Statistical Methods (ISS Statistics Paper I (Old Subjective Pattern)): Questions 40 - 45 of 72

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Question number: 40

» Statistical Methods » Tests of Significance » F-Test

Appeared in Year: 2013

Essay Question▾

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Obtain 100 (1 -α) % confidence interval for the ratio of population variances by using two independent random samples from N (µ 1, σ 12) and N (µ 2, σ 22) under the assumption that the population means are (i) known and (ii) unknown.

Explanation

Let X 1, X 2, …, X n and Y 1, Y 2, …Y m are the samples taken from independent N (µ 1, σ 12) and N (µ 2, σ 22).

In this question the hypothesis for testing is… (451 more words) …

Question number: 41

» Statistical Methods » Tests of Significance » Chi-Square

Appeared in Year: 2011

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Explain how to carry out the chi-squared test for H0:σ2=σ02 on the basis of a random sample

X 1, X 2, …, X n from N (µ, σ 2) population.

Explanation

Given that a random sample X 1, X 2, …, X n from N (µ, σ 2) population. The hypothesis is

H0:σ2=σ02Vs.HA:σ2σ02

The null hypothesis is test… (557 more words) …

Question number: 42

» Statistical Methods » Correlation Coefficient » Partial Correlation

Appeared in Year: 2009

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The two regression lines between X and Y are 8X - 10Y + 66 = 0, 40X - 18Y = 214. The variance of X is 9. Find X, Y, σ Y and ρ.

Explanation

Given that

8X10Y+66=0

40X18Y214=0

We know that the mean value of the given series satisfies the regression line that is

8X10Y=66(1… (155 more words) …

Question number: 43

» Statistical Methods » Bivariate Distributions » Bivariate Normal Distribution

Appeared in Year: 2013

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Let (X, Y) have a joint probability mass function

f(x,y)=xy36;x=1, 2,3;y=1, 2,3

= 0, elsewhere

Find the marginal mass functions of X and Y.

Explanation

Let (X, Y) have a joint probability mass function, the individual distribution of either X or Y is called the marginal distribution. So, the marginal mass functions of X is

f(x)=P(X=x)=y=13f(x… (121 more words) …

Question number: 44

» Statistical Methods » Standard Errors and Large Sample Tests

Appeared in Year: 2009

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A sample of 100 records on lengths of stay of patients in a hospital gave a standard deviation of days of stay as 4.9. In order to estimate the mean number of days of stay within 0.25 day with 95 % confidence, what should be the sample size?

Explanation

Let the patients in a hospital gave a standard deviation of days of stay as 4.9. The mean number of days of stay within 0.25 day that is margin error is the estimate fraction days between the difference of estimated and the true number of days. So, the test statistic… (60 more words) …

Question number: 45

» Statistical Methods » Association and Contingency

Appeared in Year: 2013

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Out of 200 persons appeared for an examination, 80 % were males and the rest were females. Among 30 married males, 14 were successful; 110 unmarried males were successful. In respect of 10 married females, 4 were successful; 20 unmarried females were successful. Compute Yule’s coefficient of association between success in the examination and the marital status for males as well as females.

Explanation

Out of 200 persons appeared for an examination, 160 males and 40 are females.

The frequencies for various attributes be show in this contingency table

Male

Successful (A)

unsuccessful (a)

married (B)

(AB) =14

(aB) =16

unmarried (b)

(Ab) =110

(ab) =20

Female

successful (A)

unsuccessful (a)

married (B)

(AB)… (194 more words) …

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