Question

what is the z-score with a confidence level of 95% when finding the margin of error for the mean of a normally distributed population from a sample?
0.99
1.65
1.96
2.58

Explanation:

Step1: Recall Confidence Level and Z - Scores

For a confidence level of 95%, the significance level $\alpha=1 - 0.95 = 0.05$. The two - tailed test divides $\alpha$ into two equal parts, so $\alpha/2=0.025$. We need to find the $z$ - score such that $P(Z > z_{\alpha/2})=\alpha/2 = 0.025$. Looking at the standard normal distribution table (or using a calculator with a normal - distribution inverse function), the $z$ - score corresponding to a right - tail area of 0.025 is 1.96.

Step2: Eliminate Other Options

• The $z$ - score of 0.99 is not related to a 95% confidence level.
• A $z$ - score of 1.65 is used for a 90% confidence level (since for 90% confidence, $\alpha = 0.1$, $\alpha/2=0.05$, and $z_{0.05}=1.65$).
• A $z$ - score of 2.58 is used for a 99% confidence level (since for 99% confidence, $\alpha=0.01$, $\alpha/2 = 0.005$, and $z_{0.005}=2.58$).

Answer:

1.96 (corresponding to the option with 1.96)