# Statistical Methods-Tests of Significance [ISS (Statistical Services) Statistics Paper I (Old Subjective Pattern)]: Questions 1 - 4 of 17

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## Question 1

Appeared in Year: *2013*

### Describe in Detail

Essay▾From along series of annual river flows, the variance is found out to be 49 units. For a new sample of 25 years, the variance is calculated as 81 units. Can we regard that the sample variance is significant? (Given the chi-square value at 5 % level of significance as 37.7)

### Explanation

The sample size is 25. Here we test

The null hypothesis is tested by chi-square test when the sample size is less than 30. The test statistic is

The decision criteria is reject the null hypothesis if the calculated value is greater than the tabulated value otherwise accept it.

Here n = 25, ,

The tabulated value at 5 % level of significance is 37.7

So,…

… (27 more words) …

## Question 2

Appeared in Year: *2014*

### Describe in Detail

Essay▾Define power of a test and discuss its role in selecting the best test. Describe a test procedure for testing equality of means of two independent normal populations, when standard deviations are equal but unknown for small samples.

### Explanation

**Power of a test**: The power function is the probability of rejection the null hypothesis when the alternative hypothesis is true. In statistical terms,

The quantity 1-β is called the power of the test. So, the power of the test is depends upon the difference between the parameter values specified in H_{0} and actual value of the parameter.

When we have t…

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## Question 3

Appeared in Year: *2009*

### Describe in Detail

Essay▾A manufacturer of alkaline batteries expects that only 5 % of his products are defective. A random sample of 300 batteries contained 10 defectives. Can we conclude the proportion of defectives in the entire lot is less than 0·5 at 5 % level of significance?

### Explanation

A random sample of 300 batteries contained 10 defectives. A manufacturer of alkaline batteries expects that only 5 % of his products are defective that is testing of hypothesis is

H_{0}: P = 0.05 against H_{1}: P < 0.05

Here n = 300, numbers of defectives is 10 and

The estimated value of P is

The null hypothesis can be tested by z-test for assuming the sampl…

… (67 more words) …

## Question 4

Appeared in Year: *2010*

### Describe in Detail

Essay▾You are working as a purchase manager for a company. The following information has been supplied to you by two manufacturers of electric bulbs.

Company A | Company B | |

Mean life | 1300 | 1288 |

Standard deviation | 82 | 93 |

Sample size | 100 | 100 |

Which brand of bulbs are you going to purchase if you desire to take a risk of 5 % ?

### Explanation

Here the sample size is n = 100

The standard deviation and mean of two samples is different. The null hypothesis is the mean life of bulb of company A is equal to the mean life of bulb of company B and alternative its differ.

The test statistic is

Under H_{0}

The test criteria is reject H_{0} if , otherwise accept it

Here

The tabulated value of t-test is

The …

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