Answer:
Option A) About ±2.98 minutes
Step-by-step explanation:
We are given the following information in the question:
Sample mean, [tex]\bar{x}[/tex] = 191.3 minutes
Sample size, n = 200
Population standard deviation, σ = 21.5 minutes
Alpha, α = 0.05
The leaders of the organization wish to develop an interval estimate with 95 percent confidence.
[tex]z_{critical}\text{ at}~\alpha_{0.05} = \pm 1.96[/tex]
Margin of error =
[tex]z_{\text{critical}}\times \displaystyle\frac{\sigma}{\sqrt{n}}[/tex]
Putting the values, we get:
[tex]\pm 1.96\times \displaystyle\frac{21.5}{\sqrt{200}} = \pm 2.9797 \approx \pm 2.98[/tex]
Option A) About ±2.98 minutes
