TABLE 2 Student\'s t Distribution Entries in this table provide the values oftth
ID: 3385258 • Letter: T
Question
TABLE 2 Student's t Distribution Entries in this table provide the values oftthat correspond to a given upper-tail area and a specified number of degrees of freedom df For example, for = 0.05 and df = 10, P ,2 1 .812) = 0.05. Area in Upper 3.078 6.314 2.920 2.353 2.132 31.821 6.965 4.541 12.706 63.657 4.303 5.841 4.604 0.978 0.920 0.906 0.896 3.365 2.447 1.895 2.365 3.355 3.250 0.883 0.879 1.383 1.833 1.363 12 1.356 2.681 3.055 0.870 0.868 0.866 13 1.345 1.341 1.337 1.333 2.624 15 2.602 2.947 0.863 0.862 1.328 0.860 1.725 2.845 2.080 2.831 2.074 0.858 0.858 23 2.064 25 0.856 2.485 2.787 0.856 2.779 1.703 2.473 2.045 2.042 TABLE 2 (Continued) 0.025 0.005 0.853 2.453 2.449 2.445 2.744 1.694 1.692 2.037 2.035 2.032 0.853 1.308 34 1.307 1.690 1.688 1.687 35 0.852 1.305 0.851 0.851 0.851 2.026 2.712 2.023 2.426 2.704 1.683 1.682 0.850 0.850 1.302 2.698 2.695 0.850 1.679 1.678 2.408 2.405 1.299 1.676 2.678 51 1.298 2.676 52 2.007 2.400 1.674 1.674 1.673 1.673 0.848 1.297 2.005 2.397 2.670 0.848 1.297 2.667 57 1.297 1.296 0.848 0.848 0.848 0.848 2.002 2.392 2.663 2.662 1.296 2.374 2.639 1.660 1.287 1.653 1.283 1.648 2.586 1.282 0.842 SOuRCE: t values calculated with ExcelExplanation / Answer
Mean = 154
SD =29
dF=20-1 = 19
z = (x - µ)/(/(n))
= (173 - 154)/(29/(20)) = 2.93 Answer
P(value) =0.0043
since P < 0.05
It will not exceed .
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