To calculate the 99% confidence interval for the population mean, we will use the formula for the confidence interval when the sample size $n$ is small and the population standard deviation is unknown. The formula is based on the t-distribution:

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

Where:

• $\bar{x} = 238.6$ (sample mean)
• $s = 27.8$ (sample standard deviation)
• $n = 25$ (sample size)
• $t_{\alpha/2}$ is the t-value corresponding to a 99% confidence level with $n-1 = 24$ degrees of freedom

Step 1: Find the critical t-value

For a 99% confidence interval and $n - 1 = 24$ degrees of freedom, the critical value from the t-distribution table is approximately:

$t_{\alpha/2} \approx 2.797$

Step 2: Compute the margin of error

The margin of error is:

$ME = t_{\alpha/2} \left( \frac{s}{\sqrt{n}} \right) = 2.797 \left( \frac{27.8}{\sqrt{25}} \right) = 2.797 \left( \frac{27.8}{5} \right) = 2.797 \times 5.56 \approx 15.54$

Step 3: Compute the confidence interval

The confidence interval is:

$CI = 238.6 \pm 15.54$

Thus, the 99% confidence interval is:

$(238.6 - 15.54, 238.6 + 15.54) = (223.06, 254.14)$

Conclusion:

Since the claimed grade of 250 falls within the confidence interval $(223.06, 254.14)$, we cannot accuse the manufacturer of producing tires that fail to meet the performance information on the sidewall. Therefore, the correct conclusion is:

4) No, because a grade of 250 is in the interval.

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Would you like further details or clarification on any part of this calculation?

Related Questions:

1. How does increasing the sample size affect the confidence interval?
2. What would the confidence interval be at a 95% confidence level?
3. Why do we use the t-distribution instead of the normal distribution in this problem?
4. What is the impact of a larger standard deviation on the confidence interval?
5. How could we reduce the margin of error in future tests?

Tip:

For smaller sample sizes (n < 30), always use the t-distribution to account for the added variability in estimating the population standard deviation.