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?

0.743
0.334
0.320
0.413

@John_ES

Answer 1

B ?

Answer 2

@John_ES

Answer 3

first why do u think it is B

Answer 4

gtg

Answer 5

nvm i dont know how to do this cx so forget B ..

Answer 6

Do you know something about normal distribution?

Answer 7

no idea

Answer 8

Ok. Let's remember some concepts.

Answer 9

First the confidence interval for a normal distribution,

\[(\overline{x}-Z_{\alpha/2}\frac{\sigma}{\sqrt{n}},\overline{x}+Z_{\alpha/2}\frac{\sigma}{\sqrt{n}})\]

Answer 10

Where,
\[\overline{x}\] is the mean,
\[Z_{\alpha/2}=2.575\]for a confidence level of 99%,
\[\sigma\] is the standard deviation, and
\[n\] is the size of the sample. Ok?

Answer 11

You need the margin error. I would say that the error can be obtained from the confidence interval, using this formula,
\[Error=Z_{\alpha/2}\frac{\sigma}{\sqrt{n}}\]

Answer 12

As you see, is the value that is displaced from the mean.

Answer 13

ok...

Answer 14

Now, you can substitute,
\[n=100; \sigma=5; \ Z_{\alpha/2}=2.575\]

Answer 15

Sorry, n=300

Answer 16

Then you have,

\[E=2.575\cdot 5/\sqrt{300}=0.743\] The first answer.