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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