# Statistical Methods (ISS Statistics Paper I (Old Subjective Pattern)): Questions 29 - 33 of 72

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## Question number: 29

» Statistical Methods » Tests of Significance » Chi-Square

Appeared in Year: 2010

### Describe in Detail

For 2X2 table

A | b |

C | d |

prove that Chi-square test of independence gives

### Explanation

Let the contingent table is

Class | A | α | Total |

B | a | b | a + b |

β | c | d | c + d |

Total | a + c | b + d | a + b + c + d =N |

Here A denotes the presence of any attributes and α denotes the absence of… (112 more words) …

## Question number: 30

» Statistical Methods » Bivariate Distributions » Bivariate Normal Distribution

Appeared in Year: 2014

### Describe in Detail

The marginal distributions of X and Y are given in the following table:

X | 1 | 2 | Total | |

Y | 3 4 | ? ? | ? ? | 1/4 3/4 |

Total | 1/2 | 1/2 | 1 |

If the co-variance between X and Y is zero, find the cell probabilities and see whether X and Y are independent.

### Explanation

Given that covariance between X and Y is zero

This gives that f (xy) =f (x) f (y)

So, the joint probability density function of x = 1, y = 3 is

We know that if the covariance is zero, then the conditional distribution is equal to unknown… (67 more words) …

## Question number: 31

» Statistical Methods » Association and Contingency

Appeared in Year: 2014

### Describe in Detail

A medicine supposed to have effect in preventing TB was treated on 500 individuals and their records were compared with the records of 500 untreated individuals as follows. Study the effectiveness of medicine by calculating (i) Yule’s coefficient of association (ii) Yule’s coefficient of colligation.

- | No-TB | TB |

Treated | 252 | 248 |

Untreated | 224 | 276 |

### Explanation

The frequencies for various attributes be show in this contingency table

- | No-TB (A) | TB (a) |

Treated (B) | (AB) | (aB) |

Untreated (b) | (Ab) | (ab) |

(i) Yule’s coefficient of association is a relative measure of association between two attributes, for this question the attributes are No-TB and treatment. If (AB), (aB),… (51 more words) …

## Question number: 32

» Statistical Methods » Data » Multivariate

Appeared in Year: 2011

### Describe in Detail

The following are the frequencies in the given intervals:

| (2 - 5) | (5 - 10) | (10 - 15) | (15 - 18) | (18 - 20) | |

43 | 85 | 151 | 112 | 72 | 34 |

Draw the histogram of this data. Calculate the mean of the data from the frequency table.

### Explanation

The histogram of the frequencies is given where the widths of all classes are not equal.

Class interval | mid value | Frequencies |

(0 - 2) | 1 | 43 |

(2 - 5) | 3.5 | 85 |

(5 - 10) | 7.5 | 151 |

(10 - 15) | 12.5 | 112 |

(15 - 18) | 16.5 | 72 |

(18 - 20) | 19… (10 more words) … |

## Question number: 33

» Statistical Methods » Tests of Significance » Z-Test

Appeared in Year: 2014

### Describe in Detail

Let X follow a binomial distribution B (n, P). Explain the test procedure for

H _{0}: P = P _{0} against H _{1}: P > P _{0}

when the sample size is (i) small, and (ii) large. It is desired to use sample proportion p as an estimator of the population proportion P, with probability 0.95 or higher, that p is within 0.05 of P. How large should sample size (n) be?

### Explanation

Let X follow a binomial distribution B (n, P) with mean nP and variance nP (1-P). In testing the hypothesis

H _{0}: P = P _{0} against H _{1}: P > P _{0}

The null hypothesis can be tested by z-test for assuming the sample size n is… (164 more words) …