To find the margin of error (ME) for the mean at the 95% confidence level, you can use the formula:

$\text{ME} = z \times \left( \frac{\sigma}{\sqrt{n}} \right)$

Where:

• $z$ is the z-score for a 95% confidence interval (approximately 1.96).
• $\sigma$ is the standard deviation of the sample (87 messages).
• $n$ is the sample size (53 students).

Substitute the values into the formula:

$\text{ME} = 1.96 \times \left( \frac{87}{\sqrt{53}} \right)$

First, calculate $\sqrt{53}$:

$\sqrt{53} \approx 7.28$

Then, divide the standard deviation by this result:

$\frac{87}{7.28} \approx 11.95$

Now, multiply by the z-score:

$1.96 \times 11.95 \approx 23.42$

Rounding to the nearest whole number, the margin of error is approximately $23$.