Question

Out of 500 Respondents in a recent health survey 47 reported a history of diabetes.

a. Estimate the true proportion of people with history of diabetes with 95% confidence.

b. What should the sample size be if the researchers wanted to be accurate to within 2% of the true proportion?

Answer 1

Solution :

a ) Given that,

n = 500

x = 225

= x / n = 47 / 500 = 0.094

1 - = 1 - 0.094 = 0.906

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 = 1.960

Margin of error = E = Z _{/ 2} * (( * (1 - )) / n)

= 1.960 * (((0.094 * 0.906) / 500)

= 0.025

A 95 % confidence interval for population proportion p is ,

- E < P < + E

0.094 - 0.025 < p < 0.094 + 0.025

0.068 < p < 0.119

Given that,

b ) = 0.094

1 - = 1 - 0.094 = 0.906

margin of error = E = 2% = 0.02

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 = 1.960

Sample size = n = ((Z _{/ 2}) / E)² * * (1 - )

= (1.960 / 0.02)² * 0.094 * 0.906

= 817.91

= 818

n = sample size = 818