To calculate a 98% confidence interval for the mean time workers take to get to work, we can follow these steps:

Provided Data:

• Sample mean ($\bar{x}$) = 35.06 minutes
• Sample standard deviation ($s$) = 6 minutes
• Sample size ($n$) = 15 workers
• Confidence level = 98%

Step 1: Determine the critical value ($t^*$)

Since the sample size is small ($n = 15$), we use the Student's t-distribution. The degrees of freedom ($df$) are: $df = n - 1 = 15 - 1 = 14$

For a 98% confidence level, the area in both tails of the distribution is $1 - 0.98 = 0.02$, meaning $0.01$ in each tail. Using $t$-distribution tables or a calculator, the critical value $t^*$ for $df = 14$ at 98% confidence is approximately: $t^* \approx 2.624$

Step 2: Compute the standard error (SE)

The standard error of the mean is calculated as: $SE = \frac{s}{\sqrt{n}} = \frac{6}{\sqrt{15}} \approx 1.549$

Step 3: Calculate the margin of error (ME)

The margin of error is given by: $ME = t^* \cdot SE = 2.624 \cdot 1.549 \approx 4.065$

Step 4: Determine the confidence interval

The confidence interval is calculated as: $\text{Confidence Interval} = \bar{x} \pm ME$ $\text{Confidence Interval} = 35.06 \pm 4.065$

Final Result:

$\text{Confidence Interval} = [30.995, 39.125]$

Thus, the 98% confidence interval for the mean time is approximately [30.995, 39.125] minutes.

Would you like me to elaborate on any part of this calculation?

5 Related Questions:

1. What is the importance of using the Student's t-distribution in small samples?
2. How does the confidence level affect the width of the confidence interval?
3. What happens to the margin of error if the sample size increases?
4. Why do we use a critical $t^*$ value instead of $z^*$ here?
5. How can this confidence interval help the company make decisions about worker commuting?

Tip:

To reduce the width of a confidence interval, you can increase the sample size or reduce variability in your data.