436 Chapter 9 Hypothesis Testing and 0.0025 (see Figure 9.11). We can conclude that the P-value is small enough to reject H0. Degrees of Area in the Right Tail freedom 0.40 0.25 0.10 0.05 0.025 0.01 0.005 0.0025 0.001 0.0005 1 0.325 1.000 3.078 6.314 12.706 31.821 63.657 127.321 318.309 636.619 2 0.289 0.816 1.886 2.920 4.303 6.965 9.925 14.089 22.327 31.599 3 0.277 0.765 1.638 2.353 3.182 4.541 5.841 7.453 10.215 12.924 ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ 38 0.255 0.681 1.304 1.686 2.024 2.429 2.712 2.980 3.319 3.566 39 0.255 0.681 1.304 1.685 2.023 2.426 2.708 2.976 3.313 3.558 40 0.255 0.681 1.303 1.684 2.021 2.423 2.704 2.971 3.307 3.551 50 0.255 0.679 1.299 1.676 2.009 2.403 2.678 2.937 3.261 3.496 60 0.254 0.679 1.296 1.671 2.000 2.390 2.660 2.915 3.232 3.460 80 0.254 0.678 1.292 1.664 1.990 2.374 2.639 2.887 3.195 3.416 100 0.254 0.677 1.290 1.660 1.984 2.364 2.626 2.871 3.174 3.390 200 0.254 0.676 1.289 1.653 1.972 2.345 2.601 2.839 3.131 3.340 Area = 0.0025 2.915 3.232 3.144 Area = 0.001 Figure 9.11 The P-value is the area to the right of the observed value of the test statistic, 3.144. The P-value is between 0.001 and 0.0025. Finding the P-value for a two-tailed test from a table In the previous example, what if the alternate hypothesis were H1: �� ≠ 0? The P-value would be the sum of the areas in two tails. We know that the area in the right tail is 0.0012 (see Figure 9.10). Since the t distribution is symmetric, the sum of the areas in two tails is twice as much: 0.0012 + 0.0012 = 0.0024. This is shown in Figure 9.12. Area = 0.0012 Area = 0.0012 −3.144 3.144 Figure 9.12 The P-value for a two-tailed test is the sum of the areas in the two tails. Each tail has area 0.0012. The P-value is 0.0012 + 0.0012 = 0.0024. If we are using Table A.3, we can only specify that P is between two values. We know that the area in one tail is between 0.001 and 0.0025. Therefore, the area in both tails is between 2(0.001) = 0.002 and 2(0.0025) = 0.005. This is shown in Figure 9.13.
navidi_monk_elementary_statistics_2e_ch7-9
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