I am using a T-test to compare boys and girls on the number of total absences from school. I am able to assume equal variances according to the assignment. N=10 for girls and N=10 for boys. I believe this would be a two-tailed test. I have the calculation formulas, but I am not sure which one to use. Would I need to use an independent or dependent T-test equation formula? As usual, the book I have explains this in terms as if I was a professor of math. Thanks for your help!
What kind of T-test to use for statistics class?
small sample test for the difference in means
sample X and Y from normal data sets with the sample size of X being n and the sample size of Y is m
the test statistic is a student t
t = (xbar - ybar) / sqrt(var(X) / n + var(y) / m)
the degrees of freedom are:
((var(x) / n + var(y) / m)^2) / ((var(x) / n)^2 / (n - 1) + (var(y) / m)^2 / (m - 1))
this nasty looking equation is very simple if m = n. the degrees of freedom in that case are m + n - 2.
The p-value of the test is the area under the normal curve that is in agreement with the alternate hypothesis.
H1: d %26gt; 0; p-value is the area to the right of t
H1: d %26lt; 0; p-value is the area to the left of t
H1: d ≠ 0; p-value is the area in the tails greater than |t|
where d is the difference in the means, d = μx - μy.
Consider the hypothesis as a trial against the null hypothesis. the data is evidence against the mean. you assume the mean is true and try to prove that it is not true. After finding the test statistic and p-value, if the p-value is less than or equal to the significance level of the test we reject the null and conclude the alternate hypothesis is true. If the p-value is greater than the significance level then we fail to reject the null hypothesis and conclude it is plausible. Note that we cannot conclude the null hypothesis is true, just that it is plausible.
If the question statement asks you to determine if there is a difference between the statistic and a value, then you have a two tail test, the null hypothesis, for example, would be μ = d vs the alternate hypothesis μ ≠ d
if the question ask to test for an inequality you make sure that your results will be worth while. for example. say you have a steel bar that will be used in a construction project. if the bar can support a load of 100,000 psi then you'll use the bar, if it cannot then you will not use the bar.
if the null was μ ≥ 100,000 vs the alternate μ %26lt; 100,000 then will will have a meaningless test. in this case if you reject the null hypothesis you will conclude that the alternate hypothesis is true and the mean load the bar can support is less than 100,000 psi and you will not be able to use the bar. However, if you fail to reject the null then you will conclude it is plausible the mean is greater than or equal to 100,000. You cannot ever conclude that the null is true. as a result you should not use the bar because you do not have proof that the mean strength is high enough.
if the null was μ ≤ 100,000 vs. the alternate μ %26gt; 100,000 and you reject the null then you conclude the alternate is true and the bar is strong enough; if you fail to reject it is plausible the bar is not strong enough, so you don't use it. in this case you have a meaningful result.
Any time you are defining the hypothesis test you need to consider whether or not the results will be meaningful.
Reply:In order to use a t-test, you need to assume Gaussian distribution, i.e. a bell shaped curve if you plot each data point on a graph. If you don't have a Gaussian distribution you need to use a nonparametric test such as a Mann-Whitney. A two-tailed test is appropriate if you do not have a preconceived expectation about which way the data will go. An independent test would be used if the two variables (boys versus girls) are unrelated and one does not affect the other. If boys and girls were always skipping school together and one required the other, then a dependent test would be appropriate. Stats books suck and are virtually worthless for normal people.
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