Question

A researcher for a paint company is measuring the level of a certain chemical contained in a certain type of paint. If the paint contains too much of this chemical, the quality of the paint will be compromised. On average, each can of paint contains 10 percent of the chemical. How many cans of paint should the sample contain if the researcher wants to be 98 percent certain of being within 1 percent of the true proportion of this chemical?

Answer 1

Answer: The sample should contain 4870 cans of paint.

Step-by-step explanation:

According to the given information , we have

The prior estimate of proportion= $p=0.10$

Margin of error : E=0.01

Critical value for 98% confidence interval =$z_{\alpha/2}=2.326$

Formula to find the sample size :

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

i.e. $n=0.10(1-0.10)(\dfrac{2.326}{0.01})^2$

$=4869.2484\approx4870$

Hence, the sample should contain 4870 cans of paint.