Question

According to a recent study of 7335 young people in the US, 30% had been arrested for a crime other than a traffic violation by the age of 23. Crimes included such things as vandalism, underage drinking, drunken driving, shoplifting, and drug possession.
The margin of error for the population proportion estimate, p, is 0.01 . Use this information to give a range of plausible values for the parameter.

Answer 1

Answer: $(0.29,\ 0.31)$

Step-by-step explanation:

Let p be the population proportion.

Given : According to a recent study of 7335 young people in the US, 30% had been arrested for a crime other than a traffic violation by the age of 23. i.e.p= 0.30

The margin of error for the population proportion estimate, p : E= 0.01 .

Then, the range of plausible values for the parameter will be :-

$p\pm E\\\\=0.30\pm0.01=(0.30-0.01,\ 0.30+0.01)\\\\=(0.29,\ 0.31)$

Hence, the required answer = $(0.29,\ 0.31)$

Answer 2

Answer and Step-by-step explanation:

Part - A :

$Sample~size,n=7335$

$Sample~proportion~p=0.30$

$The~sample~proportion~is~a~measure~of~simple~statistic.$

Part - B :

$Margin~of~the~error~of~the~population~is,p=0.01$

$The~lower~limit~for~plausible~values~for~the~perimeter~is:$

$p-p=0.30-0.01$

$p-p=0.29$

$The~lower~limit~for~plausible~values~for~the~parameter~is:$

$p+p=0.30+0.01$

$p-p=0.31$

Thus, the range of plausible values for the parameter = (0.29 , 0.31).