Question

"Narrative : Health and Human Services Commission (HHSC) survey of health insurance Problem statement: Kaiser Family Foundation estimated that in 2015 17% of Texans did not have health insurance. HHSC wants to update this number. Assuming a margin of error of 0.03, what should be the sample size at: a) 95% confidence level? b) 99% confidence level?"

Answer 1

Solution :

Given that,

= 17% = 0.17

1 - = 1 - 0.17= 0.83

margin of error = E = 0.03

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.96

Sample size = n = (Z_/2 / E)² * * (1 - )

= (1.96 /  0.03)² * 0.17 * 0.83

= 622

Sample size = 622

Solution :

Given that,

= 17% = 0.17

1 - = 1 - 0.17= 0.83

margin of error = E = 0.03

At 99% confidence level the z is ,

  = 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z_/2 = Z_0.005 = 2.576

Sample size = n = (Z_/2 / E)² * * (1 - )

= (2.576 /   0.03)² * 0.17 * 0.83

= 1040

Sample size =1040