Is there a difference (either way) in the average miles traveled for each of two taxi companies during a randomly selected week? Assume that the populations are normally distributed.
α = 0.05
x-bar1 = 817
σ1 = 30
n1 = 35
x-bar2 = 743
σ2 = 40
n2 = 40
H0: µ1 = µ2
H1: µ1 does not equal µ2
For difference of means where σ1 and σ2 are known.
z= (817-743)/√((30^2/35)+(40^2/40)) = 9.13
How do I get the P value for this. 9.13 is not on the chart. It doesn’t even look right…
You are not using the normal distribution here. You are using the Student-t distribution.
Knowing the behavior of these distributions will help you to see where things can go wrong. If you were meant to use a z-score, then getting a 9 should tell you something is terribly wrong as that is a score assigned to a value 9 standard deviations away from the mean - which I suppose ins't "impossible", but for practical purposes it really is.
You are using the T-distribution, and using your data I got a P-value that was essentially zero. I would recheck your numbers just to be sure you have reported them here correctly as a value of 9 is (for the t-distribution) large as well.
Thank you so much! That makes a lot of sense, but I just have one question. I have a list of formulas. It tells me to use z when σ1 and σ2 are known and t when σ1 and σ2 are unknown. They are known in this case. I suppose the reason why it is because to use the student t, you need two samples. I suppose that is tripping me up. Thank you again though!!!
Sorry, I didn't really look at your problem fully (I blame the fact that The Venture Brothers is on in the background). If the population standard deviation is known, then yes, you would use the normal distribution to calculate your standardized value (well you'd use it because they tell you to use it in the problem). As many people will tell you this rarely occurs in real life.
What an odd problem. I mean the answer is most definitely to reject the null hypothesis, but man what a wonky problem.
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