# What is the margin of error when 1,200 people took a survey and 48% said they liked android products?

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Find the margin of error when 1200 people are surveyed and the result is 48%:

We know `hat(p)=.48 ` . The standard error is given by ` `Standard Error=`sqrt((hat(p)(1-hat(p)))/n) `

Here the standard error is `sqrt(((.48)(.52))/1200)~~.0144222051 ` (Assuming that the sample is derived from a simple random sample.)

The margin of error is a factor that is used to create a confidence interval about the true population proportion. The margin of error takes into account the standard error and also the confidence level.

Thus if we want to be 68% certain, then the true population proportion lies in the interval `.48+-.014 ` . Thus we are 68% certain that the true population proportion is between .466 and .494. (This is one standard error above and below the sample proportion, using the sample proportion as a good estimate of the population proportion.)

If we want to be 95% certain (a 95% confidence level) then we use a range of `hat(p)+-1.96 ` times the standard error, or `.48+-.028 ` .

If we want to be 99% certain we use a range of `hat(p)+-2.58 ` times the standard error or `.48+-.037 ` .

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The standard error is approximately 1.4%.

For a 95% confidence level, the margin of error is about 2.8%

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The formula for margin error according to the link I added below is

z `*` `sqrt((p(1-p))/(n))`` `

P would be your percentage that is given which is 48% or 0.48. N would equal 1200 because 1200 people took the survey. When you plug in the numbers you get approximately .014 or 1.4% error.

Z would be the confidence level. If you want to be 80% certain you would multiply your value by 1.28 which gives you 1.79% error.