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Marked out of 1.00 For \( C=95 \% \) If \( \sigma=5.34 \) and \( \mathrm{n}=45 \) then \( E=1.56 \)

Ask by Coles Gough. in Saudi Arabia
Mar 18,2025

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The margin of error \( E \) is correctly calculated as 1.56 based on the given values of \( C = 95\% \), \( \sigma = 5.34 \), and \( n = 45 \).

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To determine if the provided margin of error \( E = 1.56 \) is accurate based on the given parameters, you can use the formula for the margin of error in a confidence interval for a population mean: \[ E = z \times \left( \frac{\sigma}{\sqrt{n}} \right) \] For a 95% confidence level, the z-score is approximately 1.96. Plugging in the given values: \[ E = 1.96 \times \left( \frac{5.34}{\sqrt{45}} \right) \] Calculating \( \frac{5.34}{\sqrt{45}} \): \[ \sqrt{45} \approx 6.7082 \quad \Rightarrow \quad \frac{5.34}{6.7082} \approx 0.796 \] Now substituting back: \[ E \approx 1.96 \times 0.796 \approx 1.56 \] Thus, your calculation holds true! This indicates that with a sample size of 45 and a standard deviation of 5.34, the margin of error at a 95% confidence level indeed results in 1.56, confirming the accuracy of the given information!

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