A car manufacturing company is interested in finding the average fuel consumption of one of its new models Leopard F-type. Hundred test drivers are chosen and asked to drive the car in urban conditions for 8 hours. The average fuel consumption of the 100 cars is found to be 30 miles per gallon (mpg) with a standard deviation of 6.2 mpg. Find the 99% confidence interval for the average fuel consumption of Leopard F-type. If the management says that the length of the confidence interval should not be greater than 1 mpg’s, what is the minimum number of test drivers required? If it is not possible to hire new drivers, how confident would you be about a CI with length 1 mpg?
To find the 99% confidence interval for the average fuel consumption of the Leopard F-type, we can use the formula:
Confidence Interval = Sample Mean ± (Z * Standard Deviation / √n)
Where:
- Sample Mean is the average fuel consumption of the 100 cars (30 mpg in this case)
- Z is the Z-score corresponding to the desired confidence level (99% in this case)
- Standard Deviation is the standard deviation of the sample (6.2 mpg in this case)
- n is the sample size (the number of test drivers)
To find the Z-score corresponding to the 99% confidence level, we can consult a standard normal distribution table or use a statistical calculator. The Z-score for a 99% confidence level is approximately 2.576.
Now, let's substitute the values into the confidence interval formula:
Confidence Interval = 30 ± (2.576 * 6.2 / √n)
Next, we can solve for the sample size (n) by considering the management's requirement that the length of the confidence interval should not be greater than 1 mpg:
1 = 2.576 * 6.2 / √n
Solving for n, we get:
√n = 2.576 * 6.2 / 1
√n = 15.9576
Squaring both sides of the equation, we get:
n = 254.904736
Since the number of test drivers must be a whole number, we can round up to the nearest whole number to meet the management's requirement. So, the minimum number of test drivers required is 255.
If it is not possible to hire new drivers, we can see how confident we would be about a confidence interval with a length of 1 mpg. In this case, we need to find the corresponding confidence level for a given length.
1 = 2.576 * 6.2 / √n
Solving for √n:
√n = 2.576 * 6.2 / 1
√n = 15.9576
Squaring both sides again:
n = 254.904736
For a sample size of 254, we can calculate the confidence level as follows:
Confidence Level = 1 - (1.96 * 6.2 / √254)
Using this formula, we find that the confidence level is approximately 87.61%.
Therefore, if the confidence interval has a length of 1 mpg with the given sample size of 254, we can be about 87.61% confident about it.