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This lecture is from Statistical Method. Key important points are: Sampling Distributions, Binomial Distribution, Poisson Distribution, Discrete Probability, Distribution, Probability Distribution, Frequency Distribution, Continuous Probability Distribution, Sampling Distributions, Features of Sampling Distribution

Typology: Slides

2012/2013

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Download Sampling Distributions - Statistical Method - Lecture Slides and more Slides Statistics in PDF only on Docsity! Business Statistics
Introduction to Sampling Distributions
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Types of Distribution • Frequency Distribution • Probability Distribution – Discrete Probability Distribution • Binomial Distribution • Poisson Distribution – Continuous Probability Distribution • Normal (Gaussian) Distribution – Sampling Distributions 2 Docsity.com Sampling Distribution • A sampling distribution is a distribution of the possible values of a statistic for a given size sample selected from a population 5 Docsity.com Examples of Sampling Distributions Population Item of Study Sample Sample Statistic Sampling Distribution Water in a river Mercury contamination 10-gallon containers of water Mean number of parts of mercury per million parts of water Sampling distribution of the mean All professional basketball teams Height Groups of 5 players Median Height Sampling distribution of the median All parts produced by a manufacturing process Proportion of defectives 50 parts Proportion defective Sampling distribution of the proportion 6 Docsity.com Developing a Sampling Distribution • Assume there is a population … • Population size N=4 • Random variable, x, is age of individuals • Values of x: 18, 20, 22, 24 (years) 7 A B C D Docsity.com Sampling Distribution of All Sample Means 10 1st 2nd Observation Obs 18 20 22 24 18 18 19 20 21 20 19 20 21 22 22 20 21 22 23 24 21 22 23 24 18 19 20 21 22 23 24 0 .1 .2 .3 P(x) x Sample Means Distribution 16 Sample Means _ Developing a Sampling Distribution (no longer uniform) Docsity.com Summary Measures of this Sampling Distribution: 11 Developing a Sampling Distribution 21 16 24211918 N x μ ix = ++++ == ∑ 1.58 16 21)-(2421)-(1921)-(18 N )μ(x σ 222 2 xi x = +++ = − = ∑ Docsity.com Comparing the Population with its Sampling Distribution 12 18 19 20 21 22 23 24 0 .1 .2 .3 P(x) x 18 20 22 24 A B C D 0 .1 .2 .3 Population N = 4 P(x) x _ 1.58σ 21μ xx ==2.236σ 21μ == Sample Means Distribution n = 2 Docsity.com Calculating Sampling Error • Sampling Error: The difference between a value (a statistic) computed from a sample and the corresponding value (a parameter) computed from a population Example: (for the mean) where: 15 μ - xError Sampling = mean population μ mean samplex = = Docsity.com Review • Population mean: Sample Mean: 16 N x μ i∑= where: μ = Population mean x = sample mean xi = Values in the population or sample N = Population size n = sample size n x x i∑= Docsity.com Example 17 If the population mean is μ = 98.6 degrees and a sample of n = 5 temperatures yields a sample mean of = 99.2 degrees, then the sampling error is degrees0.699.298.6μx −=−=− x Docsity.com z-value for Sampling Distribution of x • Z-value for the sampling distribution of: 20 where: = sample mean = population mean = population standard deviation n = sample size x μ σ n σ μ)x(z −= x Docsity.com Finite Population Correction • Apply the Finite Population Correction if: – the sample is large relative to the population (n is greater than 5% of N) and… – Sampling is without replacement Then 21 1N nN n σ μ)x(z − − − = Docsity.com Sampling Distribution Properties • (i.e. is unbiased ) 22 Normal Population Distribution Normal Sampling Distribution (has the same mean) x x x μμx = μ xμ Docsity.com Central Limit Theorem X 25 As sample size gets large enough (n ≥ 30) ... sampling distribution becomes almost normal regardless of shape of population