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Orlov [11]
3 years ago
11

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

Mathematics
1 answer:
Dennis_Churaev [7]3 years ago
5 0

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

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