se = \\frac{\\sigma}{\\sqrt{n}}\nthe given equation relates the standard error, se, of a sample mean to the…

se = \\frac{\\sigma}{\\sqrt{n}}\nthe given equation relates the standard error, se, of a sample mean to the population standard deviation, \\sigma, and the size of the sample, n. which of the following equations correctly gives the size of the sample in terms of the standard error and the population standard deviation?\nchoose 1 answer:\na n=(\\frac{\\sigma}{se})^2\nb n = \\frac{\\sigma^2}{se}\nc n=\\sqrt{(\\frac{\\sigma}{se})}\nd n = \\frac{\\sqrt{\\sigma}}{se}

se = \\frac{\\sigma}{\\sqrt{n}}\nthe given equation relates the standard error, se, of a sample mean to the population standard deviation, \\sigma, and the size of the sample, n. which of the following equations correctly gives the size of the sample in terms of the standard error and the population standard deviation?\nchoose 1 answer:\na n=(\\frac{\\sigma}{se})^2\nb n = \\frac{\\sigma^2}{se}\nc n=\\sqrt{(\\frac{\\sigma}{se})}\nd n = \\frac{\\sqrt{\\sigma}}{se}

Answer

Explanation:

Step1: Isolate the square - root term

Given $SE=\frac{\sigma}{\sqrt{n}}$, we can cross - multiply to get $SE\times\sqrt{n}=\sigma$. Then $\sqrt{n}=\frac{\sigma}{SE}$.

Step2: Square both sides

Squaring both sides of the equation $\sqrt{n}=\frac{\sigma}{SE}$, we have $n = (\frac{\sigma}{SE})^2$.

Answer:

A. $n = (\frac{\sigma}{SE})^2$