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Part 1

A consumer organization wants to estimate the actual tread wear index of a brand name of tires that claims​ "graded 

250250​" 

on the sidewall of the tire. A random sample of 

n equals 22n=22 

indicates a sample mean tread wear index of 

240.1240.1 

and a sample standard deviation of 

15.515.5. 

Complete parts​ (a) through​ (c).

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Part 1

a.a.

Assuming that the population of tread wear indexes is normally​ distributed, construct a 

90 %90% 

confidence interval estimate of the population mean tread wear index for tires produced by this manufacturer under this brand name.

enter your response hereless than or equals≤muμless than or equals≤enter your response here 

​(Round to two decimal places as​ needed.)

We are asked to construct a 90% confidence interval estimate for the population mean tread wear index, based on a sample of 22 tires. The formula for constructing a confidence interval for the population mean when the population standard deviation is unknown (and the sample size is less than 30) is:

\[
CI = \bar{x} \pm t_{\frac{\alpha}{2}, n-1} \left( \frac{s}{\sqrt{n}} \right)
\]

Where:
- \(\bar{x}\) = sample mean = 240.1
- \(s\) = sample standard deviation = 15.5
- \(n\) = sample size = 22
- \(t_{\frac{\alpha}{2}, n-1}\) = critical t-value for a 90% confidence interval and 21 degrees of freedom (\(n-1\))

### Step 1: Find the critical t-value
For a 90% confidence interval with 21 degrees of freedom, we need to find the t-value for \( \alpha/2 = 0.05 \). Using a t-distribution table or calculator, the critical value \( t_{0.05, 21} \) is approximately 1.721.

### Step 2: Calculate the margin of error (ME)
The margin of error is given by:

\[
ME = t_{\frac{\alpha}{2}, n-1} \left( \frac{s}{\sqrt{n}} \right)
\]

\[
ME = 1.721 \times \left( \frac{15.5}{\sqrt{22}} \right)
\]

Let's calculate the margin of error.The margin of error (ME) is approximately 5.69.

### Step 3: Calculate the confidence interval
Now, we can calculate the confidence interval as:

\[
CI = \bar{x} \pm ME
\]

\[
CI = 240.1 \pm 5.69
\]

Thus, the confidence interval is:

\[
(240.1 - 5.69, 240.1 + 5.69) = (234.41, 245.79)
\]

### Conclusion:
The 90% confidence interval for the population mean tread wear index is:

\[
234.41 \leq \mu \leq 245.79
\]

Would you like further details or have any other questions?

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Here are 5 related questions:
1. How does the confidence level affect the width of the confidence interval?
2. What is the significance of the sample size in confidence interval estimation?
3. How would the interval change if the sample standard deviation was larger?
4. What happens to the confidence interval if we increase the sample size?
5. How do we interpret a confidence interval in the context of this tire tread wear index study?

**Tip**: A larger sample size reduces the margin of error, leading to a narrower confidence interval, which provides a more precise estimate.

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=22 indicates a sample mean tread wear index of 240.1 and a sample standard deviation of 15.5. Assuming that the population of tread wear indexes is normally distributed, construct a 90% confidence interval estimate of the population mean tread wear index for tires produced by this manufacturer under this brand name.

Learn how to calculate a 90% confidence interval for the population mean tread wear index based on a sample of 22 tires, with a sample mean of 240.1 and a standard deviation of 15.5. This step-by-step solution uses t-distribution for estimating the confidence interval when the population standard deviation is unknown.

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