You want to estimate the mean exam score for students at a university. The confidence level that you use in your calculation is 90%, the standard deviation is 5, and the sample size of the population of students is 500. What is the margin of error? Round your answer to the nearest hundredth.(1 point) The margin of error is

1 answer

To calculate the margin of error for a confidence interval, you can use the formula:

\[ \text{Margin of Error} = z \times \left(\frac{\sigma}{\sqrt{n}}\right) \]

Where:

  • \( z \) is the z-score corresponding to the confidence level,
  • \( \sigma \) is the standard deviation,
  • \( n \) is the sample size.

For a 90% confidence level, the z-score (which corresponds to the critical value for a two-tailed test) is approximately 1.645.

Given:

  • Standard deviation (\( \sigma \)) = 5
  • Sample size (\( n \)) = 500

Now, let's calculate the standard error (SE):

\[ \text{Standard Error} = \frac{\sigma}{\sqrt{n}} = \frac{5}{\sqrt{500}} \]

Calculating \( \sqrt{500} \):

\[ \sqrt{500} \approx 22.36 \]

Now calculate the standard error:

\[ \text{Standard Error} = \frac{5}{22.36} \approx 0.2236 \]

Now, compute the margin of error:

\[ \text{Margin of Error} = 1.645 \times 0.2236 \approx 0.3674 \]

Finally, rounding the margin of error to the nearest hundredth:

\[ \text{Margin of Error} \approx 0.37 \]

So, the margin of error is approximately 0.37.