Find the critical value, tc, for c = 0.95 and n= 16. O A. 2.947 OB....

Question

Question

Answer 1

Answer #1

Solution :

n = 16

Degrees of freedom = df = n - 1 = 16 - 1 = 15

c = 0.95

At 95% confidence level the t is ,

$\alpha$ = 1 - c = 1 - 0.95 = 0.05

$\alpha$ / 2 = 0.05 / 2 = 0.025

t_/2,df = t_0.025,15 = 2.131 using excel ( = TINV(probability, degreed of freedom)

Critical value = 2.131

Ans: option B) 2.131 is correct

Answer 2

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