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? 

To construct a 99% confidence interval for the population mean tread wear index, we can use the formula for the confidence interval based on the sample mean in a normally distributed population:

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

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

### Step 1: Determine \(t_{\alpha/2}\)
For a 99% confidence level with \(24\) degrees of freedom, we can find the critical value \(t_{\alpha/2}\). The critical value is approximately \(t_{0.005, 24} \approx 2.797\).

### Step 2: Calculate the Margin of Error
Now, calculate the margin of error using:
\[
\text{Margin of Error} = t_{\alpha/2} \cdot \frac{s}{\sqrt{n}}
\]
\[
= 2.797 \cdot \frac{27.8}{\sqrt{25}}
\]
\[
= 2.797 \cdot \frac{27.8}{5}
\]
\[
= 2.797 \cdot 5.56
\]
\[
\approx 15.54
\]

### Step 3: Construct the Confidence Interval
Now, construct the confidence interval by adding and subtracting the margin of error from the sample mean:
\[
\text{Confidence Interval} = 238.6 \pm 15.54
\]
\[
= (238.6 - 15.54, 238.6 + 15.54)
\]
\[
= (223.06, 254.14)
\]

### Conclusion: Should the Consumer Organization Accuse the Manufacturer?
The manufacturer's claim is that the tires have a tread wear index of 250. Since 250 is within the 99% confidence interval \((223.06, 254.14)\), there is no strong statistical evidence that the true tread wear index differs significantly from the claimed 250. Therefore, the consumer organization would likely not have grounds to accuse the manufacturer of producing tires that do not meet the claimed performance standard based on this interval.

---

Would you like additional details on any step? 

Here are some related questions to explore further:

1. How does changing the confidence level affect the width of the confidence interval?
2. How would the confidence interval change with a larger sample size?
3. What is the impact of a higher standard deviation on the confidence interval width?
4. Could a different sample mean impact whether the manufacturer’s claim is supported?
5. How would this calculation differ if we did not assume normal distribution?

**Tip:** When interpreting confidence intervals, remember that the interval gives a range in which we expect the true population parameter to lie based on our sample—it's not a certainty but rather an estimate with a given confidence level.

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?

Explore how to construct a 99% confidence interval for the population mean tread wear index of a tire brand claiming a grade of 250, using a sample mean of 238.6 and a standard deviation of 27.8. Analyze whether this interval supports the manufacturer's claim.

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