Question 1197973: A small company involved in e-commerce is interested in statistics concerning the use of e-mail. A poll found that 39% of a random sample of 1076 adults, who use a computer at their home, work, or school, said they do not send or receive e-mail. Complete parts a through e.
Find the margin of error for this poll if we want 95% confidence in our estimate of the percent of American adults who do not use e-mail.
ME=
Find that margin of error in 99% confidence
ME=
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
At 95% confidence, the z critical value is roughly z = 1.96
Use a table like this
https://www.sjsu.edu/faculty/gerstman/StatPrimer/t-table.pdf
to get that value. Look at the bottom row labeled "Z" and above the 95% confidence level.
phat = sample proportion
phat = 0.39
n = sample size
n = 1076
E = margin of error for a proportion
E = z*sqrt(phat*(1-phat)/n)
E = 1.96*sqrt(0.39*(1-0.39)/1076)
E = 0.02914386568497
E = 0.029
The margin of error is roughly 0.029 for the 95% confidence interval estimating the population proportion (p).
For the 99% confidence interval, we'll repeat the same outline of steps.
However, this time we'll use z = 2.576 which is found in the table linked above.
The other values remain the same.
E = z*sqrt(phat*(1-phat)/n)
E = 2.576*sqrt(0.39*(1-0.39)/1076)
E = 0.0383033663288
E = 0.038
Answers:
Margin of error at 95% confidence = 0.029
Margin of error at 99% confidence = 0.038
Round the values however instructed.
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