Respuesta :
The sample size for the part (a) is 269, and the sample size for part (b) is 191 if a margin of error of 5 percentage points and use a confidence level of 90%
What is the margin of error(MOE)?
It is defined as an error that gives an idea about the percentage of errors that exist in the real statistical data.
The formula for finding the MOE:
[tex]\rm MOE = Z{score}\times\frac{s}{\sqrt{n} }[/tex]
Where is the z score at the confidence interval
s is the standard deviation
n is the number of samples.
We have:
MOE = 5% = 0.05 and
[tex]\alpha = 1-0.9 \Rightarrow 0.1[/tex]
Let's assume the value of p = 0.5, and q = 0.5
From the table:
[tex]\rm Z_{\alpha/2}\Rightarrow Z_{0.1/2}\Rightarrow Z_{0.05}\Rightarrow 1.64[/tex]
[tex]\rm n = p\times q \frac{Z^2_{\alpha/2}}{MOE^2}[/tex]
[tex]\rm n = 0.5\times 0.5\frac{1.64^2}{0.05^2}[/tex]
n = 268.96 ≈ 269
For part(b)
The value of p = 0.23, and q = 0.77
From the table:
[tex]\rm Z_{\alpha/2}\Rightarrow Z_{0.1/2}\Rightarrow Z_{0.05}\Rightarrow 1.64[/tex]
[tex]\rm n = p\times q \frac{Z^2_{\alpha/2}}{MOE^2}[/tex]
[tex]\rm n = 0.23\times 0.77\frac{1.64^2}{0.05^2}[/tex]
n = 190.53 ≈ 191
Thus, the sample size for the part (a) is 269, and the sample size for part (b) is 191 if a margin of error of 5 percentage points and use a confidence level of 90%
Learn more about the Margin of error here:
brainly.com/question/13990500
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