Question

On the basis of thier total scores,200 candidates of a civil service examination are divided into 2 groups ,the upper 30% and the remaining 70%. Consider the first question of this examination among the first group 40 had the correct answers where as among the second group 80 had the correct answers. On the basis of thier results can one conclude that the first question is no good at discriminating ability of the type being examined here? Test@d=0.05 level of significance.​

Answer 1

Answer:

Step-by-step explanation:

On the basis of their total scores, 200 candidates of a civil service

examination are divided into two groups, the upper 30% and the

remaining 70%. Consider the first question of the examination. Among

the first group, 40 had the correct answer, whereas among the second

group, 80 had the correct answer. On the basis of these results, can one

conclude that the first question is not good at discriminating ability of

the type being examined here

Answer 2

Answer:

The first question is good at discriminating ability of the type being examined here at 5% level of significance.

Step-by-step explanation:

This is a problem of 'Test of significance of difference between two sample proportions - large samples'.

Given,

Total population = 200 is divided into two groups.

• Let n₁ and n₂ be the sample sizes drawn from the population

n₁ = the upper 30% of 200 = 0.3 * 200

n₁ = 60

n₂ = the remaining 70% of 200 = 0.7 * 200

n₂ = 140

• Let p₁ and p₂ be the sample proportions from n₁ and n₂

p₁ = 40/60 = 2/3 = 0.667

p₂ = 80/140 = 0.571

By using the method of pooling, we pool the two sample proportions p₁ and p₂ into a single proportion,p.

p = $\frac{n_{1} p_{1} +n_{2}p_{2} }{n_{1} +n_{2} }$

p = (60*0.667 + 140*0.571)/200

p = 0.6

q = 1 - p = 1 - 0.6

q = 0.4

The Standard error of difference, S.E = $\sqrt{pq(\frac{1}{n_{1} }+\frac{1}{n_{2} } ) }$

Null Hypothesis H₀: There is no difference between sample proportions. i.e., p₁ = p₂

Alternate Hypothesis H₁: There is difference between sample proportions. i.e., p₁ ≠ p₂

Test statistic, z = $\frac{p_{1} -p_{2} }{S.E}$

z = (p₁ - p₂)/$\sqrt{pq(\frac{1}{n_{1} }+\frac{1}{n_{2} } ) }$

z = $\frac{0.667 - 0.571}{\sqrt{0.6*0.4*(\frac{1}{60}+\frac{1}{140} }) }$

z = 0.96

The level of significance, $\alpha$ = 0.05

The z - table value = 1.96

Since |z| < 1.96, we accept the null hypothesis H₀.

Therefore, the first question is good at discriminating ability of the type being examined here at 5% level of significance.

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