_ {d} Estimate the P-value of the sample test statistic. The Pvalue for the sample statistic t = 3.03? is determined using a Student's fdistribution table. Note that the table gives the positive t values, and that the area to the left of t = 3.03? is equivalent to the area to the right 0ft 2 3.03?. Since the null and alternate hypotheses are H0: 0.1 #2 = 0, H1: #1 p2 = I], we are performing a . Below is an excerpt from this table. We determined that df. = 48, but this value is skipped in the table. Instead, we round down and use the ro'wr corresponding to df. = 45. one-tail area 0.250 0.125 0.100 0.0?5 0.050 0.025 0.010 0.005 0.0005 two-tail area 0.500 0.250 0.200 0.150 0.100 0.050 0.020 0.010 0.0010 df. \\ C 0.500 0.?50 0.800 0.850 0.900 0.950 0.980 0.990 0.999 40 0.631 1.167 1.303 1.468 1.684 2.021 2.423 2.?04 3.551 45 0.680 1.165 1.301 1.465 1.679 2.014 2.412 2.690 3.520 50 0.6?9 1.164 1.299 1.462 1.6?6 2.009 2.403 2.6?8 3.496 Note that the tvalue 3.037 is not an entry in the table. However, we can say that this value falls directly between table values 2.690 and E in the row for (if. = 45. Due to the type of test we have, we will need to use the column headings for Seleoi- V to determine the P-value. Thus, the P-value is in the Following interval. 0 P-value :> 0.500 O 0.250