To find the margin of error for the mean number of text messages sent per day, we can use the formula for the margin of error (ME) in the context of a confidence interval:

$\text{Margin of Error (ME)} = z \times \frac{s}{\sqrt{n}}$

Where:

• $z$ is the z-score corresponding to the 95% confidence level.
• $s$ is the sample standard deviation.
• $n$ is the sample size.

Given:

• Sample size ($n$) = 112
• Sample mean ($\bar{x}$) = 191 (but we won't need this to calculate the margin of error)
• Sample standard deviation ($s$) = 91
• Confidence level = 95%

For a 95% confidence level, the corresponding z-score is approximately 1.96.

Now, let's calculate the margin of error:

$\text{ME} = 1.96 \times \frac{91}{\sqrt{112}}$

First, calculate the standard error ($\frac{s}{\sqrt{n}}$):

$\frac{91}{\sqrt{112}} \approx \frac{91}{10.583} \approx 8.60$

Now, calculate the margin of error:

$\text{ME} \approx 1.96 \times 8.60 \approx 16.85$

Rounding to the nearest whole number:

$\text{Margin of Error} \approx 17$

So, the margin of error is 17.

Would you like more details or have any questions? Here are some related questions to consider:

1. How would the margin of error change with a larger sample size?
2. What if the confidence level was 90% instead of 95%? How would the margin of error change?
3. Can you explain why the z-score is used in this calculation?
4. How does the standard deviation affect the margin of error?
5. What assumptions are made when calculating the margin of error in this way?

Tip: The margin of error decreases as the sample size increases, which leads to more precise estimates of the population mean.