There is no prior information about the proportion of americans who support the gun control in 2018. if we want to estimate 99% confidence interval for the true proportion of americans who support the gun control in 2018 with a 0.36 margin of error, how many randomly selected americans must be surveyed?

13

Step-by-step explanation:

If the prior population proportion is unavailable then the formula to find the sample size is given by :-

, where z* = Critical z-value

E = margin of error

Let p be the proportion of Americans who support the gun control in 2018.

As per given , we have

Confidence level = 99%

The critical z-value for 99% confidence interval is 2.576  ( BY z-table)

Margin of error : E= 0.36

Since there no  prior information about the proportion of Americans who support the gun control in 2018.

So , the required sample size  to estimate 99% confidence interval would be:

Hence,  13 Americans should be surveyed.

$450

step-by-step explanation:

first, you have to multiply 150x2=300

next, you have to add 150+300= 450