# 8.1 A single population mean using the normal distribution  (Page 7/20)

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“Metadata Description of Candidate Summary File.” U.S. Federal Election Commission. Available online at http://www.fec.gov/finance/disclosure/metadata/metadataforcandidatesummary.shtml (accessed July 2, 2013).

“National Health and Nutrition Examination Survey.” Centers for Disease Control and Prevention. Available online at http://www.cdc.gov/nchs/nhanes.htm (accessed July 2, 2013).

## Chapter review

In this module, we learned how to calculate the confidence interval for a single population mean where the population standard deviation is known. When estimating a population mean, the margin of error is called the error bound for a population mean ( EBM ). A confidence interval has the general form:

(lower bound, upper bound) = (point estimate – EBM , point estimate + EBM )

The calculation of EBM depends on the size of the sample and the level of confidence desired. The confidence level is the percent of all possible samples that can be expected to include the true population parameter. As the confidence level increases, the corresponding EBM increases as well. As the sample size increases, the EBM decreases. By the central limit theorem,

$EBM=z\frac{\sigma }{\sqrt{n}}$

Given a confidence interval, you can work backwards to find the error bound ( EBM ) or the sample mean. To find the error bound, find the difference of the upper bound of the interval and the mean. If you do not know the sample mean, you can find the error bound by calculating half the difference of the upper and lower bounds. To find the sample mean given a confidence interval, find the difference of the upper bound and the error bound. If the error bound is unknown, then average the upper and lower bounds of the confidence interval to find the sample mean.

Sometimes researchers know in advance that they want to estimate a population mean within a specific margin of error for a given level of confidence. In that case, solve the EBM formula for n to discover the size of the sample that is needed to achieve this goal:

## Formula review

$\overline{X}~N\left({\mu }_{X},\frac{\sigma }{\sqrt{n}}\right)$ The distribution of sample means is normally distributed with mean equal to the population mean and standard deviation given by the population standard deviation divided by the square root of the sample size.

The general form for a confidence interval for a single population mean, known standard deviation, normal distribution is given by
(lower bound, upper bound) = (point estimate – EBM , point estimate + EBM )
= $\left(\overline{x}-EBM,\overline{x}+EBM\right)$
= $\left(\overline{x}-z\frac{\sigma }{\sqrt{n}},\overline{x}+z\frac{\sigma }{\sqrt{n}}\right)$

EBM = $z\frac{\sigma }{\sqrt{n}}$ = the error bound for the mean, or the margin of error for a single population mean; this formula is used when the population standard deviation is known.

CL = confidence level, or the proportion of confidence intervals created that are expected to contain the true population parameter

α = 1 – CL = the proportion of confidence intervals that will not contain the population parameter

${z}_{\frac{\alpha }{2}}$ = the z -score with the property that the area to the right of the z-score is this is the z -score used in the calculation of "EBM where α = 1 – CL .

n = $\frac{{z}^{2}{\sigma }^{2}}{EB{M}^{2}}$ = the formula used to determine the sample size ( n ) needed to achieve a desired margin of error at a given level of confidence

Let x1, x2, ...,xn be a random sample of size n from N(0,σ  ), show that there exists an UMP test with significance level α for testing H0 :  2 =  2 against H1 :  2 <  2 . If n=15,  = 0.05, and  2= 3, determine the BCR
3xy^2√[x^3y^2/(12(x^3y)^2)]
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