所屬科目:研究所、轉學考(插大)◆統計學
1. Let = 3600.7 and S2 = 14655 be the sample mean and variance of the weights of nineteen neonates in grams. Suppose that there was a typewriting error in the cighth observation, the accurate weight of eighth neonate, 4320 was recorded as 4230. After we correct it, what are the accurate sample mean and variance (A) Sample mean: 3605.44, Variance: 20274.32 (B) Sample mea an: 3605.44, Varianc ce: 21175.7 (C) Sample mean: 3609.23, Variance: 22437.62 (D) Sample mean: 3609.23, Variance: 23416.3
2. Let A and B be the events such that P( A ) = 0.4, P( B ) = 0.6, and P(AC ∩ BC)=, Then P(AC∪BC) =?,P(AC ∪ BC) =? (A) (B) (C) (D)
5. Given the probabilities of some of events as follows: P(C|B) =0.95, P(C|BC)=0.3, P(AIB∩ C) = 0.2, P(A|BC∩ C) = 0.05, P(A|B ∩ C) = 0.35, P(AIBC ∩CC)= 0.2.Compute P(A|B). (A)0.2021 (B)0.2075 (C) 0.2098 (D)0.2132
7. Suppose we took a pair of randor: samples (X1,X2), with replacement frot: pogulation f(X): Find E() and P(X>1) (A) E() - 14, P(>1)=0.50 (B) E () = 1.4, P( >1)=0.49 (C) E () =1.6. P(>1)=0.69 (D) E ()= 1.6, P(> 1)=0.71
10. Consider tbe infinite population described by the probability distribution show here. Assume a rindom sumple of n = 80 measurenents is selected from the population, find the sampling distrioution of the sample mean and calculate P (0.905<<1.095) (A) = 1. P(0.905 < < 1.905) =0,7330 (B)= 1, P(0.905< <1.905) = 0.6578 (C) = 1.2. P(0.905 <<1.905)= 0.7920 (D)= 1.2, P(0.905<< 1.905)=0.9505
12. If X, Y, Z are pasitive valued and all have means of 5 and standard deviations of 3, estinnate the prooubility that the averuge of X, Y, and Z is less than J0 by Chebyshev's inequality. If X, Y and Z all have correlation coeftieients of 0.3. (A) (B) (C) (D)
13. 1f' the random vuriable X is exponentially distributed, then tbe mean of X will be: (A) greater than the median. (B) less than the median. (C) equal to tbe median. (D) Cannot tell; the answer depends on what λ is.
14. If n = 100 and the number of successes is 49 for a random sample from a binomial population, find the estimates for the variance of the sample proportion of successes (A) 0.0018 (B) 0.0025 (C)0.0033 (D) 0.0049
16. When testing H0:μ1 -μ2 =0 vs. H1=μ1 -μ2≠ 0, the observed value of the z-score was found to be -2.15. Then, the p-value for this test would be (A)0.0158 (B)0.0316 (C)0.9842 (D)0.9684
17. According to Chebyshev's inequality, the probability that in n throws of a fair die the number of fives lies between is greater than or equal to (A) (B) (C) (D)
20. A hotel chain has identically size resorts in S locations. Tive data that follow resulted from analyzing tbe Hotel occupancies on randomly selected days in the 5 locations. Analysis of Variance Referring to the above table, the value of the elerner nent in the ANOVA table that always provides an estimate of the population variance is__________.(A)9.16(B)21.0 (C) 240.9 (D) 261.9
25.1f the Type I error (a) for a given test is to be decreased, then for a fixed sample size n (A)the Type Il eror (β) will also decrease. (B) the Type Il error (β) will increase. (C) the power of the test will increase. (D)a one-tail test must be utilized. (E) None of the above.
26. Suppose we wish to test H0: μ≤45 versus H1: μ > 45. What will result if we conclude that the mean is grenter than 45 when its true value is really 52? (A) We have made a Type I error. (B) We have made a Type II error. (C) We have made a correet decision (D)None of the above are correct.
27. You have created a 95% confidence interval for p with the result 10 ≤ μ ≤ 15. What decision will you make if we test Ho: p = 17 versus H : μ≠ 17 at a = 0.10? (A)Reject H0in favor of H. (B) Do not reject H0 in favor of H. (C) Fail to reject H0 in fnvor of H. (D) We cannot tell what our dccision will be from tbe inforration given. (E) None of the abowe.
31. Testing for the existence of corelation is equivalent to (A)testing for the existence of the slope (β1). (B) testing for tbe existence of the Y-intercept (β0) (C)the contidence interval estinate for predicting Y. (D)None of the above,