Question

A poll found that 64% of a random sample of 1076 adults said they believe in ghosts.


Question 1. Find the margin of error zsqrt(((p^^\^)((1-p^^\^)))/n), abbreviated ME, for this poll if we want 90% confidence in our estimate of the proportion of adults who believe in ghosts.
ME=

(Round to 3 decimal places.)



Question 2. Find the margin of error needed to be 99% confident.
ME= (Round to 3 decimal places.)

Answer 1

Answer:

Margin of error at 90% is 0.024

Margin of error at 99% is 0.037

Step-by-step explanation:

Sample size = 1076

A poll found that 64% of a random sample of 1076 adults said they believe in ghosts.

So, No. of adults said they believe ghosts = $\frac{64}{100} \times 1076=688.64\sim 688$

So, x = 688

n = 1076

$\widehat{p} = \frac{x}{n}$

$\widehat{p} = \frac{688}{1076}$

$\widehat{p} = 0.639$

$ME=z \times \sqrt{\frac{\widehat{p}(1-\widehat{p})}{n}}$

z at 90% confidence is 1.64

$ME=1.64 \times \sqrt{\frac{0.639(1-0.639)}{1076}}$

$ME=0.024$

So, margin of error at 90% is 0.024

Find the margin of error needed to be 99% confident.

z at 99% confidence is 2.58

$ME=2.58 \times \sqrt{\frac{0.639(1-0.639)}{1076}}$

$ME=0.024$

So, margin of error at 99% is 0.037