3. The rate variance is calculated by the difference between the $16.90 actual labor rate vs the $16 budgeted rate, then we multiply the difference by the 9000 actual labor hours which gives us an $8,100 unfavorable rate variance. To figure out the efficiency variance we multiply the $16 budgeted rate by the difference between the 9,000 actual labor hours and the 10,000 budgeted hours, giving us a $16,000 favorable efficiency variance. As a result of the difference between the rate and efficiency variances we end up having a $7,900 favorable flexible budget variance. 4.
Using the 99 percent confidence interval, can the bank manager be 99 percent confident that mew is less than six minutes? Explain. d. Based on your answers to parts b and c, how convinced are you that the new mean waiting time is less than six minutes? (a) (i) 95% CIn = 100 x-bar = 5.46 s = 2.47 % = 95 Standard Error, SE = σ/√n = 0.2470 z- score = 1.9600 Width of the confidence interval = z * SE = 0.4841 Lower Limit of the confidence interval = x-bar - width = 4.9759Upper Limit of the confidence interval = x-bar + width = 5.9441The confidence interval is [ 4.976 5.944](ii) 99% CIn = 100 x-bar = 5.46 s = 2.47 % = 99 Standard Error, SE = σ/√n = 0.2470 z- score = 2.5758 Width of the confidence interval = z * SE = 0.6362 Lower Limit of the confidence interval = x-bar - width = 4.8238Upper Limit of the confidence interval = x-bar + width = 6.0962The confidence interval is [ 4.824 6.096](b) Yes, the entire 95% CI is less than 6. The manager can be 95% confident that μ is less than 6 minutes.
Construct a 95% confidence interval estimate of the mean lifetime. Show all work. Just the answer, without supporting work, will receive no credit. (8 pts) 21. Consider the hypothesis test given by 5.0: 5 .0: 10 pH p H In a random sample of 100 subjects, the sample proportion is found to be .
The numerical form of this statement is H1 < 1500. From these hypotheses we can conclude that this will be a left-sided test. Step Two The level of significance chosen is .05. This level will give us 95% confidence that our sample mean contains the population mean. The risk involved is the 5% chance that our sample mean does not contain the population mean.
Week Two Quiz 1. In a Gallup poll of 1,038 adults, 540 said that second-hand smoke is very harmful. What is the percentage of adults who said second-hand smoke is very harmful? Show the problem: 540/1038 = 0.5202 or 0.52 0.52 * 100 = 52%(Convert answer to % by times answer by 100) Answer: ___52%_______________ 2. The first class in a relative frequency table is 50–59 and the corresponding relative frequency is 0.2.
Week 4 Quiz Name: The true and false questions are worth 2 point each. 1. The quick rule would say that in a bivariate scatter plot with 25 observations, a correlation of 0.45 is significantly different than zero (but just barely). [ X ] True [ ] False 2. The least squares regression line is obtained when the sum of the squared residuals is minimized.
The maximum value that was proposed for the Drake equation, N = R x fp x ne x f1 x fi x fc x L, was N = 1,680,000. The values that have come to the conclusion are as such: * R = 7 * fp = 0.6 * ne =2.5 * f1 = 1 * fi = 1 * fc = 0.8 * L = 200,000 (7 x 0.6 x 2.5 x 1 x 1 x 0.8 x 200000) leaving N, the maximum number of communicative civilization to be 1,680,000. How does the value change if you double the lifetime of communicative civilizations? The value of ‘L’, the lifetime of communicating civilization and existence has been minimally estimated to 10,000, by doubling that, 20,000 the value of the number of communicative civilizations doubles to 4, only a vague rise in statistics. When the maximum value of L, (200,000) is doubled to 400,000- the number of communicative civilizations (N) changes to 336,000.
The mean is μ = 22.0 and the standard deviation is σ = 2.4. Find the probability that X is between 19.7 and 25.3. A) 1.0847 B) 0.7477 C) 0.4107 D) 0.3370 5. Assume that X has a normal distribution, and find the indicated probability. The mean is μ = 15.2 and the standard deviation is σ = 0.9.
The average employee "needs to work more than a month to earn what the CEO earns in one hour. " Although different from income inequality, the two are related. In Inequality for All—a 2013 documentary with Robert Reich in which he argued that income inequality is the defining issue for the United States—Reich states that 95% of economic gains went to the top 1% net worth (HNWI) since 2009 when the recovery allegedly started.  A 2011 study found that US citizens across the political spectrum dramatically underestimate the current US wealth inequality and would prefer a far more egalitarian distribution of wealth.  Wealth is usually not used for daily expenditures or factored into household budgets, but combined with income it comprises the family's total opportunity "to secure a desired stature and standard of living, or pass their class status along to one's children".
By using the confidence interval of 99% the results are as follows: The mean wage used is $30,833.46 with a standard deviation of 16.947.10 and the sample size is 100. The lower wage is 26,468.18 and the upper wage is 35,198.744. The mean education in years used is 12.73 with a standard deviation of 2.79 and sample size of 100. The lower education in years is 12.011 and the upper education in years is 13.45. The mean age is 39.11 with a standard deviation of 12.57 and a sample size of 100.