Question

g You plan to construct a confidence interval for the mean μ μ of a Normal population with known standard deviation σ . σ. Which choice will reduce the size of the margin of error? increasing the sample size using a lower level of confidence All of the answer options are correct. reducing σ

Answer 1

Answer:

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (1)  

The margin of error is given by:

$ME= z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$

And we can reduce this margin of error with:

Increasing the sample size

Step-by-step explanation:

Previous concepts  

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

$\bar X$ represent the sample mean  

$\mu$ population mean (variable of interest)  

$\sigma$ represent the population standard deviation  

n represent the sample size  

The confidence interval for the mean is given by the following formula:  

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (1)  

The margin of error is given by:

[tex] ME= z_{\alpha/2}\frac{\sigma}{\sqrt{n}}

And we can reduce this margin of error with:

Increasing the sample size