Question

A study of three hundred teenagers found that the number of hours they spend on instant messaging sites each week is normally distributed with a mean of 13 hours. The population standard deviation is 5 hours. What is the margin of error for a 99% confidence interval?

0.745

0.334

0.320

0.413

I just need to know what to plug in not the answer

Answer 1

@Hero @texaschic101 @wolfe8

Answer 2

@shamil98

Answer 3

@DemolisionWolf

Answer 4

do you have the equation for marginal error?

Answer 5

yes i know it but i dont understand it

Answer 6

1.96 * a/√n right?

Answer 7

the equation of varies based on how the teacher teaches is.. so.. i can't really say if that is the right equation for your class.. BUT!

so from the information given, pick out the
population (or sample size) = ?
confidence interval (CI) = ?
standard deviation = ?

Answer 8

population (or sample size) = 13
confidence interval (CI) = .99
standard deviation = 5

Answer 9

population (or sample size) = 300
mean of sample = 13
confidence interval (CI) = .99
standard deviation = 5

Answer 10

1.96 * a/√n is the equation from your class?

Answer 11

yes
its not exactly an a just a weird symbol close to an a
1.96 • sigma over square root n

Answer 12

its a sigma

Answer 13

σ = the population standard deviation and n = the sample size

Answer 14

You there @DemolisionWolf

Answer 15

what is the 1.96 mean anything or is just equation?

Answer 16

part of the formula so you dont change that

Answer 17

@shamil98

Answer 18

Then Margin of Error = 1.96 * (sigma/sqrt(n) )
so Margin of Error = 1.96 * (5/sqrt(300) )

Answer 19

i got .565 not a answer

Answer 20

ya, i'm not sure what equation is for you guys to use..

Answer 21

is it 0.07?