Asked by kim

Ala rge fleet of cars is maintained with an average of 26 mpg. 50 recent trips displayed a mean of 25.02 mpg and standard deviation of 4.1 mpg. At the 5% level of significance, is their evidence the company has failed to keep their fuel goals? I did the math and got .03380315... but when I put it in calc I got z = 1.69015... It does not compute correctly! I know I do not need p value and .o5 level of sig. = +/- 1.96...but am at a loss how to put this together!
Answered by PsyDAG
Z = (mean1 - mean2)/standard error (SE) of difference between means

SEdiff = √(SEmean1^2 + SEmean2^2)

SEm = SD/√n (n = # of trips)

If only one SD is provided, you can use just that to determine SEdiff.

Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to the Z score.

However, from the question asked, don't you want a one-tailed test?
Answered by kim
what test are you using? This is very confusing!
Answered by PsyDAG
It is the Z test (google it). Where is your confusion centered?
Answered by kim
whether it is one-tailed or two tailed?
and if one tailed...I do agree with you now that this is the left tail test. However I am not sure how to handle negative value on t. Here is what I did.

I used formula for hypothesis test with mu unknow, t=(xbar-mu)/(sd/sqr n)=(26-25.02)/(4.1/sqr 50)=1.69015..

I am not sure what to do next?
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