Question

Use the information from the polling agency to determine the sample size needed to construct a 95% confidence interval with a margin of error of no more than 3.6%. For consistency, use the reported sample proportion for the planning value of p* (rounded to 4 decimal places) and round your Z* value to 3 decimal places. Your answer should be an integer.

Answer 1

Answer:

the sample size n= 771

sample proportion is p = 1/2

Step-by-step explanation:

Step1:-

marginal error formula is $\frac{2S.D}{\sqrt{n} }$

but if 'n' is sample size then $n = \frac{1}{(margin error)^{2} }$

Given margin error is 36% = 0.036

$n= \frac{1}{(0.036)^2}$

on simplification , we get n= 777 polling agencies

Step 2:-

$Margin of error = \frac{2\sqrt{p(1-p)} }{\sqrt{n} }$

substitute values , we get

$\frac{36}{100} = \frac{2\sqrt{p(1-p)} }{\sqrt{771} }$

after simplification, we get

$\sqrt{p(1-p)} = \frac{1}{2}$

squaring on both sides, we get

$p(1-p) = \frac{1}{4}$

multiply 'p' inside

$p-p^{2} =\frac{1}{4}$

cross multiplication '4' and simplify

$4p^{2} -4p +1 =0$ ........(1)

finding 'p' value

The equation (1) is of the form $(a-b)^{2} = a^{2} -2ab+b^{2}$

$(2p)^{2} -2(2p)(1)+1^{2} = (2p-1)^2$

$(2p-1)^2 =0$

$p=\frac{1}{2}$

conclusion:-

The sample size is 771

the sample proportion is p = $\frac{1}{2}$