A consumer organization wants to estimate the actual tread wear index of a brand name of tires that claims​ "graded 250​" on the sidewall of the tire. A random sample of n=25 indicates a sample mean tread wear index of 238.6 and a sample standard deviation of 27.8.

Assuming that the population of tread wear indexes is normally​ distributed, construct a 99% confidence interval estimate of the population mean tread wear index for tires produced by this manufacturer under this brand name.

Based on the interval you calculated, do you think that the consumer organization should accuse the manufacturer of producing tires that do not meet the performance information on the sidewall of the​ tire?

1)No, because a grade of 250 is not in the interval.
2)Yes, because a grade of 250 is in the interval.
3)Yes, because a grade of 250 is not in the interval.
4)No, because a grade of 250 is in the interval.

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.**

---

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.

A consumer organization wants to estimate the actual tread wear index of a brand name of tires that claims​ 'graded 250' on the sidewall of the tire. A random sample of n=25 indicates a sample mean tread wear index of 238.6 and a sample standard deviation of 27.8.

Assuming that the population of tread wear indexes is normally​ distributed, construct a 99% confidence interval estimate of the population mean tread wear index for tires produced by this manufacturer under this brand name.

Based on the interval you calculated, do you think that the consumer organization should accuse the manufacturer of producing tires that do not meet the performance information on the sidewall of the​ tire?

1) No, because a grade of 250 is not in the interval.
2) Yes, because a grade of 250 is in the interval.
3) Yes, because a grade of 250 is not in the interval.
4) No, because a grade of 250 is in the interval.

Calculate the 99% confidence interval for the tread wear index of tires with a claimed grade of 250. Learn step-by-step how to determine if the manufacturer's claim is supported, using sample data with a mean of 238.6 and a standard deviation of 27.8, based on the t-distribution for small samples.

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