left-tail t test

Question

left-tail t test

H0: b1>= 0.3

H1:b1<0.3

t*= 5.5857

tc=1.860

should i reject H0???

My decision is do not reject H0 since t*>- tc.

But the notes said should reject H0, since t*> tc.

I don't understand why should reject H0.

If you konw which one is correct, please tell me....since i'm frustrated and i

have test on tue

Thank you very much!

更新:

the calculated t* is positive, not negative, that's why i feel confused.

is that t* in left tail t test cannot be positive?

t*= (0.5091-0.3)/ 0.0357 = 5.857

thank you vey much!

Answer 1

最愛解答

The notes seems correct since the H_0 is >=, you should use left tail test with the critical value t_c=-1.860. Since t_calculate = - 5.5857 which is smaller than 1.860 (some people like the format |t_calculate|>t_c), reject H_0 and conclude that b1 < 0.3.

2010-03-08 17:30:43 補充：

is smaller than -1.860

2010-03-08 17:33:12 補充：

應該0甘講﹐t_c是1.86﹐但其實您是在看-1.86﹐因為t_calculate =-5.5857 < -1.86﹐因此reject H_0