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The t-table gives the percentage points of t-distribution. To use the table it is important to note that the t-distribution is symmetrical.
The values in the main body of the table are percentage points (we call them the critical values in hypothesis testing), and they are the values that run along the horizontal line of the t distribution curve. Note they are not probabilities, which is obviously as the values takes values above 1.
The probabilities in the upper part of the tail i.e. area to the right of the critical value p(Tdf > b) -the alphas - are in the 1st 2 rows. The degrees of freedom (df) run down the 1st column.
The 1 tail is concerned with p(T > b) = α; the 2 tail is concerned with p( T > b) = α/2.
A note on the row df = infinity: The percentage points correspond to those in the Z table.
|1 tail α =||0.1||0.05||0.025||0.01||0.005|
|2 tails α =||0.2||0.1||0.05||0.02||0.01|
1. Find b such that p(T > b) = 0.1 where T is a Student's t random variable with 5 degrees of freedom, and α is 0.1.
Run down 1st column until df = 5; now run along this row until it intersects with column of α = 0.1.
The answer is b = 1.476
2. Suppose you have calculated a one-tailed t-test. The value of the test statistic is 1.8. The degrees of freedom is 20. Is there evidence to reject the null hypothesis?
Solution.We are told this is a one-tailed test. We need to look up the critical values, and to do this week need both df and the significance level (alpha). The df is given (df=20). We are not told what sig. level to use. It is common to check at levels 10%, 5% and 1%. (We do not have to report all the results)
At alpha = 0.1 (10%), the critical value is 1,325. Since the test stat exceeds the critical value, we reject the null at 10% sig level.
At alpha = 0.05 (5%), the critical value is 1.725. We reject the null at the 5% sig. level.
At alpha = 0.01 (1%) , the critical value is 2.528. We do not reject the null at 1%.
When it comes to reporting this, we need only report the result for 5% and 1%. Knowing that the test is rejected at 5% automatically means it is rejected at 10% sig. level, so there is no point in mentioning it.
3. Consider Question 2, but this time let's suppose we are conducting a 2 tailed test.
Solution.With df = 20, the critical value at 10% sig. level is 1.325 (which is 10% in each of lower and upper tail). The critical value at 5% sig. level is 2.086 and 2.845 at 1% sig level.
So reject null only at 10% significance level.
4. Find the value of b such that p(T119.< b) = 0.05
There are 2 things to note. 1st the df = 119 is not tabulated. In such a case it makes sense to pick the closer df =120 rather than df=60. (You could use interpolation instead, but it's messier and not likely to be important.) The other thing is now we are dealing with "less than".
A sketch may help here.Using the fact that the t-distribution is symmetrical, the answer is b = -1.658
5. Find b such that p(-b < T300 < b) = 0.05
2 things to note: df is above 120 so we look at the infinity row; we are looking at 2 tails.
So, b = 1.96
This would be the value you would get if you used the Z table.
If you have any difficulties using the table that has not been addressed here, discuss your problem in theforum