Question

Of 150 patients randomly selected by a hospital, 90 of them said that their health insurance was not adequate.

(a) Construct a 99% confidence interval for the proportion of patients who would say that their health insurance was not adequate.

(b) Can we say that more than 50% of all patients of the hospital did not have adequate health insurance? Test at the 1% significance level.

Answer 1

a)

Standard error of the mean = SEM = √x(N-x)/N³ = 0.040

α = (1-CL)/2 = 0.005

Standard normal deviate for α = Z_α = 2.576

Proportion of positive results = P = x/N = 0.600

Lower bound = P - (Z_α*SEM) = 0.497

Upper bound = P + (Z_α*SEM) = 0.703

b) No, because interval contains values less than 50% also