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## Normal Distributions

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**Normal Distributions**Wikipedia**Standard Normal Distribution**• = 0 • s=1 • To convert a normal distribution to standard normal: • z = (x - m)/s**Normal Approximation of Binomial**When there are a large number of trials calculation of a probability using the binomial formula may be very difficult. The normal probability distribution provides an easy-to-use approximation of binomial probabilities where: n > 20, np> 5, and n(1 - p) > 5**Normal Approximation of Binomial**• The mean and standard deviation of a binomial distribution are: • = np**Normal Approximation of Binomial**The continuity correction factor is used because a continuous distribution is being used to approximate a discrete distribution. Examples: P(x = 3) is approximated by P(2.5 < x < 3.5) P(x > 3) is approximated by P(x > 3.5) P(x > 3) is approximated by P(x > 2.5)**Normal Approximation of Binomial**When there are a large number of trials calculation of a probability using the binomial formula may be very difficult. The normal probability distribution provides an easy-to-use approximation of binomial probabilities where: n > 20, np> 5, and n(1 - p) > 5**Normal Approximation of Binomial Applet**http://onlinestatbook.com/stat_sim/normal_approx/index.html**Normal Approximation of Binomial**A software firm’s technical support staff employs 200 people. On any randomly chosen day 5 percent of the staff are not at work. What is the probability that less than 15 people will not be at work?**Normal Approximation of Binomial**• Step 1: Find mean and standard deviation • = np = (.05)(200) = 10 • Step 2: Find z score (remember continuity correction factor) • z = (14.5 – 10)/3.08 = 1.46 • Step 3: Find corresponding value in standard normal table • P(z < 1.46) = .9279**Normal Approximation of Binomial**What is the probability that at least 8 people will not be at work?**Normal Approximation of Binomial**• Step 1: Find mean and standard deviation • = np = (.05)(200) = 10 • Step 2: Find z score (remember continuity correction factor) • z = (7.5 – 10)/3.08 = -.81 • Step 3: Find corresponding value in standard normal table • P(z > -.81) = 1 - .2090 = .7910**Chapter 7**Sampling and Sampling Distributions**Population**The set of all the elements of interest in a study Sample A subset of the population**Parameter**A numerical characteristic of the population Statistic A numerical characteristic of a sample**Point Estimator**The value of the sample statistic when it is used to estimate the value of the population parameter**Deviations from the Population Value**• The statistic calculated from a sample may different from the parameter because of: • Systemic error – The sample was collected in a way that increased the probability that observations with certain characteristics would be selected • Random error – Differences between the characteristics of the sample and the population due to the “luck of the draw”**Simple Random Sample from a Finite Population**A sample selected such that each observation in the population has an equal probability of being chosen • Simple Random Sample from an Infinite Population • A sample selected such that: • Each element selected comes from the population • Each element is selected independently**Interval Estimate**An interval around a point estimate designed to have a given likelihood of including the population parameter**Sampling Distribution**The distribution of a statistic for all possible samples of a given size.**Sampling Distribution, Example**Assume we have a population consisting of five observations with the following values of x: Obsx 2 3 3 3 4**Sampling Distribution, Example**Assume we are going to draw a sample of 3 observations to estimate the population mean. How many samples could we draw? (5!)/[3!(5-3)!] = [(5)(4)]/[(2)(1)] = 10**Sampling Distribution, Example**Possible samples: Obs.X valuesSample Mean 1,2,3 2,3,3 2 2/3 1,2,4 2,3,3 2 2/3 1,2,5 2,3,4 3 1,3,4 2,3,3 2 2/3 1,3,5 2,3,4 3 1,4,5 2,3,4 3 2,3,4 3,3,3 3 2,3,5 3,3,4 3 1/3 2,4,5 3,3,4 3 1/3 3,4,5 3,3,4 3 1/3**Sampling Distribution, Example**The resulting sampling distribution would be: 2 2/3 3/10 3 4/10 4 3/10 One way to think about a sampling distribution is that is the proportion of the time we would draw some value (or range of values) for the sample statistic.**Sampling Distribution of the Mean**It is not necessary to use the finite population correction factor even when the population is finite if the sample size is less than or equal to 5% of the population (n/N < .05)**Properties of the Standard Error**• The greater the dispersion in the variable the larger the standard error • The larger the sample size the small the standard error • The size of the standard error decreases at a decreasing rate as observations are added • Assume s = 100: • nStandard errorD • 500 4.5 • 1,000 3.2 -1.3 • 1,500 2.6 -0.6 • 2,000 2.2 -0.4**Shape of the Sampling Distribution of the Mean**If the variable has a normal distribution the sampling distribution will have a normal distribution**Shape of the Sampling Distribution of the Mean**If the variable does not have a normal distribution the Central Limit Theorem applies. Central Limit Theorem In selecting simple random samples of size n from a population, the sampling distribution of the sample mean can be approximated by a normal distribution as the sample size becomes large (typically n > 30).**Central Limit Theorem**http://onlinestatbook.com/stat_sim/sampling_dist/index.html See page 273