1. Gas Mileage Problem
The EPA efficiency rating Ina particular model of new car says that the car is expected to get 22.7 miles per gallon (mpg) of gasoline. Assume that the standard deviation is 1.4 mpg and that the actual mileage are normally distributed about 22.7. The factory produces 2000 of this model.
a. How many of the cars would be expected to get above 23 mpg?
My answer: 830 cars
b. How many would be expected to get below 20 mpg?
My answer: 54 cars
c. How many would be expected to get mileages in the interval from 21 to 24 mpg?
My answer: 1422 cars
d. In what interval about 22.7 could you expect 80% of the mileage to be?
My answer: between 20.9 and 24.5
e. If you purchase one of the cars at random, what is the probability that it will get at least 21.5 mpg?
My answer: 80.4%
f. Suppose you purchased a car of this model and it got only 17.2 mpg. If dealer's service manager told you that this low a mileage was normal for this model car, how would you refute his claim based on statistic? I am stunt on this!
(a)-(d) correct
(e) 17.2 is 3.93 std below the mean
There's almost zero chance of any car getting mileage that low. Most Z tables stop at 3.4 std below the mean, which is 0.03%
Run far, run fast from that dealer!!!
To answer question f, you would need to use statistical reasoning to refute the service manager's claim. Here's how you can approach it:
1. Define the null and alternative hypotheses:
- Null hypothesis (H0): The average mileage of this model car is 22.7 mpg or higher.
- Alternative hypothesis (H1): The average mileage of this model car is lower than 22.7 mpg.
2. Calculate the z-score for the observed mileage of 17.2 mpg:
The z-score is a measure of how many standard deviations an observation is from the mean. The formula to calculate the z-score is:
z = (X - μ) / σ
Where:
X = observed mileage (17.2 mpg)
μ = mean mileage (22.7 mpg)
σ = standard deviation (1.4 mpg)
Plugging in the values, we get:
z = (17.2 - 22.7) / 1.4 ≈ -3.93
3. Determine the critical value for the desired level of significance (α):
The level of significance, denoted as α, determines the threshold for rejecting the null hypothesis. Commonly used values for α are 0.05 or 0.01. Let's assume α = 0.05.
4. Look up the critical value in a standard normal distribution table:
For α = 0.05, the critical value is approximately -1.645. Since this is a one-tailed test (looking for evidence of the average being lower), we only need to consider the left tail.
5. Compare the z-score to the critical value:
If the z-score is less than the critical value, we reject the null hypothesis. In this case, since the z-score (-3.93) is less than the critical value (-1.645), we can reject the null hypothesis.
6. Interpretation:
Based on the statistical analysis, we have evidence to refute the service manager's claim that a mileage as low as 17.2 mpg is normal for this car model. The data suggests that the average mileage is higher than 17.2 mpg.
Remember, this analysis assumes that the population follows a normal distribution, the sample size is large enough, and the data is representative of the population.