

### Describe in Detail

Essay▾x_{1} , x_{2} , … , x_{n} be a random sample from the following distribution

### Explanation

Let x_{1} , x_{2} , … , x_{n} be a random sample from f (x, α) and let L (α| x) denote the likelihood function. Then

The log-likelihood function is

We do not differentiable log L with respect to α because this is free from parameter. So, according to maximum likelihood principle log L has to be maximum when is minimum. For this choose α when the is minimum.…

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## Question 15

Appeared in Year: *2014*

### Describe in Detail

Essay▾x_{1} , x_{2} , … , x_{n} is a random sample from N (θ, σ ^{2}) (σ ^{2} not specified) . Derive likelihood ratio test of testing H_{0}: θ = θ_{0} against H_{1}: θ ≠ θ_{0} .

### Explanation

Under the given model, the parameter space is

Under H_{0} , the parameter space is

The likelihood function is

Under the whole space, the unrestricted MLE is

Under H_{0} , the restricted MLE is

The statistic is

The likelihood ratio test is reject H_{0} if λ (__x__) ⩽c is equivalent to

Under H_{0}

And these two random variables are independent. We can therefore construct a …

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## Question 16

Appeared in Year: *2014*

### Describe in Detail

Essay▾Obtain a g-inverse of A given below and verify that AA ^{-} A = A

### Explanation

Let A be a full rank m n matrix. By full rank we mean rank (A) = min {m , n} .

If m < n, then A has a right inverse given by

If m > n , then A has a left inverse given by

Our matrix A is 2 × 3 with rank two, so A has a right inverse given by

Putting A ^{-} = then gives

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## Question 17

Appeared in Year: *2014*

### Describe in Detail

Essay▾A single observation of a r. v. having a geometric distribution with pmf

The null hypothesis is H_{0}: θ = 0.5 against the alternative hypothesis H_{1}: θ = 0.6 is rejected if the observed value of r. v. is greater than equal to 5. Find the probabilities of type I error and type II error.

### Explanation

The type I error is the probability that reject H_{0} when it is true denote by α and type II error is the probability that accept H_{0} when it is false denote by β

Given that the null hypothesis is rejected if the observed value of r. v. is greater than equal to 5 that means X ⩾ 5.

For type I error,

For type II error,

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## Question 18

Appeared in Year: *2014*

### Describe in Detail

Essay▾chart is used to control the mean of normally distributed characteristic. It is know that . The center line is 200. If the process mean shifts to 188, find the probability that this shift is detected on the first subsequent sample.

### Explanation

Type II error is the probability of saying that the process is under control when it is not under control, that is, the probability of a point falling within control limits after the shift the mean. This is also known as the probability of not detecting the shift after shift the mean. Thus,

The probability of detecting the shift on the first subsequ…

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