SAMPLING AND THE LAW OF LARGE NUMBERS The Law of Large Numbers, LLN, tells us it‟s possible to estimate certain information about a population from just the data measured, calculated, or observed from a sample of the population. Sampling saves the project manager time and money, but introduces risk. How much risk for how much savings? The answer to these questions is the subject of this paper. Examples for the Project Manager A whitepaper by John C. Goodpasture, PMP Managing Principal Square
Words: 3019 - Pages: 13
concerns about the differences in the mean prices and mean mpg across vehicle types and the possible causes depending on whether it’s a mpg or price table. The distribution of price and mpg shapes across each vehicle type and what it means. Confidence intervals are to be used to talk about prices in the population of cars. Several differences of means tests on both prices and mpg are to be taken where there is no much difference between the means of different car types, in both prices and mpg. To
Words: 3076 - Pages: 13
270 3.353 3.400 3.411 3.437 3.477 (a) Construct a 90 percent confidence interval for the true mean weight. (b) What sample size would be necessary to estimate the true weight with an error of } 0.03 grams with 90 percent confidence? (c) Discuss the factors which might cause variation in the weight of Tootsie Rolls during manufacture. (Data are from a project by MBA student Henry Scussel.) Tootsie Answers: a) Confidence intervals are used to find a region in which we are 100 * ( 1 - α )% confident
Words: 917 - Pages: 4
We will again make use of the concept of a random variable. We have reviewed discrete random variables already. Before moving on, we need to review continuous random variables. Continuous random variables take on continuous or interval values (there are an infinite number of possibilities). If you are measuring, the distribution of the result will almost always be continuous. For example, the width of an extruded bar is a continuous random variable. The distribution
Words: 2939 - Pages: 12
the difference between a. and a. The graph shows the simulation of 25 intervals from 25 samples with a 95% confidence level. One interval missed the population parameter (p=0.88). What is the formula for the confidence interval for a sample proportion? The formula for calculating the confidence interval for a sample proportion is: CI = p z*(p(1-p)/n) where p is the sample percentage, z is the z-score for the specified confidence level, and n is the sample size (Gravetter & WaLLnau, 2013). b) What
Words: 579 - Pages: 3
Solutions to Practice Problems for Part IV 1. Five inspectors are employed to check the quality of components produced on an assembly line. For each inspector, the number of components that can be checked in a shift can be represented by a random variable with mean 120 and standard deviation 16. Let X represent the number of components checked by an inspector in a shift. Then the total number checked is 5X, which has mean 600 and standard deviation 80. What is wrong with this argument? Assuming
Words: 3946 - Pages: 16
| 25 | | | 34 | | | 35 | A sample size of 14 is used to find a confidence interval. The number of degrees of freedom needed is 13 A small sample confidence interval formula for a population mean differs from the corresponding large sample formula in that the small sample formula contains a ____________ The sample standard deviation S may be used in place of σ in the large sample confidence interval for N provided that n is at least _____________ . Consider a large population
Words: 290 - Pages: 2
Task 1 explores a little bit into confidence interval and how the width can be affected by the sample size. Students are expected to acquire the mean (average) of a normal population with a population variance of 1 and a confidence interval of 95%. In this task, we will see the relationship between sample size and width in detail by plotting graphs. We will also explore on the accuracy of an interval and how it can be increased or decreased by varying the value of the width. Variance, σ^2 = 1 Standard
Words: 1786 - Pages: 8
95%) = 2.045 Margin of Error = t*(Sd Error) = 11.0392 Confidence Interval = (119.0667-11.0392, 119.0667+11.0392) = (108.0275, 130.1059) 4. One Sample t-test data: friends[, 1] t = -2.0254, df = 29, p-value = 0.05212 alternative hypothesis: true mean is not equal to 130 95 percent confidence interval: 108.0262 130.1071 sample estimates: mean of x 119.0667 Confidence Interval = (108.0262, 130.1071) The confidence interval obtained by the software is close to the one obtained
Words: 768 - Pages: 4
several years in the UK, with the point estimate for the sample standard deviation of 4.028. B) 95% Confidence Interval of Amount Stolen: Sample Size (n): 364 raids Sample Mean (x): 20,330.5 Std. Dev. (σ): 53,510.2 α=1-.95= 0.05→zα/2 is 1.96 x - zα/2( σn) to x + zα/2( σn) 20,330.5 - (-1.96)53,510.2364 = 14,833.295 20,330.5 + (-1.96)53,510.2364 = 25,827.705 Thus the 95% confidence interval is from 14,833.295 to 25,827.705. I am 95% confident that the mean amount stolen, μ, from all 364 raids
Words: 773 - Pages: 4
