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2 months ago

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

Mathematics

1 answer:

AnnZ [12.3K]2 months ago

5 0

Answer:

The margin of error for a 90% confidence level is 0.024

The margin of error for a 99% confidence level is 0.037

Step-by-step explanation:

Sample size = 1076

A poll indicated that 64% of a randomly selected group of 1076 adults believe in ghosts.

Thus, the number of adults expressing belief in ghosts equals $\frac{64}{100} \times 1076=688.64\sim 688$

So, x = 688

n = 1076

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

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

$\widehat{p} = 0.639$

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

For a 90% confidence interval, z is 1.64

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

$ME=0.024$

Thus, margin of error at 90% is calculated as 0.024

To identify the margin of error for 99% confidence

z for this confidence level is 2.58

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

$ME=0.024$

Therefore, the margin of error at 99% amounts to 0.037

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