Please notice, if your calculated value that you took to the t-table matches exactly one
of the data on that d.f. data line, then the top tail value corresponding to this matched
value would be your p-value, and there is no need to average anything else, because
you had a value in the table that matched the calculated value.
Example to clarify the table values
How to use t-table for t-p-values
You need a degree of freedom and the significant level to find the t-values from the
table [critical values].
The table can be used in different ways as described below:
1- If we are only given the confidence level [the central area probability, which is
the researcher’s confidence], then we can use the top row values.
2- If we are given significant level, alpha, then we can use the next top rows for
either one side or 2 sides tests.

5. Use the data in
BUSI1013 Bank Dataset.xlsx
(from Unit 1 Exercise Question 2)
to answer this question. (
4 points
)
BUSI
1013:
S
TATISTICS FOR
B
USINESS
11
Here are some examples on how to find values from the t-table for your
information:
Let us say for
d.f .= 9
at 95% confidence the C.V [table value] = 2.262 which is the
same for 2-tails test at level of significance, alpha = 0.05.
Now if on the other hand you know your test statistics value [calculated value], and
you wish to find the p-value for this calculated value, you need to take this value to the
table and find the associated alpha value, which will be your p-value. For example:
If in a one side test your calculated [test statistics] value was 1.68 at d.f. = 9, then your
p-value for this t-value is the tail area from the calculated value [alpha value]: look at
the table in the row of d.f.= 9. Check if the value 1.68 exists in this row. You can see
that it does not exist. The closest values to this t=1.68 are the 1.383 & 1.833. So, your
value of 1.68 falls between these 2 values, which means the alpha value for 1.68 is
also falling between the 2 alpha values corresponding to these 2 borderline values
[1.383 & 1.833]. The 2 alpha values associated with 1.383 & 1.833 for a one side test
ant d.f. = 9 are 0.05 and 0.1.
So, the alpha value associated with 1.68 is average of these 2 alphas, i.e., [0.05 +
0.1]/2 = 0.075, which is the tail area after the line 1.68, which is your p-value.
Thank you and
Best Regards Zarbi

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