Margin of error

Surveys conducted in American high schools concluded that 90% of the students in a sample of 400 students had more than one active email account. What would the margin of error be for the population proportion?

1.5%
3%
4.5%
9%

I think 1.5% but just checking. @perl

Question

Margin of error

Surveys conducted in American high schools concluded that 90% of the students in a sample of 400 students had more than one active email account. What would the margin of error be for the population proportion?

1.5%
3%
4.5%
9%

I think 1.5% but just checking. @perl

perl · Answer

are you given a signficance level?

anonymous · Answer

no

perl · Answer

we can plug in signficance levels. lets start with 95% significance

perl · Answer

The margin of error for a sample proportion is  Zc*SE .
Zc is the critical z score , which depends on level of significance.
SE = sqrt(p*q/n) where p is the population proportion q is 1-p, and n is the sample size.

Typically, for a confodence interval, the sample proportion and its complement are substituted for p and q.)

perl · Answer

Margin of Error  = Zc * √(.90 * .10 / 400 )

anonymous · Answer

I got 0.015. What is the Zc?

perl · Answer

Margin of Error  = Zc * √(.90 * .10 / 400 )

Zc is the critical z value 

the higher significance you want, the higher the z score (the bigger the margin of error)

perl · Answer

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