Question

The proportion of Hispanics in the population of the USA is 0.164 (=16.4%). Calculate (a, b, c, d, e, f) the 90%, 95%, and 99% confidence intervals of p for samples of n=1500. What (g, h, i) are the margins of error, E, of each of these three confidence intervals? What (j) is the correct interpretation of a confidence interval?

Answer 1

Solution:-

p = 0.164

n = 1500

a) 90% confidence interval for the proportion is C.I = (0.148, 0.1797).

C.I = 0.164 + 1.645 × 0.00956

C.I = 0.164 + 0.01573

C.I = (0.148, 0.1797)

b) 95% confidence interval for the proportion is C.I = (0.1453, 0.1827).

C.I = 0.164 + 1.96 × 0.00956

C.I = 0.164 + 0.01874

C.I = (0.1453, 0.1827)

c) 99% confidence interval for the proportion is C.I = ( 0.1394, 0.1886).

C.I = 0.164 + 2.576 × 0.00956

C.I = 0.164 + 0.02463

C.I = ( 0.1394, 0.1886)

d) 90% Margin of error for the proportion is 0.01573.

M.E = 1.645 × 0.00956

M,E = 0.01573

e) 95% Margin of error for the proportion is 0.01874.

M.E = 1.96 × 0.00956

M.E = 0.01874

f) 99% Margin of error for the proportion is 0.02463.

M.E = 2.576 × 0.00956

M.E = 0.02463