Question

ch 16 you plan to construct a confidence interval for the mean (mu) 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. increase the sample size. increase the standard deviation. increase z*.

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 $E = z^*\frac{\sigma}{\sqrt{n}}$, where $z^*$ is the critical value, $\sigma$ is the population standard deviation, and $n$ is the sample size.

Step2: Analyze effect of confidence level

A lower confidence level corresponds to a smaller $z^*$ value. Since $E$ is directly proportional to $z^*$, a smaller $z^*$ reduces $E$.

Step3: Analyze effect of sample size

As $n$ increases, the denominator $\sqrt{n}$ increases. Since $E=\frac{z^*\sigma}{\sqrt{n}}$, an increase in $n$ leads to a decrease in $E$.

Step4: Analyze effect of standard deviation

Since $E$ is directly proportional to $\sigma$, increasing $\sigma$ will increase $E$.

Step5: Analyze effect of $z^*$

Increasing $z^*$ (which occurs with a higher confidence level) will increase $E$ as $E$ is directly proportional to $z^*$.

Answer:

Use a lower level of confidence, Increase the sample size.