PART A: i) Male: Female: The mean value of life satisfaction for male is about 7.7459 while for female is 7.7101‚ which proves there is no significant different life satisfaction between male and female‚ thus gender does not affect life satisfaction a lot. But when it comes to sample variance‚ for male is 2.5684 while for female is 3.0081. From this pair of figures it is obvious that the life satisfaction for female is more flexible than male. Man’s life satisfactions are easy to be affected
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PBL PROBLEM WORKSHEET Title of Problem: Judges at the Olympics Module: B218 Business Statistics 1. A running coach is researching how the pace of runners changes during the course of a race. a. He takes a simple random sample of 20 athletes running an 800m race. The 800m race consists of two laps round a 400m track. What statistical test should he perform to determine whether there is a difference between population average completion times for the first lap and the last lap? b. He takes
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success if the students think they are inferior. These data were gathered in 2013 to help community college increase graduation rate. Interpreting the data article I come to the conclusion that it is a type II error because it more bias toward the null hypothesis rejecting the alternative
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A SYUDY ON EMPLOYEE ENGAGEMENT IN VASAN PUBLICATIONS PRIVATE LIMITED‚CHENNAI J Swaminathan‚ AVCCE ABSTRACT Employee plays a vital role in each and every organization; the interest of employee will help to achieve the organization’s objectives. Successful employee engagement strategy creates a community at a work place and not just a work force. When the employees are effectively and positively engaged with
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1. Industrial situations that show the usefulness of analysis of covariance Analysis of covariance (ANCOVA) gives evaluation of whether the population means on the dependent variable (DV) adjusted for differences on the covariate(s)‚ are different across the independent variable (IV) levels. The variability in the DV due to the control variable (concomitant variable or a covariate) is removed. The covariate increases the opportunity to find statistical significance for the factors‚ fixed or
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9Name: Brielle Cantagallo Class: Statistics Date: 3/17/13 □ EXERCISE 29 Questions to be Graded * 1.Were the groups in this study independent or dependent? Provide a rationale for your answer. The groups in this study were independent because the two sets of data were not taken from the same subjects. The subjects were in one group: female and the second group: male. * 2.t = −3.15 describes the difference between women and men for what variable in this study? Is this value significant
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satisfaction ratings to provide evidence supporting the claim that the mean composite satisfaction rating for the XYZ-Box exceeds 42. a. Letting u represent the mean composite satisfaction rating for the XYZ-Box‚ set up the null hypothesis H0 and the alternative hypothesis Ha needed if we wish to attempt to provide evidence supporting the claim that µ exceeds 42. H0: mu 42 b. The random sample of 65 satisfaction rating yields a sample mean of x = 42.954. Assuming that s equals 2.64‚ use critical
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Chapter 8: Restricting Rows and Sorting Data TRUE/FALSE 1. Retrieving only certain rows from a table is known as projection. ANS: F PTS: 1 REF: 223 2. Retrieving only specific rows from a table is a process known as selection. ANS: T PTS: 1 REF: 223 3. Although Oracle10g is not case sensitive in regards to keywords‚ table names‚ and column names‚ it is case sensitive when comparing search conditions to the data contained in a database table. ANS: T PTS: 1 REF: 225 4. Search conditions
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Confidence Intervals and Hypothesis Testing Name: Math 256‚ Nelson 1. A soft-drink machine is regulated so that the amount of drink dispensed is approximately normally distributed with a standard deviation equal to 1.5 deciliters. Find a 95% confidence interval for the mean of all drinks dispensed by this machine if a random sample of 36 drinks had an average content of 2.25 dl. 2. The heights of a random sample of 50 college students showed a mean of 174.5 centimeters. Assume the
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significance‚ can we conclude that those joining Weight Reducers on average will lose less than 10 pounds? Determine the p-value. Answer: H0: = 10 pounds H1: < 10 pounds Reject the null hypothesis if Z < -1.65 9.0 – 10.00 Z= ---------------- = -2.53 2.8/sqrt(50) The null hypothesis is rejected‚ the average weight loss is less than 10 pounds. p-value = .5000 - .4943 = .0057 32. Dole Pineapple‚ Inc. is concerned that the 16-ounce can of sliced pineapple is being
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