σ σ x n = µ µx = n ↑ Docsity.com If the Population is not Normal • We can apply the Central Limit Theorem: – Even if the population is not normal, – …sample means from the population will be approximately normal as long as the sample size is large enough – …and the sampling distribution will have and 26 μμx = n σσx = Docsity.com If the Population is not Normal 27 Population Distribution Sampling Distribution (becomes normal as n increases) Central Tendency Variation (Sampling with replacement) x x Larger sample size Smaller sample size Sampling distribution properties: μμx = n σσx = xμ μ Docsity.com Example Solution (continued): 30 x 0.31080.4)zP(-0.4 36 3 8-8.2 n σ μ- μ 36 3 8-7.8P 8.2) μ P(7.8 xx =<<= <<=<< z 7.8 8.2 -0.4 0.4 Sampling Distribution Standard Normal Distribution .1554 +.1554 x Population Distribution ? ? ? ? ? ? ? ? ? ? ? ? Sample Standardize 8μ = 8μx = 0μz = Docsity.com Population Proportions, p p = the proportion of population having some characteristic • Sample proportion ( p ) provides an estimate of p: • If two outcomes, p has a binomial distribution 31 size sample sampletheinsuccessesofnumber n xp == Docsity.com Sampling Distribution of p • Approximated by a normal distribution if: – where and 32 (where p = population proportion) Sampling Distribution P( p ) .3 .2 .1 0 0 . 2 .4 .6 8 1 p pμ p = n p)p(1σp − = 5p)n(1 5np ≥− ≥ Docsity.com Example • if p = .4 and n = 200, what is P(.40 ≤ p ≤ .45) ? 35 .03464 200 .4).4(1 n p)p(1σp = − = − = 1.44)zP(0 .03464 .40.45z .03464 .40.40P.45)pP(.40 ≤≤= −≤≤ − =≤≤ Find : Convert to standard normal: pσ Docsity.com Example • if p = .4 and n = 200, what is P(.40 ≤ p ≤ .45) ? 36 z .45 1.44 .4251 Standardize Sampling Distribution Standardized Normal Distribution Use standard normal table: P(0 ≤ z ≤ 1.44) = .4251 .40 0 p Docsity.com Case Study 3.1 Suppose the weights of male adults are known to have a normal distribution with mean 70 kg and standard deviation 12 kg. a. Describe the population of interest in this problem. b. Suppose we are to take a random sample of 10 adults from this population. Explain why we can safely ignore the Finite Population Correction Factor in our calculation of the standard deviation of the sample mean, even if the population is finite. c. Calculate the expected value and the standard error for the mean of the above sample of 10. d. An old elevator has a posted carrying capacity of 750 kg. If ten randomly selected adults from this population enter the elevator, what is the probability that their total weights exceed the elevator's capacity? e. If the distribution of weights was not normal, would the expected value and the Standard Error in part (c) be still valid? How about the probability in part (d), would it still be approximately correct? Explain your answer. 37 Docsity.com Solution Case Studv 3.2
a. No, the sample size n= 50 is greater than 5% of the population size N = 100. We must use the FPCF.
b. The expected value of the sample proportion is E(p)= p= 0.6 - And the standard deviation is
o; = — Jed —pyin = ro foC750 = 0.0492
ec. np = 50(0.6) = 30, and n(1-p) = 50(0.4) = 20 are both greater than 5. The sample size is large enough to apply the CLT.
d. Getting between 30 to 40 blue marbles is equivalent to observing a value of the sample proportion p that is between
30/50 = 0.6 and 40/50 = 0.8. The desired probability can then be approximated by
6-6 5. 8-6
0492 -0492
PL6<5<8)x a(
- POO < 2 24.07) 20.5
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Case Study 3.3 Suppose the sampling in Case Study 3.2 is done with replacement. ie., one marble is picked at random, its colour recorded, and then returned into the bag before the next marble is picked. How will this change affect the above calculations? 41 Docsity.com Solution Case Study 3.3
If sampling is with replacement, the population is treated as ifit was infinite. The FPCF should be dropped from the Standard Error formula.
Without the FPCF, we have SE =. fp(l—p)in = .f-604)/50 = 0.0693 .
