Q 8. You fit a Pareto distribution to a sample of 200 claim amounts and use the likelihood ratio test to test the hypothesis that a = 1.5 and 0 = 7.8. You are given: (i) The maximum likelihood estimates are a = 1.4 and @= 7.6. (ii) The natural logarithm of the likelihood function evaluated at the maximum likelihood estimates is -817.92. (iii) Eln (x + 7.8) = 607.64. Determine the result of the test.

The distance, X, between flaws on a long cable has an exponential distribution with a parameter λ = f(x, 2) = 2 e-ix for 0 ≤ x ≤ 00, λ = 12 a) What is the expected value of X? Hint: This is a problem involving the exponential distribution. Knowing the parameter for the distribution allows you to easily answer parts a,b,c and use the built-in R functions for the exponential distribution (dexp(), pexp(), gexp()) for other parts. Or (not recommended) you should be able to use the R integrate command with f(x) defined as above or with dexp() for all parts. b) What is the variance of X? c) What is the standard deviation of X? is what R calls rate. d) What is the probability that X is larger than its expected value? e) What is the probability that X is > 5? f) What is the probability that X is > 10? . Thus, the density of X is: g) What is the probability that X > 10 given that X > 5? h) What is the median of X?

Please solve within 20 minutes

For the situation described below, state the null and alternative hypotheses to be tested. (Enter != for ≠ as needed.) A new variety of pearl millet is expected to provide an increased yield over the variety presently in use, which has a mean yield of about 55 bushels per acre.

Stoaches are fictional creatures, sometimes mistaken for mome raths. Suppose a two-tailed t-test is conducted to test for a difference between the population means of adult female stoach weights and adult male stoach weights. The female sample has 13 weights, and the male sample has 7 weights. The t-statistic for the test is -0.20542. What is the p-value? (Give your answer to 4 decimal places.) p-value:

How would you report the results? The derived t = 0.02 was not significant at p = .05 with df = 10. Therefore, Ho was not rejected, and it was concluded that the mean pretest score for students at the beginning of the semester (18.27) was not different than the mean posttest score for students at the end of the semester (22.00), t(10) -= 0.02, p .05. In terms of the research question, it appears that the curriculum was not effective in promoting learning.

1. You are interested in the effect of a new learning technique. The performance score on the test before applying the new learning technique is 50 (u=50), the standard deviation is 4 (o=4).  You obtained a sample of 64 subjects and applied the new learning technique. After that, their performance score on the test increased by 2 (x=52). Did the new learning technique improve/had any effect on the performance? The answer, using 4 steps of hypothesis testing.  Please use APA style to report your result.  Assume that a-level=.05 2. You are interested in the effect of a new learning technique. The performance score on the test before applying the new learning technique is 50 (u=50), the standard deviation is 24 (o=24).  You obtained a sample of 64 subjects and applied the new learning technique. After that, their performance score on the test increased by 2 (x=52). Did the new learning technique improve/had any effect on the performance? The answer, using 4 steps of hypothesis testing.…

Can you help me with this please! Thank you.

A large automobile insurance company selected samples of single and married male policyholders and recorded the number who made an insurance claim over the preceding three-year period. Single Policyholders Married Policyholders 721 = 575 n2 = 740 Number making claims = 115 Number making claims = 111 a. Use α = 0.05. Test to determine whether the claim rates differ between single and married male policyholders. z-value p-value We can conclude (to 2 decimals) X (to 4 decimals) that there is the difference between claim rates. b. Provide a 95% confidence interval (to 4 decimals) for the difference between the proportions for the two populations. Enter negative answer as negative number.