If n = 520 and ˆ p (p-hat) = 0.35, construct a 99% confidence interval.

Question

Help please ASAP! Need help on this statistics homework

Answer 1

Answer: 0.296 < p < 0.404

"We are 99% confident that the interval from 0.296 to 0.404 actually does contain the true value of p"

Data provided: C = 0.99

α = 1-0.99 = 0.01

(α/2) = 0.01/2 = 0.005

+/-Z (α/2) =+/- 2.59 from Z-score chart

p-hat = 0.35

1 minus p-hat = 0.65

n = sample size = 520

p-hat - E < p < p-hat + E

Formula:

+/-E (error) = Z (α/2)*sq rt [((p-hat)*(1-p)-hat))/n]

Procedure:

Plug in: +/- 2.59* sq rt [(0.2275)/520]

= +/- 2.59* 0.0209165

=+/- 0.054173737

So, p-hat - E < p < p-hat + E

0.35 - 0.054173737 < p < 0.35 + 0.054173737

Round your answer to 3 places or one more digit than the number of digits (2 places) in the input data

0.296 < p < 0.404