A 2010 study finds that in a random sample of 3000 American adults aged 18 and over, 1350 owned an MP3 player such as an iPod. Give a 95% large-sample confidence interval for the proportion of all American adults aged 18 and over who own an MP3 player.

0.45 ± 0.018

Step-by-step explanation:

In this question, we are asked to find the proportion of American citizens owning a mp3 using a confidence interval of 95%

We proceed as follows;

p=1350/3000=0.45

margin of error=1.96* √(0.45*0.55/3000)=

so 95% cI= mean+/- margin of error

=0.45 ± 0.018

3 qrs + 3/2 xy hope this : /

step-by-step explanation:

look at photo