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tino4ka555 [31]
3 years ago
9

A local board of education conducted a survey of residents in the community concerning a property tax levy on the coming local b

allot. 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
Mathematics
1 answer:
IrinaK [193]3 years ago
8 0

Answer:

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

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