To calculate the 95% confidence interval for the mean attention span, we can use the following formula for the confidence interval of the mean when the population standard deviation is unknown and the sample size is small (less than 30). The formula is:

$\text{Confidence Interval} = \bar{x} \pm t_{\alpha/2} \left( \frac{s}{\sqrt{n}} \right)$

Where:

• $\bar{x}$ is the sample mean.
• $t_{\alpha/2}$ is the critical value from the t-distribution for a 95% confidence level and $n - 1$ degrees of freedom.
• $s$ is the sample standard deviation.
• $n$ is the sample size.

Step 1: Identify the given values

• Sample mean ($\bar{x}$) = 48 minutes
• Sample standard deviation ($s$) = 24 minutes
• Sample size ($n$) = 14
• Degrees of freedom ($df$) = $n - 1 = 14 - 1 = 13$

Step 2: Find the critical t-value

We need the critical value for a 95% confidence level and 13 degrees of freedom. Consulting a t-distribution table or using statistical software, we find that:

t_{\alpha/2} \text{ for 13 degrees of freedom and 95% confidence} = 2.160

Step 3: Calculate the margin of error

The margin of error is given by:

$\text{Margin of Error} = t_{\alpha/2} \times \left( \frac{s}{\sqrt{n}} \right)$

Substitute the values:

$\text{Margin of Error} = 2.160 \times \left( \frac{24}{\sqrt{14}} \right)$

Now, calculate the margin of error.

$\text{Margin of Error} = 2.160 \times \left( \frac{24}{3.742} \right)$

$\text{Margin of Error} = 2.160 \times 6.414 = 13.854$

Step 4: Calculate the confidence interval

Now that we have the margin of error, we can calculate the confidence interval:

$\text{Lower limit} = \bar{x} - \text{Margin of Error} = 48 - 13.854 = 34.1$ $\text{Upper limit} = \bar{x} + \text{Margin of Error} = 48 + 13.854 = 61.9$

Final Answer:

The 95% confidence interval for the mean attention span is:

$(34.1 \text{ minutes}, 61.9 \text{ minutes})$

Do you want any more details on these calculations or have any questions?

Here are 5 related questions you could explore:

1. How does the confidence level affect the width of the confidence interval?
2. What would happen to the confidence interval if we increased the sample size?
3. Why do we use the t-distribution instead of the z-distribution for this problem?
4. How would you interpret the confidence interval in a real-world context?
5. How would a higher sample standard deviation impact the confidence interval?

Tip: Increasing the sample size decreases the margin of error, leading to a narrower confidence interval, assuming the same confidence level.