What is the difference between a Z Test and a T Test?

is_919

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What is the difference between a Z Test and a T Test? I am taking Elementary Statistics online and it really is burdensome that basic concepts like this will not be explained. Anyway, I have been solving problems involving normal and discrete distributions, inferences on means and proportions, hypothesis testing, as well as estimates and samples. I am able to answer most of them but only AFTER clicking the "HELP ME SOLVE THIS" button. This online resource tells me which formula and/or table (whether T-Distribution, Chi-Distribution or Normal Distribution) to use. But I am helpless in figuring them out on my own. The problems all sort of read the same to me. THey all seem to be asking or looking for the same answers (such as test statistic, margin of error, P-value, etc.), and yet, require different methods and tables to be solved. I cannot seem to determine which table or TI-84 formula to use, especially during exam, when the questions from these different topics will be randomly arranged. Is there simple ways to figure out which is which? For example, if the problem is asking for the confidence interval,I know right away that this problem involves two tailed-test. I need clues and/or key words or information like that. Please advise. Thank you!
 
Z assumes a normal distribution.
t assumes a distribution with fatter tails. More or less, your data don't quite suggest Normal, but it's getting there.
 
As far as question and/or problem construction is concerned, is there any distinct characteristics I should look for to determine whether to use a Z test or a T test to solve a problem?
 
Most fundamentally, Sample Size. There are other considerations, but not nearly as important.
 
Also, how will I know when to use the 2-SampZInt and the 2-SampTInt on TI-84 calculator? Thanks again
 
This is a sample from my assignment:

A randomized trial tested the effectiveness of diets on adults. Among 43 subjects using Diet 1, the mean weight loss after a year was 3.1 lb with a standard deviation of 5.3 lb. Among 43 subjects using Diet 2, the mean weight loss after a year was 2.0 lb with a standard deviation of 6.7 lb. Construct a 95% confidence interval estimate of the difference between the populations means, assuming the population standard deviations are equal. Does there appear to be a difference in the effectiveness of the two diets?


Question # 1:
Why do I need to use the 2-SampTInt function here instead of the 2-SampZInt function? I started computing this problem using the latter but while my answer was almost similar to the correct answer, it was marked as incorrect because the turn out values were not exact (almost but not quite). I got the correct and EXACT answers only when I enter the given data using the 2-SampTInt function. Please advise.

Question # 2:
Since I need to use the T Distribution to get the EXACT values, what option do I need to choose when asked whether to pool or not to pool?

Question # 3:
What difference does the statement: "ASSUME THE POPULATION STANDARD DEVIATIONS ARE EQUAL" make as far as solving the problem? What if they are not equal? Please advise. Thanks again for your time
 
It's not so much what is "exact" as it is what is appropriate.

Consideration #1 - Sample Size. This is plenty big enough for most purposes to use Z, but with modern computing devices, such as the TI-84, accurate values for 't' can be used for much larger smaples. I don't have a problem using 'Z', just from the Sample Size consideration. In the old days, anything over 30ish samples HAD to be Z, since the tables for 't' didn't exist after that. Results for Z and t should be extremely similar for a sample size of 43, and you have indicated that they are. This is encouraging.

Consideration #2 - The idea of "Independent and Identically Distributed" is a ruling concept. It is often abbreviated simply "IID", it is so common. Things are simplified if these conditions exist. In this case, we are wondering if the Means are different. In other words, they may be the same. While we are doing this, we rely on the designation of DIFFERENT variances. This certainly violates IID! Lean toward 't', rather than 'Z'. Having said that, what difference does it make if we ignore the standard deviation information given and assume that they have equal variance? We nudge a little closer to IID, don't we?

I cannot address the issue of pooling.
 
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