Question 297954: A company with a large fleet of cars hopes to keep gasoline costs down and sets a goal of attaining a fleet average of at least 26 miles per gallon. To see if the goal is being met, they check the gasoline usage for 50 company trips chosen at random, finding a mean of 25.02 mpg and a standard deviation of 4.83 mpg. Is this strong evidence that they have failed to attain their fuel economy goal?
a) write appropriate hypothesis =
b) Are the necessary assumptions to perform inference satisfied?
c) Test the hypothesis and find the P-value
d) Explain what the P-value means in this context
e) State an appropriate conclusion
Answer by alanc(27)   (Show Source):
You can put this solution on YOUR website! Part a.)
Start with your significance level. The standard is usually a 5% significance level.
Null Hypothesis is: H_ : u >= 26
Alternate Hypothesis: H_a : u < 26
Part b.)
I am going to assume this sample set follows a Normal Distribution. One with a normal density function.
we want to find P(u >= 26) = 100% - 5%
Standardize this normal distribution so we can look it up in the Z-scores table.
Z = (x - 26)/s , where Z is a random variable with a standard normal distribution. and s is the standard deviation. s = 4.83
For a particular case we have Z = (25.02 - 26)/4.83 = 0.202898
Part c) To test the hypothesis I found P(Z >= 0.20289) which is our P-value. We have to compare the P-value to our significance level. and since the P-value here is greater than 5%, I cannot reject the null hypothesis.
Part d) The P-value means the probability that the average of the miles per gallon variable is at least 26 mpg. In this cas the probability is 41.3%
Part e)
In the Z-scores table this corresponds to P(Z >= 0.20289) = 0.4129
Which is 41.3% > 5%, Z is not in the rejection region, so therefore I cannot reject the null hypothesis. I would have to conclude that the average mpg is 26 miles per gallon with a 95% confidence interval.
Sincerely Alan C,
CSU Math Graduate 07
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