Gamma corporation wants to estimate the mean difference in fuel consumption with a margin of error of estimate of 2 mpg with a 90% confidence level. assume that the differences in fuel consumption are normally distributed and that previous studies suggest that an estimate of sd=3sd=3 mpg is reasonable. how many cars should be tested?

This is not the whole story, but may help you.

Basically, you need to divide by sqrt(n), n = number of cars.

For a 90% confidence interval, this means the error is 1.645 sigma (the 1.645 is probably the part of find by error and trial). Sigma is the standard deviation in your problem.

So we basically want 2 mpg to be 1.645 * sigma, so sigma is 2/1.645 = 1.216

Now the initial standard deviation for one car is 3 mpg, and if you measure it in 'n; cars, it will be

3/sqrt(n), so 3/sqrt(n) = 1.216, n = ( 3/1.216)^2 = 6.08 cars

So, the answer may be *6* cars

Hope it helps

so sigma

Answer: c. x y2 04 66 16