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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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Answer

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

Solution

To determine if the given information is correct, we can calculate the margin of error \( E \) using the formula for the margin of error in a confidence interval for a population mean: \[ E = z \cdot \frac{\sigma}{\sqrt{n}} \] Where: - \( E \) is the margin of error, - \( z \) is the z-score corresponding to the desired confidence level, - \( \sigma \) is the population standard deviation, - \( n \) is the sample size. ### Step 1: Find the z-score for \( C = 95\% \) For a 95% confidence level, the z-score can be found using standard normal distribution tables or calculators. The z-score for 95% confidence is approximately: \[ z \approx 1.96 \] ### Step 2: Substitute the known values into the formula Given: - \( \sigma = 5.34 \) - \( n = 45 \) We can now calculate \( E \): \[ E = 1.96 \cdot \frac{5.34}{\sqrt{45}} \] ### Step 3: Calculate \( \sqrt{n} \) First, we calculate \( \sqrt{45} \): \[ \sqrt{45} \approx 6.708 \] ### Step 4: Calculate \( \frac{\sigma}{\sqrt{n}} \) Now, we calculate \( \frac{5.34}{\sqrt{45}} \): \[ \frac{5.34}{6.708} \approx 0.796 \] ### Step 5: Calculate the margin of error \( E \) Now we can calculate \( E \): \[ E = 1.96 \cdot 0.796 \approx 1.56 \] ### Conclusion The calculated margin of error \( E \) is approximately 1.56, which matches the given value. Therefore, the information provided is correct.

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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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