Question
For a sample of 9 automobiles, the mileage (in 1000s of miles) at which the original front brake pads were worn to 10% of their original thickness was measured, as was the mileage at which the original rear brake pads were worn to 10% of their original thickness. The results were as follows:
Car 1: Rear: 41.3 , Front: 32.0
Car 2: 36.2 / 27.0
3: 47.1/35.9
4: 46.9/37.3
5: 39.4/30.0
6: 52.2/40.3
7: 52.0/41.0
8: 47.0/34.6
9: 47.4/36.3
Construct a 99.5% confidence interval for the difference in mean lifetime between the front and rear brake pads. Let d represent the mileage of the rear pads minus the mileage of the front ones. Round the answers to two decimal places.
Car 1: Rear: 41.3 , Front: 32.0
Car 2: 36.2 / 27.0
3: 47.1/35.9
4: 46.9/37.3
5: 39.4/30.0
6: 52.2/40.3
7: 52.0/41.0
8: 47.0/34.6
9: 47.4/36.3
Construct a 99.5% confidence interval for the difference in mean lifetime between the front and rear brake pads. Let d represent the mileage of the rear pads minus the mileage of the front ones. Round the answers to two decimal places.
Answers
GPT-4o mini
To construct the 99.5% confidence interval for the difference in mean lifetime between the front and rear brake pads, we will follow these steps:
1. **Calculate the differences**: We'll find the difference \(d\) (rear mileage - front mileage) for each car.
2. **Determine the sample mean \( \bar{d} \)** and sample standard deviation \( s_d \) of the differences.
3. **Find the critical t-value** for a 99.5% confidence level with \( n-1 \) degrees of freedom.
4. **Calculate the margin of error**.
5. **Construct the confidence interval**.
### Step 1: Calculate the differences
| Car | Rear Mileage (d) | Front Mileage (f) | d - f |
|-----|------------------|------------------|---------|
| 1 | 41.3 | 32.0 | 9.3 |
| 2 | 36.2 | 27.0 | 9.2 |
| 3 | 47.1 | 35.9 | 11.2 |
| 4 | 46.9 | 37.3 | 9.6 |
| 5 | 39.4 | 30.0 | 9.4 |
| 6 | 52.2 | 40.3 | 11.9 |
| 7 | 52.0 | 41.0 | 11.0 |
| 8 | 47.0 | 34.6 | 12.4 |
| 9 | 47.4 | 36.3 | 11.1 |
#### Differences:
- \( d_1 = 9.3 \)
- \( d_2 = 9.2 \)
- \( d_3 = 11.2 \)
- \( d_4 = 9.6 \)
- \( d_5 = 9.4 \)
- \( d_6 = 11.9 \)
- \( d_7 = 11.0 \)
- \( d_8 = 12.4 \)
- \( d_9 = 11.1 \)
### Step 2: Calculate the mean and standard deviation of \( d \)
1. **Mean \( \bar{d} \)**:
\[
\bar{d} = \frac{9.3 + 9.2 + 11.2 + 9.6 + 9.4 + 11.9 + 11.0 + 12.4 + 11.1}{9} = \frac{104.1}{9} \approx 11.57
\]
2. **Sample Standard Deviation \( s_d \)**:
First, calculate the squared differences:
\[
s_d = \sqrt{\frac{\sum (d_i - \bar{d})^2}{n-1}}
\]
Calculating the squared differences:
- \( (9.3 - 11.57)^2 \approx 5.3 \)
- \( (9.2 - 11.57)^2 \approx 5.63 \)
- \( (11.2 - 11.57)^2 \approx 0.13 \)
- \( (9.6 - 11.57)^2 \approx 3.82 \)
- \( (9.4 - 11.57)^2 \approx 4.68 \)
- \( (11.9 - 11.57)^2 \approx 0.11 \)
- \( (11.0 - 11.57)^2 \approx 0.32 \)
- \( (12.4 - 11.57)^2 \approx 0.73 \)
- \( (11.1 - 11.57)^2 \approx 0.22 \)
Now sum these:
\[
\sum = 5.3 + 5.63 + 0.13 + 3.82 + 4.68 + 0.11 + 0.32 + 0.73 + 0.22 = 21.79
\]
Now, plug into the formula for \( s_d \):
\[
s_d = \sqrt{\frac{21.79}{8}} \approx \sqrt{2.7225} \approx 1.65
\]
### Step 3: Find the critical t-value
Using a t-table for \( \alpha = 0.005 \) and \( df = 8 \):
The critical t-value \( t_{0.005} \) for \( df = 8 \) is approximately 3.355.
