Question

A tyre manufacturer found that the sample mean tread life of 16 radial tyres tested was 60000 km with standard deviation 12000 km. Construct a 90% confidence interval estimate for the true mean tread life of all radial tyres manufactured. Assume that the population tyres manufactured is normal distributed.
Give the upper limit.

Answer 1

To construct a 90% confidence interval for the true mean tread life of all radial tyres, we can use the formula for the confidence interval which is given by:

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

Where:

- \(\bar{x}\) = sample mean
- \(s\) = sample standard deviation
- \(n\) = sample size
- \(z_{\alpha/2}\) = z-value that corresponds to the desired confidence level

### Given:
- Sample mean (\(\bar{x}\)) = 60000 km
- Sample standard deviation (\(s\)) = 12000 km
- Sample size (\(n\)) = 16

### Step 1: Find \(z_{\alpha/2}\) for a 90% confidence level
For a 90% confidence interval, we have:

\[
\alpha = 1 - 0.90 = 0.10
\]
\[
\alpha/2 = 0.05
\]

From the z-table, the z-value that corresponds to 0.05 in the upper tail is approximately:

\[
z_{\alpha/2} \approx 1.645
\]

### Step 2: Calculate the standard error (SE)
\[
SE = \frac{s}{\sqrt{n}} = \frac{12000}{\sqrt{16}} = \frac{12000}{4} = 3000
\]

### Step 3: Construct the confidence interval
Now we can calculate the confidence interval:

\[
\text{Margin of error} = z_{\alpha/2} \times SE = 1.645 \times 3000 = 4935
\]

Therefore, the confidence interval is:

\[
\text{CI} = \bar{x} \pm \text{Margin of error} = 60000 \pm 4935
\]

### Step 4: Calculate the upper limit
Upper limit = \(60000 + 4935 = 64935\)

Thus, the **upper limit** of the 90% confidence interval estimate for the true mean tread life of all radial tyres manufactured is:

\[
\boxed{64935} \text{ km}
\]