Question

PLEASE HELP!!! I'll give you the brainliest! a random sample of 3000 people was chosen for a survey. 47% of them had children under 18 living at home. find the 95% confidence interval - see picture

Answer 1

Answer: 0.4521 - 0.4879

Step-by-step explanation:

1) Find the standard deviation with the given information:

n=3000

p=47% ⇒ 0.47

1-p= 1 - 0.47 = 0.53

$\sigma =\sqrt{\dfrac{p(1-p)}{n}}\\\\\\.\ =\sqrt{\dfrac{0.47(0.53)}{3000}}\\\\\\.\ =\sqrt{\dfrac{0.2491}{3000}}\\\\\\.\ =\sqrt{0.00008303}\\\\\\.\ =0.009112$

2) Find the margin of error (ME) with the given information:

C=95% ⇒ Z = 1.960

σ=0.009112

ME = Z × σ

= 1.96 (0.009112)

= 0.01786

3) Find the confidence interval with the given information:

p = 0.47

ME = 0.01786

CI = p ± ME

= 0.47 ± 0.01786

= (0.4521, 0.4879)