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

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Answer:

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

The margin of error is given by:

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

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".  

[tex]\bar X[/tex] represent the sample mean  

[tex]\mu[/tex] population mean (variable of interest)  

[tex]\sigma[/tex] represent the population standard deviation  

n represent the sample size  

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

[tex]\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex] (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