Answer:
The standard error (SE) of the sample proportion is 0.0218.
Step-by-step explanation:
The standard error (SE) of a sample measures of spread, and can be determined by dividing the standard deviation by the square root of the sample size.
So that with respect to proportion, SE can be calculated by:
SE = [tex]\sqrt{\frac{p(1-p)}{n} }[/tex]
Where: p is the sample proportion and n is the sample size.
From the question, n= 500, while 304 complained of the unemployment rate.
Thus, the sample proportion is;
p = [tex]\frac{304}{n}[/tex]
p = [tex]\frac{304}{500}[/tex]
=0.608
∴ p = 0.608
Substituting the values of p and n in SE, we have:
SE = [tex]\sqrt{\frac{0.608(1 - 0.608)}{500} }[/tex]
= [tex]\sqrt{\frac{0.238336}{500} }[/tex]
= [tex]\sqrt{4.76672* 10^{-4} }[/tex]
= 0.021833
∴ SE = 0.0218
The standard error (SE) of the sample proportion is 0.0218.
The standard error of the sample proportion is similar to the standard deviation the only difference is standard error use static data and in standard deviation we use parameter.
The standard error of the sample proportion is 0.0218.
Given:
The random sample is [tex]n=500[/tex].
The adult who complain the unemployment is 304.
Calculate the sample proportion.
[tex]p=\dfrac{304}{n}\\p=\dfrac{304}{500}\\p=0.608[/tex]
Calculate the standard error (SE).
[tex]SE=\sqrt{\dfrac{p(1-p)}{n}[/tex]
Substitute the value.
[tex]SE=\sqrt{\dfrac{0.608(1-0.608)}{500}}\\SE= \sqrt{4.7662\times 10^-4}\\SE=0.021833[/tex]
Thus, the standard error of the sample proportion is 0.0218.
Learn more about standard error here:
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