Question

An insurance company wants to conduct a poll to find an admitted patient's average length of stay in a hospital. A researcher estimates that, the standard deviation is 2. The insurance company wants the error to be less than 1%. How large a sample should the company take to obtain a 98% confidence level?

Answer 1

Solution :

Given,

= 2 . .Population SD

E = 0.01 Margin of error

c = 98% = 0.98 ...confidence level

Find sample size required.

c = 0.98

= 1- c = 1- 0.98 = 0.02

  /2 = 0.01

Using Z table ,

= 2.326

Now, sample size (n) is given by,

=  {( 2.326 * 2 ) / 0.01 }2

= 216411.04

= 216412 ..(round to the next whole number)

Answer : Required Sample size is n = 216412