past experience indicates that the time required for high school seniors to complete a standardized test is a normal random variable with a standard deviation of 6 minutes. Test the hypothesis that
after i calculated the chi-square values σ =6 against the alternative that σ<6 if a random sample of 20 high school seniors has a standard deviation
s = 4.51.Use a 0.05 level of significance.
after i calculated the x^2 = (19)(4.51)^2/36 = 10.74
the P value is: P[X19^2 >10.74] = 0.0678
how can i use 10.74 to calculate the 0.0678?