### Step 4: Calculate the margin of error
\[
\text{Margin of Error} = t \cdot \frac{s_d}{\sqrt{n}} = 3.355 \cdot \frac{1.65}{\sqrt{9}} = 3.355 \cdot \frac{1.65}{3} \approx 1.85
\]
### Step 5: Construct the confidence interval
\[
\text{Confidence interval} = \left( \bar{d} - \text{Margin of Error}, \bar{d} + \text{Margin of Error} \right) = (11.57 - 1.85, 11.57 + 1.85) = (9.72, 13.42)
\]
### Final Answer:
Thus, the 99.5% confidence interval for the difference in mean lifetime between the front and rear brake pads is:
\[
\boxed{(9.72, 13.42)}
\]
1. **Calculate the differences**: We'll find the difference \(d\) (rear mileage - front mileage) for each car.
2. **Determine the sample mean \( \bar{d} \)** and sample standard deviation \( s_d \) of the differences.
3. **Find the critical t-value** for a 99.5% confidence level with \( n-1 \) degrees of freedom.
4. **Calculate the margin of error**.
5. **Construct the confidence interval**.
### Step 1: Calculate the differences
| Car | Rear Mileage (d) | Front Mileage (f) | d - f |
|-----|------------------|------------------|---------|
| 1 | 41.3 | 32.0 | 9.3 |
| 2 | 36.2 | 27.0 | 9.2 |
| 3 | 47.1 | 35.9 | 11.2 |
| 4 | 46.9 | 37.3 | 9.6 |
| 5 | 39.4 | 30.0 | 9.4 |
| 6 | 52.2 | 40.3 | 11.9 |
| 7 | 52.0 | 41.0 | 11.0 |
| 8 | 47.0 | 34.6 | 12.4 |
| 9 | 47.4 | 36.3 | 11.1 |
#### Differences:
- \( d_1 = 9.3 \)
- \( d_2 = 9.2 \)
- \( d_3 = 11.2 \)
- \( d_4 = 9.6 \)
- \( d_5 = 9.4 \)
- \( d_6 = 11.9 \)
- \( d_7 = 11.0 \)
- \( d_8 = 12.4 \)
- \( d_9 = 11.1 \)
### Step 2: Calculate the mean and standard deviation of \( d \)
1. **Mean \( \bar{d} \)**:
\[
\bar{d} = \frac{9.3 + 9.2 + 11.2 + 9.6 + 9.4 + 11.9 + 11.0 + 12.4 + 11.1}{9} = \frac{104.1}{9} \approx 11.57
\]
2. **Sample Standard Deviation \( s_d \)**:
First, calculate the squared differences:
\[
s_d = \sqrt{\frac{\sum (d_i - \bar{d})^2}{n-1}}
\]
Calculating the squared differences:
- \( (9.3 - 11.57)^2 \approx 5.3 \)
- \( (9.2 - 11.57)^2 \approx 5.63 \)
- \( (11.2 - 11.57)^2 \approx 0.13 \)
- \( (9.6 - 11.57)^2 \approx 3.82 \)
- \( (9.4 - 11.57)^2 \approx 4.68 \)
- \( (11.9 - 11.57)^2 \approx 0.11 \)
- \( (11.0 - 11.57)^2 \approx 0.32 \)
- \( (12.4 - 11.57)^2 \approx 0.73 \)
- \( (11.1 - 11.57)^2 \approx 0.22 \)
Now sum these:
\[
\sum = 5.3 + 5.63 + 0.13 + 3.82 + 4.68 + 0.11 + 0.32 + 0.73 + 0.22 = 21.79
\]
Now, plug into the formula for \( s_d \):
\[
s_d = \sqrt{\frac{21.79}{8}} \approx \sqrt{2.7225} \approx 1.65
\]
### Step 3: Find the critical t-value
Using a t-table for \( \alpha = 0.005 \) and \( df = 8 \):
The critical t-value \( t_{0.005} \) for \( df = 8 \) is approximately 3.355.
