A local board of education conducted a survey of residents in the community concerning a property tax levy on the coming local ballot. They randomly selected 850 residents in the community and contacted them by telephone. Of the 850 residents surveyed, 410 supported the property tax levy. Let p represent the proportion of residents in the community that support the property tax levy. How large a sample n would you need to estimate p with margin of error 0.04 with 95% confidence? Assume that you don't know anything about the value of p. n = 601 n = 423 n = 1037 n = 256

a. [ 0.454,0.51]

b. 599.472 ~ 600

Step-by-step explanation:

a)

Confidence Interval For Proportion

CI = p ± Z a/2 Sqrt(p*(1-p)/n)))

x = Mean

n = Sample Size

a = 1 - (Confidence Level/100)

Za/2 = Z-table value

CI = Confidence Interval

Mean(x)=410

Sample Size(n)=850

Sample proportion = x/n =0.482

Confidence Interval = [ 0.482 ±Z a/2 ( Sqrt ( 0.482*0.518) /850)]

= [ 0.482 - 1.645* Sqrt(0) , 0.482 + 1.65* Sqrt(0) ]

= [ 0.454,0.51]

b)

Compute Sample Size ( n ) = n=(Z/E)^2*p*(1-p)

Z a/2 at 0.05 is = 1.96

Samle Proportion = 0.482

ME = 0.04

n = ( 1.96 / 0.04 )^2 * 0.482*0.518

= 599.472 ~ 600

the answer is d. (1.2, -4.7) on edgn

hope this !