Suppose you have decided to gather some fixed number n of ob

Suppose you have decided to gather some fixed number n of observations, and create a confidence interval at down confidence level c, where the variance of the underlying population is known to be o^2. Before you even gather your n numbers, you already know how high your margin of error will be. Explain why.

because the margin of error depends of significance level that is c

if you have a 95% confidence level, you know your significance level that is 1 - 0.95 = 0.05

but as you will have a confidence interval that is 2 tailed, upper and lower

significance level / 2 = 0.025

so your magin of error will be closer to 0.025

because the margin of error not only depends of alpha ( significance level)

Margin of error for a normal distribution with sigma is

Z* sigma / srqt(n)

but before you have \"n\" you alaready know how high your amrgin of error will be, as I explain

is because a confidence interval with 95% of confidence, we are saying that we will tolerate 0.05 of error

and that is for all confidence interval

so for this question , in general

we have a confidence level of \"C\"

so we are saying that we will tolerate 1 - C of error

Suppose you have decided to gather some fixed number n of observations, and create a confidence interval at down confidence level c, where the variance of the

Suppose you have decided to gather some fixed number n of ob

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