If p-value of a statistical 2 Sample t test conducted by a Black Belt is 0.032 at a Significance level of 0.05, what is the Degree of confidence with the Black Belt with which he can reject the conclusion that there is no statistically significant difference between the two groups?
OPTIONS
- 0.95
- 0.9
- 0.968
- 0.96
ANSWER
0.968
EXPLANATION
P-value of 0.032, 3.2% of risk, which means with 96.8% confidence, the Black Belt can reject the null, so 0.968 is correct.
