To find the margin of error for the given values of the confidence level [tex]\( c \)[/tex], the population standard deviation [tex]\( \sigma \)[/tex], and the sample size [tex]\( n \)[/tex], we need to follow these steps:
1. Determine the Z score for the given confidence level. The confidence level [tex]\( c = 0.95 \)[/tex] corresponds to a 95% confidence interval. This means that [tex]\( 95\% \)[/tex] of the data lies within a certain number of standard deviations (Z score) from the mean in a standard normal distribution. For a [tex]\( 95\% \)[/tex] confidence level, the Z score is commonly found in statistical tables or using software and is approximately 1.96.
2. Compute the standard error of the mean using the population standard deviation [tex]\( \sigma \)[/tex] and the sample size [tex]\( n \)[/tex]. The formula for the standard error (SE) is:
[tex]\[
SE = \frac{\sigma}{\sqrt{n}}
\][/tex]
Substituting the given values ([tex]\( \sigma = 677 \)[/tex] and [tex]\( n = 40 \)[/tex]):
[tex]\[
SE = \frac{677}{\sqrt{40}}
\][/tex]
3. Calculate the margin of error using the Z score and the standard error. The formula for the margin of error (MoE) is:
[tex]\[
MoE = Z \times SE
\][/tex]
Putting it all together, we get the following steps:
1. Find the standard error:
[tex]\[
SE = \frac{677}{\sqrt{40}} \approx 107.068259
\][/tex]
2. Use the Z score of 1.96 for a 95% confidence level.
3. Compute the margin of error:
[tex]\[
MoE = 1.96 \times 107.068259 \approx 209.8006184350941
\][/tex]
Therefore, the margin of error for the given values is approximately [tex]$210.
The correct answer is:
D. $[/tex]\[tex]$ 210$[/tex]
