To calculate the margin of error for a 98% confidence interval, we use the formula:

$\text{Margin of Error} = z \times \frac{\sigma}{\sqrt{n}}$

where:

• $z$ is the z-score corresponding to a 98% confidence level,
• $\sigma$ is the standard deviation,
• $n$ is the sample size.

Step 1: Identify the values

• Confidence level: 98%, so $z \approx 2.33$ (using a z-score table or calculator),
• $\sigma = 4.74$,
• $n = 37$.

Step 2: Plug values into the formula

$\text{Margin of Error} = 2.33 \times \frac{4.74}{\sqrt{37}}$

Step 3: Calculate

1. Calculate $\sqrt{37} \approx 6.08$,
2. Then, $\frac{4.74}{6.08} \approx 0.78$,
3. Finally, $2.33 \times 0.78 \approx 1.82$.

Answer

The margin of error for a 98% confidence interval is approximately 1.82 mph.

Would you like further explanation on this process or any other details?

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1. What would the margin of error be for a 95% confidence interval?
2. How does the sample size affect the margin of error?
3. How does changing the standard deviation impact the margin of error?
4. What would the margin of error be with a confidence level of 99%?
5. Why do we use a z-score in this calculation?

Tip: For higher confidence levels, the margin of error increases because we are aiming to capture a broader range around the mean.