when finding the margin of error for the mean of a normally distributed population from a sample, what is…

when finding the margin of error for the mean of a normally distributed population from a sample, what is the critical probability, assuming a confidence level of 86%?\n0.14\n0.86\n0.93\n0.99

when finding the margin of error for the mean of a normally distributed population from a sample, what is the critical probability, assuming a confidence level of 86%?\n0.14\n0.86\n0.93\n0.99

Answer

Explanation:

Step1: Recall the relationship between confidence - level and critical probability

The confidence level $C$ and the critical probability $z_{\alpha/2}$ are related. The total area under the normal - distribution curve is 1. The confidence level $C$ is the area in the middle of the distribution, and the remaining area $\alpha$ is split evenly between the two tails. So, $\alpha=1 - C$. $C = 0.86$, then $\alpha=1 - 0.86=0.14$.

Step2: Calculate the critical probability

The critical probability is the area to the left of the upper - tail critical value. Since the area in the two tails is $\alpha = 0.14$, the area in the upper tail is $\frac{\alpha}{2}=0.07$. The critical probability $p$ is $1-\frac{\alpha}{2}$. $p = 1-0.07 = 0.93$.

Answer:

C. 0.93