contestada

In a survey of 200 employees, 65% agreed that the management of the cafeteria should be changed.

If the confidence interval is 95%, the margin of error for this survey is
A) +/- 6.75
B) +/- 5.06
C) +/- 3.37
and the maximum number of employees who could have agreed is about
A) 144
B) 140
C) 136

Respuesta :

conairtownsend0113

Answer:

first one is c becuase of the way i added and then subtactied

second one is a i divide then ad sub and ad it again bye 3

srry if i am wrong

brainlyest would be nice ( ;

Step-by-step explanatnation

We are given the following in the question:

Sample size, n = 200

\hat{p} = 65\% = 0.65

Margin of error =

\pm 6.75\% = \pm 0.0675

\hat{p}\pm z_{stat}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}

\hat{p}\pm \text{Margin of error}

Putting values:

(0.65 \pm 0.0675)\\=(0.5825,0.7175)

Answer:

C) +/- 3.37

A). 144

Step-by-step explanation:

Otras preguntas