7. Suppose a sample has a standard deviation of sigma =21.26 and a sample size of n=500. Find the margin of error to the nearest tenth. square

To find the margin of error, we need to use the formula:<br /><br />\[ \text{Margin of Error} = z \times \frac{\sigma}{\sqrt{n}} \]<br /><br />where:<br />- \( z \) is the z-score (for a 95% confidence level, \( z = 1.96 \))<br />- \( \sigma \) is the standard deviation<br />- \( n \) is the sample size<br /><br />Given:<br />- \( \sigma = 21.26 \)<br />- \( n = 500 \)<br /><br />First, calculate the standard error:<br /><br />\[ \text{Standard Error} = \frac{\sigma}{\sqrt{n}} = \frac{21.26}{\sqrt{500}} \]<br /><br />Calculate \( \sqrt{500} \):<br /><br />\[ \sqrt{500} \approx 22.36 \]<br /><br />Now, calculate the standard error:<br /><br />\[ \text{Standard Error} = \frac{21.26}{22.36} \approx 0.95 \]<br /><br />Next, calculate the margin of error:<br /><br />\[ \text{Margin of Error} = 1.96 \times 0.95 \approx 1.87 \]<br /><br />Rounding to the nearest tenth:<br /><br />\[ \text{Margin of Error} \approx 1.9 \]<br /><br />Therefore, the margin of error to the nearest tenth is \( \boxed{1.9} \).