The probability of getting between 30 to 60 blue marbles is then approximated as:
6-.6 8-.6
B25
0693 0693
POS BSR) af
) = P(O< Z<2.89)=0.4081
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Case Study 3.5 Refer to Case Study 3.4. What is the probability that in a random sample of 200 readings, 10 or fewer exceed 26.2 km/h? Hint: first define the applicable parameter and statistic. 45 Docsity.com Solution Case Study 3.5 In this problem, we are interested only in whether or not the wind speed is greater than 26.2 km/h. This means we must first convert the wind speed data from a quantitative one (measured in km/h) to a categorical one (with two categories: Yes that it exceed 26.2 km/h, or No it does not) Accordingly the applicable parameter and statistics are: p = the proportion of time that the wind speed exceed 26.2 km/h, and the sample proportion of readings of wind speed exceeding 26.2 km/h. From part (a) in Case 3.4, we know that p = .0778. Now, in a sample of 200 readings we want to find the probability that Using the CLT for sample proportion, the desired probability can then be approximated by: 46 Docsity.com Case Study 3.6
A bottling company uses a filling machine to fill plastic bottles with a popular cola. The
bottles are supposed to contain 300 milliliters. In fact, the contents vary according to a
normal distribution with mean p. = 298 inl and standard deviation o = 3 ml.
a. What is the probability that an individual bottle contains less than 295 ml?
b. What is the probability that the mean contents of the bottles m a six-pack is less than 295
ml?
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Solution Case Study 3.7
Find L such that P( Y > L)=0.01.
This is equivalent to finding L such that P( X <L)=1- 0.01=0.99.
And, this is equivalent to finding z, such that P(Z < z, )=1- 0.01= 0.99. Using Table III,
the normal distribution table, we find that z, = 2.33.
Now, we just need to transform this back to.Y .
. X-U : v a
Since Z —, we can rearrange this so that Y = jt +z.) —].
alyn
That is, Y= 1.4 + 2.33(0.3/11.18) = 1.463.
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Case Study 3.8
The number of flaws per square yard in a type of carpet material varies, with mean 1.6 flaws
per square yard and standard deviation 1.2 flaws per square yard. The distribution is not
notmal--in fact, it is discrete. An inspector studies 200 random square yards of the material,
records the number of flaws found in each square yard, and calculates the mean number of
flaws per square yard inspected. Use the central lunit theorem to find the approximate
probability that the mean number of flaws exceeds 2 per square yard.
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Solution Case Study 3.8
Since the sample size is larger than 30 (n = 200), the Central Limit Theorem applies. So,
o- - & - 12% ~ oogs
© Vn 200
So, P(X > 2) = P[Z > (2- 1.6)/0.085] = P(Z > 4.71) = 0.
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Case Study 3.A According to Nielsen Media Research, the average number of hours of TV viewing per household per week in the US is 50.4 hours. Suppose the standard deviation is 11.8 hours and a random sample of 42 US households is taken. a. What is the probability that the sample average is more than 52 hours? b. What is the probability that the sample average is less than 47.5 hours? c. What is the probability that the sample average is less than 40 hours? If the sample average actually is less than 40 hours, what would it mean in terms of the Nielsen Media Research figures? d. Suppose that the population standard deviation is unknown. If 71% of all sample means are greater than 49 hours and the population mean is still 50.4 hours, what is the value of the population standard deviation? 55 Docsity.com Case Study 3.B If 10% of a population of parts is defective, what is the probability of randomly selecting 80 parts and finding that 12 or more parts are defective? Solution Here, p=.10, phat = 12/80 = .15 and n = 80. Entering these values in the z formula yields Z = (.15-.10)/sqrt(.10*.90/80) = .05/.0335 = 1.49 From z table, P(1.49) = .4319, therefore P(phat .= .15) = .5 - .4319 = .0681 Thus, about 6.81% of the time, 12 or more defective parts would appear in a random sample of 80 parts when the population proportion is .10. If this result actually occurred, the 10% proportion for population defects would be open to question. 56 Docsity.com