a) How many independent network servers would be needed if each has 99 percent reliability? independent failure rate = 0.01 combined failure rate = 0.00001 0.01^n = 0.00001 10^(-2n) = 10^-5 -2n = -5 n = 2.5 servers needed = 3 b) If each has 90 percent reliability? independent failure rate = 0.1 combined failure rate = 0.00001 0.1^n = 0.00001 10^(-n) = 10^-5 -n = -5 n = 5 servers needed = 5 5.74 Refer to the contingency table shown below. (a) Calculate each probability (i–vi) and explain in words what it means. (b) Do you see evidence that smoking and race are not independent?
(5 pts) --- 6. A random sample of STAT200 weekly study times in hours is as follows: 2 15 15 18 30 Find the sample standard deviation. (Round the answer to two decimal places. Show all work.Just the answer, without supporting work, will receive no credit.) (10 pts) --- image501Refer to the following information for Questions 7, 8, and 9.
Practice Test 4 –Bus 2023 Directions: For each question find the answer that is the best solution provided. There is only one correct answer. 1. For a sample size of 30, changing from using the standard normal distribution to using the t distribution in a hypothesis test, a. will result in the rejection region being smaller b. will result in the rejection region being larger c. would have no effect on the rejection region d. Not enough information is given to answer this question. ANSWER: a -recall that all the t-values are larger than the z-values so it makes it more difficult to reject the null.
If you choose 40 random employees from the corporation, the standard error would equal 6/Square root of 40 = .95 days. The 12 days in this department corresponds to (12-8.2)/.95 = 4 standard errors above the corporation average of 8.2. This is much higher than two or three standard errors, and it appears to be beyond chance variation. Chapter 9 Exercise 3 The p- value tells you how likely it would be to get results at least as extreme as this if there was no difference in the taste and only chance variation was operating. In this problem, p-value of 0.02 means that, if there is no difference in taste, then there is only 2% chance that 70% or more people would declare one drink better than the
We can conclude that the data are Poisson distributed. Chi-Square test of independence Problem 12.12 Use the following contingency table to determine whether variable 1 is independent of variable 2. Let α = .01 | Variable 2 | Variable1 | 24 | 13 | 47 | 58 | | 93 | 59 | 187 | 244 | Step 1 Ho: the two classifications are independent Ha: the two classifications are dependent Step 2 d.f = (r – 1) (c – 1) Step 3 α = 0.01 x 2 0.01, 3df = 11.3449 Step 4 Reject Ho if x 2 > 11.3449 | Variable 2 | Total | Variable1 | 24 (22.92) | 13 (14.10) | 47 (45.83) | 58 (59.15) | 142 | | 93 (94.08) | 59 (57.90) | 187 (188.17) | 244 (242.85) | 583
57). To figure out the contribution ratio one must divide the largest revenue source by the total revenues (Martin, 2001, p. 57). Valley of the Sun United Way’s largest funding source is from grants totaling 30,903,947 and the total revenue in 2012 was 63,588,104 (Ernst & Young, 2012, p. 4). The ratio is calculated at .48 which is still on the safe side. An agency should not be .5 or higher because then they would be too dependent on one source (Martin, 2001, p.
Some of the similarities to a closely held corporation is each shareholder's liability is limited to the amount of their investment. There is also a limited supply of shares for the market, which make these shares less desirable to the market. The legal limit of shares in a S-corporation is 100 shares and in a C-corporation it is unlimited. S-corporations cannot sale their shares on the foreign market since foreigners don't pay individual income taxes. Shareholders vote to becoming a s-corporation and one or more shareholder holding 50 percent or more of the shares can revoke becoming a s-corporation.
Lin Article Critique: Part 3 Anne Hermanns Liberty University Lin Article Critique: Part 3 Critique or Result: Numbers 1, 2, 4, 5, and 10 Comparing the result study composed by Lin, Enright, Krahn, Mack, and Baskin (2004) with Pyrczak’s (2008) evaluation, elements are missing from Lin et al.’s (2004) article. For example, Pyrczak (2008) indicates the importance of percentages reported to be accurate without misleading the reader. Lin et al. (2004) determined the end number of participants in the follow up study decreased from 40 to six. The information given is misleading the audience because the data is not consistent throughout the entire study.
Grading Element #5: “A” work: Exceeds number of required references; all references high quality choices. “C” work: Does not meet the required number of references; some or all references poor quality choices. All newspapers and political sites are biased and are not “high quality choices.” It is hard to avoid them, but keep them below 50% of your
According to Drucker, (1985, p. 22) their batting average is no better than .333: at most one-third of such decisions turn out right; one-third are minimally effective; and one-third are outright failures. Unfortunately, the statistics haven’t changed much in twenty years. Dr. John Sullivan (2010, p. 3) supplements this with an additional, rather scary statistic: a bad promotional decision costs 150-300% more than the annual salary of that single employee, for each year they remain with the organisation. According to the Peter Principle, ‘Occupational incompetence is everywhere, (Peter & Hull, 1969, p. 10). Originated by Dr Lawrence J Peter, the rule posits that ‘in a hierarchy every employee tends to rise to his level of incompetence.’ Meaning, it is only when you don’t look so smart that the promotions will end; thus everyone finishes their promotion curve at their point of incompetence.