Question

A car insurance company would like to determine the proportion of accident claims covered by the company. According to a preliminary estimate, 60 percent of the claims are covered. How large a sample should be taken to estimate the proportion of accident claims covered by the company if we want to be 98 percent confident that the sample percentage is within ±3 percent of the actual percentage of the accidents covered by the insurance company?

Answer 1

Answer: 1443

Step-by-step explanation:

Formula we use to find the sample size , if prior proportion is available :-

$n=p(1-p)(\dfrac{z_{\alpha/2}}{E})^2$

Given : Percent of the claims are covered : p= 60% =0.60

Margin of error : E=3% = 0.03

Critical value for 98 % confident : $z_{\alpha/2}=2.326$

Then , the required sample will be :

$n=(0.6)(1-0.6)(\dfrac{2.326}{0.03})^2=1442.74026667\approx1443$

Hence, the required sample size = 1443

Answer 2

How large a sample should be taken to estimate the proportion of accident claims covered by the company is 1,442.

Sample size

z-score for 98% confidence interval=2.325

Using this formula

Sample size=(% of covered claims)×(1- % of covered claims)×(z-score/margin of error)²

Let plug in the forumla

Sample size=(.6)×(1-.6)×(2.325/.03)²

Sample size= (.6)×(.4)×(2.325/.03)²

Sample size= .24×(77.5)²

Sample size= 1441.5

Sample size=1,442 (Approximately)

Inconclusion the sample size is 1,442.

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