### Step 4: Calculate the margin of error
\[
\text{Margin of Error} = t \cdot \frac{s_d}{\sqrt{n}} = 3.355 \cdot \frac{1.65}{\sqrt{9}} = 3.355 \cdot \frac{1.65}{3} \approx 1.85
\]
### Step 5: Construct the confidence interval
\[
\text{Confidence interval} = \left( \bar{d} - \text{Margin of Error}, \bar{d} + \text{Margin of Error} \right) = (11.57 - 1.85, 11.57 + 1.85) = (9.72, 13.42)
\]
### Final Answer:
Thus, the 99.5% confidence interval for the difference in mean lifetime between the front and rear brake pads is:
\[
\boxed{(9.72, 13.42)}
\]
(b) An automotive engineer claims that the mean lifetime for rear brake pads is more than 10000 miles more than the mean lifetime for front brake pads. Does the confidence interval contradict this claim?
GPT-4o mini
To determine whether the confidence interval derived in part (a) contradicts the engineer's claim that the mean lifetime for rear brake pads is more than 10,000 miles (which is 10.0 in units of 1000s of miles) more than the mean lifetime for front brake pads, we need to analyze the confidence interval.
### Confidence Interval Recap:
The 99.5% confidence interval for the difference in mean lifetime (rear - front) we calculated was:
\[
(9.72, 13.42)
\]
### Evaluating the Engineer's Claim:
The engineer claims that the difference in mean lifetime is greater than 10.0 (which corresponds to more than 10,000 miles). This means we want to evaluate whether 10.0 falls within our confidence interval.
1. **Interval Analysis**:
- Lower bound of the interval: **9.72**
- Upper bound of the interval: **13.42**
### Conclusion:
- The lower bound of the confidence interval, 9.72, is less than 10.0.
- The upper bound of the confidence interval, 13.42, is greater than 10.0.
Since the interval (9.72, 13.42) includes values below and above 10.0, it suggests that we cannot definitively ascertain that the difference is greater than 10.0 based on this confidence interval.
### Final Answer:
The confidence interval **does contradict** the engineer's claim because it includes some values less than 10.0 (or 10,000 miles). Thus, we cannot conclude with confidence that the mean lifetime for rear brake pads is more than 10,000 miles greater than that for front brake pads.
### Confidence Interval Recap:
The 99.5% confidence interval for the difference in mean lifetime (rear - front) we calculated was:
\[
(9.72, 13.42)
\]
### Evaluating the Engineer's Claim:
The engineer claims that the difference in mean lifetime is greater than 10.0 (which corresponds to more than 10,000 miles). This means we want to evaluate whether 10.0 falls within our confidence interval.
1. **Interval Analysis**:
- Lower bound of the interval: **9.72**
- Upper bound of the interval: **13.42**
### Conclusion:
- The lower bound of the confidence interval, 9.72, is less than 10.0.
- The upper bound of the confidence interval, 13.42, is greater than 10.0.
Since the interval (9.72, 13.42) includes values below and above 10.0, it suggests that we cannot definitively ascertain that the difference is greater than 10.0 based on this confidence interval.
### Final Answer:
The confidence interval **does contradict** the engineer's claim because it includes some values less than 10.0 (or 10,000 miles). Thus, we cannot conclude with confidence that the mean lifetime for rear brake pads is more than 10,000 miles greater than that for front brake pads.