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STEVENS MA 331 - Lecture 3 Sampling Distributions

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Lecture 3Slide Number 2Sampling Distribution for Counts and Proportions: Binomial Distribution for Sample Counts:EXAMPLES (Bin or not Bin)Finding Binomial ProbabilitiesSlide Number 7Slide Number 8Slide Number 9Slide Number 10ExampleBinomial Mean and Standard DeviationSample Proportions Slide Number 14Normal approximation for counts and proportionsSampling distribution of p^Example:The continuity correction:The continuity correction(cont.):Slide Number 20Section 5.2: Sampling distribution of the sample meanExample: Distribution of individual stocks (up) vs. distribution of mutual funds (down)Sample mean: Slide Number 24Central Limit Theorem:Example: Slide Number 27Example: 5.34 Example: 5.67Practical noteThe central limit theoremIncome distributionHow large a sample size?Sampling distributionsFurther propertiesWeibull distributionsSlide Number 37Lecture 3Sampling distributions. Counts, Proportions, and sample mean.• Statistical Inference: Uses data and summary statistics (mean, variances, proportions, slopes) to draw conclusions about a population or process.• Statistic: Any random variable measured from a random sample or in a random experiment.• Sampling distribution of a statistic: shows how a statistic varies in repeated measurements of an experiment. The probability distribution of a statistic is called its sampling distribution.• Population distribution of a statistic: distribution of values for all members of the population. Unknown, but estimable using laws of statistics.Sampling Distribution for Counts and Proportions:• In a survey of 2500 engineers, 600 of them say they would consider working as a consultant. Let X = the number who would work as consultants.• X is a count: • Sample Proportion of people who would work as consultants:Distinguish count from sample proportion, they have different distributions.Binomial Distribution for Sample Counts:• Distribution of the count, X, of successes in a binomial setting with parameters n and p• n = number of observations• p = P (Success) on any one observation• X can take values from 0 to n • Notation: X ~ Bin (n, p)• Setting:1. Fixed number of n observations2. All observations are independent of each other3. Each observation falls into one of two categories: Success or Failure4. P (Success) = P (S) = pEXAMPLES (Bin or not Bin)• Toss a fair coin 10 times and count the number X of heads. What about a biased coin?• Deal 10 cards from a shuffled deck of 52.X is the number of spades. Suggestions??• Number of girls born among first 100 children in a (large) hospital this year.• Number of girls born in this hospital so far this year.Finding Binomial ProbabilitiesUse Table C: page T-6• (How to: - find your n = number of observations• find your p = probability of success• find the probability corresponding to k = number of successes you are interested in)• You can use R as well to evaluate probabilities: » pbinom(4,size=10,prob=0.15) (calculates P(Bin(10,0.15)<=4) )» [1] 0.990126• If you want the entry in the table do:» pbinom(4,size=10,prob=0.15)-pbinom(3,size=10,prob=0.15)» [1] 0.04009571ExampleYour job is to examine light bulbs on an assembly line. You are interested in finding the probability of getting a defective light bulb, after examining 10 light bulbs.– Let X = number of defective light bulbs– P (defective) = .15– N = 101. Is this a binomial set up?2. What is the probability that you get at most 2 defective light bulbs?3. What is the probability that the number of defective light bulbs you find is greater than eight?4. What is the probability that you find between 3 and 5 defective light bulbs?Binomial Mean and Standard DeviationnpX=μ)1(2pnpx−=σ)1( pnpx−=σExample: Find the mean and standard deviation of the previous problemsSample Proportions• Let X be a count of successes in n = total number of observations in the data set.• Then the sample proportion:– NOTE!!!!• We know that X is distributed as a Binomial, however is NOT distributed as a Binomial. nXp =ˆˆpNormal approximation for counts and proportions• If X is B(n,p), np≥10 and n(1-p)≥10 then: is approximately ( , (1 ) )(1 )ˆ is approximately ( , )XNnp np ppppNpn−−Sampling distribution of p^The sampling distribution of is never exactly normal. But as the sample size increases, the sampling distribution of becomes approximately normal. The normal approximation is most accurate for any fixed n when p is close to 0.5, and least accurate when p is near 0 or near 1.pˆpˆExample:• In a survey 2500 engineers are asked if they would consider working as consultants. Suppose that 60% of the engineers would work as consultants. When we actually do the experiment 1375 say they would work as consultantsFind the mean and standard deviation of .What is the probability that the percent of to be consultants in the sample is less than .58? Between .59 and .61?pˆThe continuity correction:• Example: According to a market research firm 52% of all residential telephone numbers in Los Angeles are unlisted. A telemarketing company uses random digit dialing equipment that dials residential numbers at random regardless of whether they are listed or not. The firm calls 500 numbers in L.A.1. What is the exact distribution of the number X of unlisted numbers that are called?2. Use a suitable approximation to calculate the probability that at least half the numbers are unlisted.The continuity correction(cont.):• In the previous problem if we compute the probability that exactly 250 people had unlisted numbers using the normal approximation we would have find this probability equals zero. • That is obviously not right because this number has to have some probability (small but still not zero).• The problem comes from the fact that we use a continuous distribution (Normal Distribution) to approximate a discrete one (Binomial Distribution). • So to improve the approximation we use a correction: • Whenever we compute a probability involving a count we will move the interval we compute 0.5 as to include or exclude the endpoints of the interval depending on the type of interval (closed or open) we compute in the problem. • Then we use the normal approximation to compute the probability of this new interval.• Example: In the previous problem find:( 250)PX≥( 250)PX>( 250)PX=( 250)PX<(248 251)PX<<(248 251)PX<≤(248 251)PX≤≤Section 5.2: Sampling distribution of the sample mean•


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