# ISS (Statistical Services) Statistics Paper IV: Questions 42 - 46 of 92

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## Question 42

### Describe in Detail

Essay▾Compute the crude and standardized death rates of the populations A and B, regarding A as standard population, from the following data:

Age groups (years) | A | B | ||

Population | Deaths | Population | Deaths | |

Under 10 | 20000 | 600 | 12000 | 372 |

10 - 20 | 12000 | 240 | 30000 | 660 |

20 - 40 | 50000 | 1250 | 62000 | 1612 |

40 - 60 | 30000 | 1050 | 15000 | 525 |

Above 60 | 10000 | 500 | 3000 | 180 |

### Explanation

From the given data, the crude death rate for any population is

For population A

For population B

The standardized death rate is

Where p_{x}^{s} = population of standard group of x and D_{x} = Age specific death rate of group x.

Age groups | Standard population (P_{x}) | D_{x} for population A | D_{x} for population B | P_{x} ⚹ D_{x} for A | P_{x} ⚹ D_{x} for B |

… (82 more words) …

## Question 43

### Describe in Detail

Essay▾Following data gives the diameter of the rings manufactured by a company. Obtain the Control limits for R-chart.

Sample No. | Observations | ||||

1 | 5.02 | 5.01 | 4.94 | 4.99 | 4.96 |

2 | 5.01 | 5.03 | 5.07 | 4.95 | 4.96 |

3 | 4.99 | 5.00 | 4.93 | 4.92 | 4.99 |

4 | 5.03 | 4.91 | 5.01 | 4.98 | 4.89 |

5 | 4.95 | 4.92 | 5.03 | 5.05 | 5.01 |

6 | 4.97 | 5.06 | 5.06 | 4.96 | 5.03 |

7 | 5.05 | 5.01 | 5.10 | 4.96 | 4.99 |

8 | 5.09 | 5.10 | 5.00 | 4.99 | 5.08 |

9 | 5.14 | 5.10 | 4.99 | 5.08 | 5.09 |

10 | 5.01 | 4.98 | 5.08 | 5.07 | 4.99 |

### Explanation

From the tables, for n = 10, we have , ,

**Control limits for** :

5.01 + 0.0354 = 5.045

And 5.01 + 0.0354 = 4.97

**chart shows the process is out of control**.

**Control limits for R chart** :

= (0.223) (0.115) = 0.0256

R chart **shows the process is under control**.

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## Question 44

Appeared in Year: *2013*

### Describe in Detail

Essay▾For a PBIB design with parameters , b, r, k, , show that

### Explanation

A PBIB design with m associate classes based on the association scheme is an arrangement of treatments into b blocks such that

(i) Each block contains k distinct treatments (k <

(ii) Each treatment occurs in r blocks

(iii) Any two treatments, which are ith associates occur together in blocks (i = 1,2, . . , m) are called the parameters of the PBIB …

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## Question 45

Appeared in Year: *2012*

### Describe in Detail

Essay▾Discus symmetrical BIBD. For an SBIBD , under usual notations, show that

=

### Explanation

**Symmetrical BIBD**:

A BIBD is said to be symmetric if b = υ and r = k.

υ, r, b, k and λ are called the parameters of the BIBD

υ = number of varities or treatments

b = number of blocks

r = number of replicates for each treatment

k = block size

λ = number of blocks in which any pair of treatments occurs together

**Proof**: **=**

Let N be the incident matrix of a SBIB…

… (79 more words) …

## Question 46

Appeared in Year: *2015*

### Describe in Detail

Essay▾Discuss how you would proceed with the analysis if data on one plot is missing in a Latin Square design.

### Explanation

To analysis the data of a Latin square design having missing values, we first estimate the missing value by minimizing the error mean square and the rest missing values by guess estimate which is generally the mean of all available values. The estimated value of x is obtain by to minimize the error variance, differentiate the expression for it with…

… (200 more words) …