A researcher wishes to estimate the proportion of adults who have​ high-speed Internet access. What size sample should be obtained if she wishes the estimate to be within with ​% confidence if ​(a) she uses a previous estimate of ​? ​(b) she does not use any prior​ estimates?

732 samples ;

752 samples

Step-by-step explanation:

Given :

α = 90% ; M.E = 0.03 ; p = 0.58 ; 1 - p = 1 - 0.58 = 0.42

Using the relation :

n = (Z² * p * (1 - p)) / M.E²

Zcritical at 90% = 1.645

n = (1.645² * 0.58 * 0.42) / 0.03²

n = 0.65918769 / 0.0009

n = 732.43076

n = 732 samples

B.)

If no prior estimate is given, then p = 0.5 ; 1 - p = 1 - 0.5 = 0.5

n = (Z² * p * (1 - p)) / M.E²

Zcritical at 90% = 1.645

n = (1.645² * 0.5 * 0.5) / 0.03²

n = 0.67650625 / 0.0009

n = 751.67361

n = 752 samples

i would say the third one.

step-by-step explanation: liner would mean in a line and the only one that doesn't look like it's even thinking about forming a line would be the third one

amount of oatypop that contains the prize = 40%

amount   of oatypop which doesn't contain the prize = (100-40)=60%

probability that hannah will win a prize

    = s +f s +f f s+ff   f s+ff f f s+ infinity.where f= failure,s= success

=   or 60%

probability that hannah only has to buy 3 or less boxes before getting a prize= (s) and ( f s) and (ff s) and (f f f s)

        = = 87.04 %  

step-by-step explanation: