Question
5. A tyre manufacturer found that the sample mean tread life of 49 radial tyres tested was 52345 km with standard deviation 12943 km. Construct a 99% confidence interval estimate for the true mean tread life of all radial tyres manufactured.
5.1 Give the critical value.
5.2 Give the upper limit.
5.3 Give the lower limit.
5.1 Give the critical value.
5.2 Give the upper limit.
5.3 Give the lower limit.
Answers
GPT-4o mini
To construct a 99% confidence interval for the true mean tread life of all radial tyres, we can use the following formula for the confidence interval:
\[
\text{Confidence Interval} = \bar{x} \pm z^* \left( \frac{s}{\sqrt{n}} \right)
\]
Where:
- \(\bar{x}\) = sample mean
- \(s\) = sample standard deviation
- \(n\) = sample size
- \(z^*\) = critical value from the standard normal distribution for the specified confidence level
Given:
- Sample mean (\(\bar{x}\)) = 52345 km
- Sample standard deviation (\(s\)) = 12943 km
- Sample size (\(n\)) = 49
### Step 1: Find the critical value (\(z^*\)) for a 99% confidence level
For a 99% confidence level, the critical value can be found using the Z-table or standard normal distribution. The critical value for a 99% confidence interval is approximately 2.576.
### Step 2: Calculate the standard error of the mean (SEM)
\[
\text{SEM} = \frac{s}{\sqrt{n}} = \frac{12943}{\sqrt{49}} = \frac{12943}{7} \approx 1849
\]
### Step 3: Construct the confidence interval
Now, we can compute the margin of error (ME):
\[
\text{ME} = z^* \times \text{SEM} = 2.576 \times 1849 \approx 4755.424
\]
### Step 4: Calculate the upper and lower limits of the confidence interval
- **Lower Limit**:
\[
\bar{x} - \text{ME} = 52345 - 4755.424 \approx 47589.576
\]
- **Upper Limit**:
\[
\bar{x} + \text{ME} = 52345 + 4755.424 \approx 57099.424
\]
### Final Answers:
5.1 **Critical value**:
- \(z^* \approx 2.576\)
5.2 **Upper limit**:
- Upper limit \( \approx 57099.424 \, \text{km}\)
5.3 **Lower limit**:
- Lower limit \( \approx 47589.576 \, \text{km}\)
So, the 99% confidence interval for the true mean tread life of all radial tyres is approximately (47589.576 km, 57099.424 km).
\[
\text{Confidence Interval} = \bar{x} \pm z^* \left( \frac{s}{\sqrt{n}} \right)
\]
Where:
- \(\bar{x}\) = sample mean
- \(s\) = sample standard deviation
- \(n\) = sample size
- \(z^*\) = critical value from the standard normal distribution for the specified confidence level
Given:
- Sample mean (\(\bar{x}\)) = 52345 km
- Sample standard deviation (\(s\)) = 12943 km
- Sample size (\(n\)) = 49
### Step 1: Find the critical value (\(z^*\)) for a 99% confidence level
For a 99% confidence level, the critical value can be found using the Z-table or standard normal distribution. The critical value for a 99% confidence interval is approximately 2.576.
### Step 2: Calculate the standard error of the mean (SEM)
\[
\text{SEM} = \frac{s}{\sqrt{n}} = \frac{12943}{\sqrt{49}} = \frac{12943}{7} \approx 1849
\]
### Step 3: Construct the confidence interval
Now, we can compute the margin of error (ME):
\[
\text{ME} = z^* \times \text{SEM} = 2.576 \times 1849 \approx 4755.424
\]
### Step 4: Calculate the upper and lower limits of the confidence interval
- **Lower Limit**:
\[
\bar{x} - \text{ME} = 52345 - 4755.424 \approx 47589.576
\]
- **Upper Limit**:
\[
\bar{x} + \text{ME} = 52345 + 4755.424 \approx 57099.424
\]
### Final Answers:
5.1 **Critical value**:
- \(z^* \approx 2.576\)
5.2 **Upper limit**:
- Upper limit \( \approx 57099.424 \, \text{km}\)
5.3 **Lower limit**:
- Lower limit \( \approx 47589.576 \, \text{km}\)
So, the 99% confidence interval for the true mean tread life of all radial tyres is approximately (47589.576 km, 57099.424 km).