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Mathematics 15 Online
OpenStudy (anonymous):

CAN ANYONE HELP ME FIND A MARGIN OF ERROR?!

OpenStudy (anonymous):

TV World sold 1200 television sets this year. 240 of the sets were sold with extended warranties. What is the proportion of sets sold with extended warranties? Write your answer as a decimal. If the decimal is less than 1, write a 0 before the decimal point. Enter your answer to 1 decimal place.

OpenStudy (anonymous):

this was the first question and the answer is 0.2

OpenStudy (anonymous):

i need to now find the margin of error

OpenStudy (bahrom7893):

I'm not sure what a margin of error is.. let me read up on it

OpenStudy (anonymous):

alright

OpenStudy (bahrom7893):

@TuringTest

OpenStudy (bahrom7893):

Sorry i couldn't follow what they meant

OpenStudy (anonymous):

you don't know the answer?

OpenStudy (anonymous):

or?

OpenStudy (anonymous):

i am just unsure on how to calculate this... I've tried several times

OpenStudy (turingtest):

If I am interpreting it correctly they first mean the maximum error possible after the 0.2 as in it could be anywhere between 0.2x and 0.1x depending on if we round up or down

OpenStudy (turingtest):

so what is the maximum that the x could be?

OpenStudy (turingtest):

I also may have misinterpreted the problem

OpenStudy (anonymous):

the margin of error comes from the first question i posted....

OpenStudy (anonymous):

the answer to that first question gave me 0.2

OpenStudy (anonymous):

Margin of error depends on confidence level.\[ \hat{p}\pm z^*{SE}_\hat{p} \]

OpenStudy (anonymous):

but i have no idea what the confidence level is

OpenStudy (anonymous):

the first question above is what i am supposed to get the margin of error from

OpenStudy (anonymous):

from the 0.2

OpenStudy (anonymous):

The standard error can be found to be as follows.\[{SE}_\hat{p}=\sqrt\frac{\hat{p}(1-\hat{p})}{n}\]

OpenStudy (anonymous):

to be honest I'm so confused... do u know the answer?

OpenStudy (anonymous):

You need to have a confidence level :-) This is AP Statistics right?

OpenStudy (anonymous):

no no... this is algebra 2

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