Respuesta :
Answer:
sample size if he uses a previous estimate of 32% is 523
sample size if he does not use any prior estimates is 601
Step-by-step explanation:
estimate E = 4 percentage = 0.04
confidence Cl = 95% = 0.95
previous estimate p = 32% = 0.32
q = 1 - 0.32 = 0.68
to find out
size of sample
solution
first we calculate z value for 95% confidence E = 0.04 is 1.96
from probability P(-1.96 < z < 1.96) = 0.95)
here z = 1.96
so in 1st part size of sample we know
E = z × [tex]\sqrt{pq/n}[/tex]
put all value E, z p and q and we get n
n = (z/E)² ×p×q
n = (1.96/0.04)² ×0.32×0.68
n = 523
sample size if he uses a previous estimate of 32% is 523
now in 2nd part we take p = 0.5
so n will be
n = (z/E)² ×p×q
n = (1.96/0.04)² ×0.5×0.5
n = 601
sample size if he does not use any prior estimates is 601
Sample size if he uses a previous estimate of [tex]32[/tex] % is [tex]523[/tex]
Sample size if he does not use any prior estimates is [tex]601[/tex]
Estimate [tex]E=4[/tex], Percentage = [tex]0.04[/tex] and Confidence [tex]CI=95[/tex] % [tex]=0.95[/tex]
Previous Estimate [tex]p=32[/tex] % [tex]=0.32[/tex] and [tex]q = 1 - 0.32 = 0.68[/tex]
To find out size of sample solution first we calculate [tex]Z[/tex] value that is [tex]1.96[/tex]
from Probability, [tex]P(-1.96 < z < 1.96) = 0.95)[/tex]
So, in first part size of sample we know
[tex]E=Z\times \sqrt{\frac{pq}{n}}\; \Rightarrow \; n=\left ( \dfrac{Z}{E} \right )^2\times p\times q\\ \\ n=\left ( \frac{1.96}{0.04} \right )^2\times 0.32\times 0.68\\ \\ n=523[/tex]
Therefore, Sample size if he uses a previous estimate of [tex]32[/tex] % is [tex]523[/tex]
Now in second part we take [tex]p = 0.5[/tex]
[tex]E=Z\times \sqrt{\frac{pq}{n}}\; \Rightarrow \; n=\left ( \dfrac{Z}{E} \right )^2\times p\times q\\ \\ n=\left ( \frac{1.96}{0.04} \right )^2\times 0.5\times 0.5\\ \\ n=601[/tex]
Therefore, Sample size if he does not use any prior estimates is [tex]601[/tex]
Learn more about sample size.
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