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

(1)  

The margin of error is given by:

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".  

represent the sample mean  

population mean (variable of interest)  

represent the population standard deviation  

n represent the sample size  

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

(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

b

step-by-step explanation:

the word superfluous means unnecessary

step-by-step explanation:

because 1/3 is a fraction. we can tell because there is a divisor symbol between the 1 and the 3. if we were to add 2 fractions together, we would add the numerators, which is the number on top, but not the denominators, which is the number on the bottom

1/3+1/3=2/3

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