.A study of two hundred teens found that the number of hours they spend on social networking sites each week is normally distributed with a mean of 12 hours. The population standard deviation is 4 hours. What is the margin of error for a 98% confidence interval?

0.122

0.659

0.295

0.313

Question

.A study of two hundred teens found that the number of hours they spend on social networking sites each week is normally distributed with a mean of 12 hours. The population standard deviation is 4 hours. What is the margin of error for a 98% confidence interval?

 0.122

 0.659

 0.295

 0.313

anonymous · Answer

@amistre64

amistre64 · Answer

what is your z score for 98% CI,  then adjust it by the standard error:$$Z*SE=z(sd/\sqrt n)$$

amistre64 · Answer

in other words:$$CI=mean\pm z(sd/\sqrt n)$$
so your margin of error is the second term

anonymous · Answer

Okay, i cant remember the formula for z score.

amistre64 · Answer

there is no formula, there is a messy integration, which people did long time ago and made tables from

amistre64 · Answer

since we want a 98% CI about the mean, then a left tail area will give us 98 + 1, or 99% to look up in a left tail table

amistre64 · Answer

or if you have a ti83,  invnorm(.9900)

amistre64 · Answer

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