Question

ch 16 you plan to construct a confidence interval for the mean, μ, of a normal population with (known) standard deviation σ. which of the following will reduce the size of the margin of error? use a lower level of confidence. reduce sigma. increase the sample size.

Explanation:

Step1: Recall margin - of - error formula

The margin of error $E$ for a confidence interval of the mean of a normal population with known standard deviation $\sigma$ is $E = z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$, where $z_{\alpha/2}$ is the z - score corresponding to the level of confidence, $\sigma$ is the population standard deviation, and $n$ is the sample size.

Step2: Analyze effect of confidence level

Lowering the confidence level decreases the value of $z_{\alpha/2}$. Since $E = z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$, a lower $z_{\alpha/2}$ reduces $E$.

Step3: Analyze effect of $\sigma$

Reducing $\sigma$ directly reduces the value of the fraction $\frac{\sigma}{\sqrt{n}}$. Since $E = z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$, a smaller $\sigma$ leads to a smaller $E$.

Step4: Analyze effect of sample size

Increasing the sample size $n$ increases the denominator $\sqrt{n}$ of the fraction $\frac{\sigma}{\sqrt{n}}$. Since $E = z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$, a larger $n$ makes the value of $E$ smaller.

Answer:

All of the above (Use a lower level of confidence, Reduce sigma, Increase the sample size)