Question

imize File Preview ZOOM+ 5. (a) If T has a t-distribution with 9 degrees of freedom, Use Table IV in the book to find tsch that P(T < t) = 0.99 (b) If X has a Chi-Square distribution with 15 degrees of freedom, Use Table V in the 0.95 book to find z such that P(X <z) (c) Suppose a random sample of size 12 is taken from a Normal Population. If the population variance is 8, what is P(S214.3091). Use Table V in the book. (d) Suppose a random sample of size 10 is taken from a Normal Population. If the population standard deviatión is 4, what is (S > 5.815382). Use Table V in the book. (e) If Si and S2 are the standard deviations of independent random samples of sizes 31 from normal populations with σ., 12 and σ/-18, find n1 = 61 and n2 P(ST/S >1.16) Use Table VI in the book * Previous Next 参" MacBook Air F5 F7

Answer 1

• The T value associated with probability value of 0.99 with 9 degrees of freedom is 2.821.
• b) the chi square x value associated with the probability value of 0.95 with 15 degrees of freedom is 7.261
• c)The sampling distribution of S² when sampling from a normal population, with
  mean $\mu$ and variance $\sigma ^{2}$ , is
• $\frac{(n-1)*S^{2}}{\sigma ^{2}}\sim \chi ^{2}_{n-1}$
• therefore probability value association with chisquare valueP(S²>14.309)=1-P(S²<14.309) with 11 degrees of freedom is 1-0.7670=0.233
• d) similarly probability value association with chisquare value with 9 degrees of freedom P(S>5.81)=1-P(S<5.81)=1-0.2621=0.7379