Assuming that the samples come from normal distribution, find the margin of error E given the following:
n = 10 and x̄ = 28 with s = 4.0, 90% confidence
n=10, "\\bar{x}" =28, s=4, "\\alpha" =100-90=10%=0.1
"Z_{\\frac{\\alpha}{2}}=Z_{0.05}=2.575"
Therefore, The critical value is 2.575.
The standard deviation is 2.0, but as this is a sample, we need the standard error for the mean.
The formula for the SE of the mean "\\sigma= \\dfrac{s}{\\sqrt{n}}" ,
so: "\\sigma=\\dfrac{2}{\\sqrt{10}}=0.6325."
Margin of Error="2.575 \\times 0.6325 = 1.629."
Therefore, the margin of error is 1.629.
Comments
Leave a comment