SOLUTION: The standard deviation of fuel consumption of a manufacturer's sport utility vehicle is hypothesized to be 3.3 miles per gallon. A random sample of 18 vehicles has a standard dev

Algebra ->  Probability-and-statistics -> SOLUTION: The standard deviation of fuel consumption of a manufacturer's sport utility vehicle is hypothesized to be 3.3 miles per gallon. A random sample of 18 vehicles has a standard dev      Log On


   

Question 1207161: The standard deviation of fuel consumption of a manufacturer's sport utility vehicle is
hypothesized to be 3.3 miles per gallon. A random sample of 18 vehicles has a standard
deviation of 2.8 miles per gallon. At ᵯ =0.10, is the claim believable? You must show all 5
steps of a hypothesis test for full credit
Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
the standard deviation is hypothesized to be 3.3 miles per gallon.

a sample of 18 vehicles has a standard deviation of 2.8 miles per gallon.

at two tail alpha = .10, the critical t-score, with degrees of freedom of 17, will be plus or minus 1.7396, rounded to 4 decimal places.

the test t-score formula is t = (x-m)/s
t is the test t-score
x is the test raw score
m is the mean
s is the standard error.

the test t-score becomes t = (2.8 - 3.3) / s

s = standard deviation / sqrt(sample size) = 2.8 / sqrt(18) = .6600, rounded to 4 decimal places.

the standard error is used when you are looking at a distribution of sample means.

apparently, it can also be used when you are looking at a distribution of sample standard deviations.

this is because each sample has a unique mean and a unique standard deviation, allowing each to be tested in the same say.

that's what i think, anyway.

test t-score becomes (2.8 - 3.3) / .66 = -.7576 rounded to 4 decimal places.

that is well within the critical t-score limits of plus or minus 1.7396.

this means the probability is pretty high that the 2.8 standard deviation is quite possibly due to random variations in the standard deviation of multiple samples of 18 elements.

the conclusion is that there is not sufficient evidence that the standard deviation is other than 3.3.

here's what it looks like on a t-score graphing calculator.



the graph showos that the test t-score is between the two critical t-scores.
the green is within the confidence interval.
the red is outside the confidence interval.