Find the margin of error for the given values of c, $\sigma _ { , }$ and n. $\mathrm { c } = 0.90 , \sigma = 11.5 , \mathrm { n } = 120$
A)1.73
B)0.16
C)1.05
D)0.94

Question

Quizplus · Accepted Answer

The margin of error (E) for a given confidence level (c), standard deviation ($\sigma$), and sample size (n) can be calculated using the formula $E = Z \cdot \frac{\sigma}{\sqrt{n}}$, where Z is the Z-score corresponding to the confidence level. For a 90% confidence level, the Z-score is approximately 1.645. Plugging in the given values: $E = 1.645 \cdot \frac{11.5}{\sqrt{120}} \approx 1.73$.

Find the Margin of Error for the Given Values of C