The estimated model is
Ahe = −13.8666 + 1.0217Age, R2= 0.121
(0.368)
Where the number in parentheses is the homoskedastic standard error for the regression coefficient β1. Using the sample of high school graduates, suppose you wanted to test the hypothesis that β1 equals zero at the 5% significance level. That is,
H0: β1=0 vs. H1: β1≠0
Report the t-statistic and p-value for this test. The t-statistic is ____
College Distance | ||
Ahe | Age | College |
12.6102 | 34 | 1 |
17.2572 | 25 | 1 |
4.0307 | 32 | 1 |
26.146 | 33 | 1 |
33.352 | 27 | 1 |
8.9155 | 27 | 1 |
64.6712 | 26 | 1 |
9.3487 | 26 | 1 |
44.0298 | 30 | 1 |
8.1751 | 31 | 1 |
24.4249 | 26 | 1 |
16.0365 | 31 | 1 |
31.7643 | 26 | 1 |
32.8926 | 33 | 1 |
25.9257 | 29 | 1 |
26.8499 | 30 | 1 |
22.6707 | 30 | 1 |
14.4586 | 27 | 1 |
15.6864 | 27 | 1 |
55.2604 | 32 | 1 |
27.1889 | 29 | 1 |
38.0335 | 31 | 1 |
28.1463 | 26 | 1 |
14.3612 | 31 | 1 |
15.8709 | 30 | 1 |
29.2048 | 28 | 1 |
45.106 | 34 | 1 |
23.4888 | 27 | 1 |
18.44 | 33 | 1 |
27.2437 | 25 | 1 |
39.4441 | 27 | 1 |
14.2008 | 30 | 1 |
22.896 | 33 | 1 |
31.9338 | 27 | 1 |
17.7686 | 28 | 1 |
25.2584 | 27 | 1 |
36.9844 | 33 | 1 |
17.4559 | 32 | 1 |
24.827 | 31 | 1 |
25.9448 | 34 | 1 |
9.3946 | 31 | 1 |
24.5044 | 34 | 1 |
36.0618 | 29 | 0 |
13.5713 | 28 | 0 |
8.0061 | 26 | 0 |
20.0158 | 29 | 0 |
45.4987 | 33 | 0 |
7.2858 | 27 | 0 |
11.9818 | 33 | 0 |
18.1214 | 25 | 0 |
7.6007 | 32 | 0 |
12.5471 | 29 | 0 |
16.9855 | 28 | 0 |
15.1409 | 29 | 0 |
6.7083 | 27 | 0 |
22.2952 | 33 | 0 |
12.148 | 30 | 0 |
8.2473 | 28 | 0 |
15.0424 | 27 | 0 |
6.7796 | 25 | 0 |
16.8188 | 25 | 0 |
14.8258 | 30 | 0 |
15.7019 | 25 | 0 |
2.6229 | 30 | 0 |
23.8469 | 31 | 0 |
12.2451 | 31 | 0 |
21.8335 | 33 | 0 |
15.4616 | 26 | 0 |
32.2909 | 29 | 0 |
10.7363 | 28 | 0 |
11.2132 | 34 | 0 |
14.9038 | 28 | 0 |
24.3312 | 29 | 0 |
12.9998 | 30 | 0 |
17.0164 | 26 | 0 |
9.7228 | 25 | 0 |
10.2584 | 28 | 0 |
18.4956 | 29 | 0 |
11.1666 | 33 | 0 |
18.4905 | 29 | 0 |
15.8924 | 34 | 0 |
17.4147 | 29 | 0 |
17.99 | 29 | 0 |
12.807 | 25 | 0 |
9.8222 | 29 | 0 |
4.0649 | 25 | 0 |
7.177 | 25 | 0 |
9.2321 | 31 | 0 |
19.2232 | 30 | 0 |
12.0358 | 25 | 0 |
22.8428 | 29 | 0 |
20.2687 | 26 | 0 |
14.0444 | 30 | 0 |
8.7165 | 25 | 0 |
31.0064 | 33 | 0 |
7.6344 | 26 | 0 |
16.8722 | 31 | 0 |
12.7747 | 28 | 0 |
23.2383 | 30 | 0 |
16.8357 | 30 | 0 |
H0: β1=0 vs. H1: β1≠0
As we can see that B1^ = 0.74879
T statistics =( B1^ - B1)/ SE(B1)
T statistics =( 0.74879 - 0 )/ 0.399
T statistics = 0.74879 / 0.399 = 1.87
Also P value of B1^ is given by 0.06399 as shown in table . So P value is greater than 5 % significance level . So we fail to reject the null hypothesis .
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