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

hello from mrbilldoesmath!

answer:

x - 3

discussion:

x^2 - 9 = (x-3) (x+3)   so

(x^2-9)/ (x + 3) = x-3

you,

mrb