To address this problem, we'll first calculate the standard deviation and the standard error of the sample, then use this to estimate the confidence interval around the mean. From there, we'll determine how many tires need to be tested to achieve a margin of error of 200 miles with a desired level of confidence (usually 95%).

Step 1: Convert data into miles

The data in thousands of miles is: 54.5, 55.4, 56.9, 54.7, 55.9, 55.6, 53.2, 54.0, 52.6, 53.4. Converted to miles: 54,500, 55,400, 56,900, 54,700, 55,900, 55,600, 53,200, 54,000, 52,600, 53,400.

Step 2: Calculate the sample mean

The sample mean is given as 54,510 miles.

Step 3: Calculate the standard deviation

Using the formula for standard deviation for a sample: $s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n - 1}}$

I'll compute this now.import numpy as np

Data in miles

data = np.array([54500, 55400, 56900, 54700, 55900, 55600, 53200, 54000, 52600, 53400])

Sample mean

mean = 54510

Standard deviation (sample)

std_dev = np.std(data, ddof=1)

std_dev1349.73248378253741349.7324837825374successThe standard deviation of the sample is approximately 1,349.73 miles.

Step 4: Calculate the standard error

The standard error of the mean (SE) is calculated using: $SE = \frac{s}{\sqrt{n}}$ where $s$ is the standard deviation and $n$ is the sample size. Let me calculate this next.# Sample size n = len(data)

Standard error

std_error = std_dev / np.sqrt(n)

std_error426.82288806690974426.82288806690974successThe standard error of the mean is approximately 426.82 miles.

Step 5: Confidence interval and margin of error

To determine how close the sample mean is to the population mean, we can calculate a 95% confidence interval: $\text{Margin of Error} = z \cdot SE$ where $z$ for a 95% confidence level is 1.96. Let me compute the margin of error and the confidence interval.# Z-value for 95% confidence level z = 1.96

Margin of error

margin_of_error = z * std_error

Confidence interval

lower_bound = mean - margin_of_error upper_bound = mean + margin_of_error

margin_of_error, lower_bound, upper_bound(836.572860611143, 53673.427139388856, 55346.572860611144)(836.572860611143, 53673.427139388856, 55346.572860611144)successThe margin of error for the sample mean at a 95% confidence level is approximately 836.57 miles. The confidence interval for the population mean is [53,673.43, 55,346.57 miles].

Step 6: Sample size needed for a 200-mile margin of error

To calculate the required sample size ($n$) for a margin of error ($E$) of 200 miles: $n = \left(\frac{z \cdot s}{E}\right)^2$ Let me calculate this.# Desired margin of error desired_margin_of_error = 200

Required sample size

required_sample_size = (z * std_dev / desired_margin_of_error) ** 2

required_sample_size174.96353777777773174.96353777777773successTo achieve a margin of error of approximately 200 miles with 95% confidence, you would need to test 175 tires.

Recap:

1. The sample mean is accurate within a margin of 836.57 miles for the 95% confidence interval.
2. To achieve a smaller margin of error (200 miles), a larger sample size of 175 tires is required.

Would you like further details on these calculations? Here are 5 related questions to delve into:

1. How does increasing the sample size reduce the margin of error?
2. What happens if a different confidence level (e.g., 99%) is used?
3. Can we calculate the exact population mean from this data?
4. How does the standard deviation impact required sample size?
5. What assumptions are we making about the tire lifetimes?

Tip: Larger sample sizes improve accuracy but may require significant resources; balance precision with feasibility.