A random sample of 20 items is selected (from a population with an unknown population standard deviation). When computing a conf
idence interval for the population mean, what number of degrees of freedom should be used to determine the appropriate t-value?
1 answer:
Answer:
degree of freedom is 19
Step-by-step explanation:
Given data
sample = 20
to find out
degree of freedom
solution
we know degree of freedom formula
that is
degree of freedom = sample -1
we have given sample 20
so
degree of freedom = 20 -1
so degree of freedom is 19
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![4^\frac{1}{3}\cdot4^\frac{1}{6}\\\\\text{use}\ a^n\cdot a^m=a^{n+m}\\\\=4^{\frac{1}{3}+\frac{1}{6}}=4^{\frac{1\cdot2}{3\cdot2}+\frac{1}{6}}=4^{\frac{2}{6}+\frac{1}{6}}=4^\frac{2+1}{6}=4^\frac{3}{6}=4^\frac{3:3}{6:3}=4^{\frac{1}{2}}\\\\\text{use}\ a^\frac{m}{n}=\sqrt[n]{a^m}\\\\=\sqrt[2]{4^1}=\sqrt4=2](https://tex.z-dn.net/?f=4%5E%5Cfrac%7B1%7D%7B3%7D%5Ccdot4%5E%5Cfrac%7B1%7D%7B6%7D%5C%5C%5C%5C%5Ctext%7Buse%7D%5C%20a%5En%5Ccdot%20a%5Em%3Da%5E%7Bn%2Bm%7D%5C%5C%5C%5C%3D4%5E%7B%5Cfrac%7B1%7D%7B3%7D%2B%5Cfrac%7B1%7D%7B6%7D%7D%3D4%5E%7B%5Cfrac%7B1%5Ccdot2%7D%7B3%5Ccdot2%7D%2B%5Cfrac%7B1%7D%7B6%7D%7D%3D4%5E%7B%5Cfrac%7B2%7D%7B6%7D%2B%5Cfrac%7B1%7D%7B6%7D%7D%3D4%5E%5Cfrac%7B2%2B1%7D%7B6%7D%3D4%5E%5Cfrac%7B3%7D%7B6%7D%3D4%5E%5Cfrac%7B3%3A3%7D%7B6%3A3%7D%3D4%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D%5C%5C%5C%5C%5Ctext%7Buse%7D%5C%20a%5E%5Cfrac%7Bm%7D%7Bn%7D%3D%5Csqrt%5Bn%5D%7Ba%5Em%7D%5C%5C%5C%5C%3D%5Csqrt%5B2%5D%7B4%5E1%7D%3D%5Csqrt4%3D2)
LO = MN because LMNO is the isosceles trapezod
so <LON = <ONM = 100
Answer
<LON = 100 degrees
