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Mathematics 12 Online
OpenStudy (anonymous):

I WILL MEDAL! A study of six hundred adults found that the number of hours they spend on social networking sites each week is normally distributed with a mean of 17 hours. The population standard deviation is 6 hours. What is the margin of error for a 95% confidence interval?

OpenStudy (anonymous):

Ive been working on this for over hours and I cant figure it out, please solve step-by-step and i will medal you

OpenStudy (kropot72):

The margin of error is the distance from x-bar, the center of the Confidence Interval, to the end of the interval. The margin of error can be found as follows: \[\large Margin\ of\ Error=z\frac{\sigma}{\sqrt{n}}\ .........(1)\] z = 1.96 for a 95% Confidence Interval. Sigma = 6. n = 600 Plugging these values into equation (1) we get: \[\large Margin\ of\ Error=\frac{1.96\times6}{\sqrt{600}}=you\ can\ calculate\]

OpenStudy (anonymous):

Thanks, but where did the (1.96) come from

OpenStudy (anonymous):

am i supposed to memorize 95% = 1.96?

OpenStudy (anonymous):

THANK YYOU I GET IT NOW

OpenStudy (anonymous):

I WISH I COULD MEDAL MORE THAN ONCE

OpenStudy (kropot72):

There is a good explanation of how the critical values of z are calculated here: http://www.stat.yale.edu/Courses/1997-98/101/confint.htm

OpenStudy (kropot72):

You're welcome :)

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