Answer to Question #173587 in Statistics and Probability for ANJU JAYACHANDRAN

Question

3b) In a locality of 18,000 families, a random sample of 840 families was taken. Of these

840 families, 206 families were found to have a monthly income of Rs. 500 or less.

Give the confidence interval for the families having income Rs. 500 or less.

Answer 1

Percentage of families that have income as 500 or less are 20% and 28%.

Let the proportion of families that have income as 500 or less be = p

Thus,

$p = \dfrac{x}{n} = \dfrac{206}{840} = 0.24$

$q = 1-p = 1 - 0.24 = 0.75$

Therefore,

Expected number of families that have income as 500 or less = 18000 × p

$= 18000 × 0.24 = 4410$

$\text{ S.e of p} =\dfrac{p(1-p)}{n} =\dfrac{ 0.24 \times 0.75 }{ 840} = 0.014$

Thus, the most probable limit for p will be - $0.24 + 3 ( 0.014)$

$- 0.200 \text{ and } 0.289$

Therefore, the percentage of families that have income as 500 or less are 20% and 28%.