Quantitative Methods in Business - Lecture (2)
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The document discusses sampling distribution in quantitative methods, focusing on population and sample parameters. It highlights the calculation of mean, variance, and standard error using examples with dice throws to illustrate the concepts. Additionally, it introduces the central limit theorem, explaining how the sampling distribution approaches a normal distribution with larger sample sizes.
Quantitative Methods in Business - Lecture (2)
- 1. Quantitative Methods Dr. Mohamed Ramadan Moh_Ramadan@icloud.com
- 3. Lecture (2) Sampling Distribution Population Sample Fact Estimate Population Mean (𝝁)= 𝟏𝟎#𝟐𝟎#𝟑𝟎 𝟑 = 𝟐𝟎 tons/day Average of all Possible Sample Averages (𝝁&𝒙)= 𝟏𝟓#𝟐𝟎#𝟐𝟓 𝟑 = 𝟐𝟎 tons/day Population Parameter Sample Parameters Population Sample Fact Estimate
- 4. Lecture (2) Sampling Distribution Population Sample Fact Estimate Population Mean (𝝁)= 𝟏𝟎#𝟐𝟎#𝟑𝟎 𝟑 = 𝟐𝟎 tons/day Population Parameter Sample Parameters Population Sample Fact Estimate 𝝁 = 𝟐𝟎 𝜎 ̅" 𝜎 ̅"
- 5. Lecture (2) Sampling Distribution What are the Main Distribution Parameters? -1- Mean -2- Variance = (Std. Dev)2 0.0 0.1 0.1 0.2 0.2 0.3 0.3 0.4 40 41 42 43 44 45 P(x) X = Age at absolute year 𝒙 = 𝑴𝒐𝒏𝒕𝒉𝒍𝒚 𝑺𝒂𝒍𝒂𝒓𝒚 𝒇𝒙
- 6. - !"" # 𝑃 𝑥 = 1.0 Lecture (2) Sampling Distribution Sampling Distribution of the Mean? One Dice (x=number of spots showing on any one throw) 1 2 3 4 5 6 0.00 0.02 0.04 0.06 0.08 0.10 0.12 0.14 0.16 0.18 1.0 2.0 3.0 4.0 5.0 6.0 The population is created by throwing a fair dice infinitely many times # Spots 𝒙 Freq. 𝑃 𝑥 1 1 1/6 = 0.17 2 1 1/6 = 0.17 3 1 1/6 = 0.17 4 1 1/6 = 0.17 5 1 1/6 = 0.17 6 1 1/6 = 0.17 Sum 6 1.0 1.0 0.0 ≤ 𝑃 𝑥 ≤ 1.0
- 7. Lecture (2) Sampling Distribution Sampling Distribution of the Mean? One Dice (x=number of spots showing on any one throw) 1 2 3 4 5 6 # Spots 𝒙 Freq. 𝑃 𝑥 1 1 0.17 2 1 0.17 3 1 0.17 4 1 0.17 5 1 0.17 6 1 0.17 Sum 6 1.0 𝜇 = ∑$%& ' 𝑥$ 𝑁 = - !"" # 𝑥𝑃 𝑥 𝜎( = - !"" # 𝑥( 𝑃 𝑥 − 𝜇( Population Mean: Population Variance: The population is created by throwing a fair dice infinitely many times
- 8. Lecture (2) Sampling Distribution Sampling Distribution of the Mean? One Dice (x=number of spots showing on any one throw) 1 2 3 4 5 6 # Spots 𝒙 Freq. 𝑃 𝑥 𝒙𝑃 𝑥 𝒙 𝟐 𝒙 𝟐 𝑃 𝑥 1 1 0.17 0.2 1 0.2 2 1 0.17 0.3 4 0.7 3 1 0.17 0.5 9 1.5 4 1 0.17 0.7 16 2.7 5 1 0.17 0.8 25 4.2 6 1 0.17 1.0 36 6.0 Sum 6 1.0 3.5 15.2 𝜇 = ∑$%& ' 𝑥$ 𝑁 = - !"" # 𝑥𝑃 𝑥 = 3.5 𝜎( = - !"" # 𝑥( 𝑃 𝑥 − 𝜇( = 15.2 − 3.5 ( = 2.92 Population Mean: Population Variance: 𝜎 = 2.92 = 1.71 The population is created by throwing a fair dice infinitely many times
- 9. Lecture (2) Sampling Distribution Sampling Distribution of the Mean? Two Dice (x=number of spots showing on any one throw) Samples Samples Samples Samples Samples Samples 1,1 2,1 3,1 4,1 5,1 6,1 1,2 2,2 3,2 4,2 5,2 6,2 1,3 2,3 3,3 4,3 5,3 6,3 1,4 2,4 3,4 4,4 5,4 6,4 1,5 2,5 3,5 4,5 5,5 6,5 1,6 2,6 3,6 4,6 5,6 6,6 The sampling distribution is created by drawing all possible samples of size 2 from the population
- 10. Lecture (2) Sampling Distribution Sampling Distribution of the Mean? Two Dice (x=number of spots showing on any one throw) Samples Samples Samples Samples Samples Samples 1,1 1.00 2,1 1.50 3,1 2.00 4,1 2.50 5,1 3.00 6,1 3.50 1,2 1.50 2,2 2.00 3,2 2.50 4,2 3.00 5,2 3.50 6,2 4.00 1,3 2.00 2,3 2.50 3,3 3.00 4,3 3.50 5,3 4.00 6,3 4.50 1,4 2.50 2,4 3.00 3,4 3.50 4,4 4.00 5,4 4.50 6,4 5.00 1,5 3.00 2,5 3.50 3,5 4.00 4,5 4.50 5,5 5.00 6,5 5.50 1,6 3.50 2,6 4.00 3,6 4.50 4,6 5.00 5,6 5.50 6,6 6.00 The sampling distribution is created by drawing all possible samples of size 2 from the population &𝒙 &𝒙 &𝒙 &𝒙 &𝒙 &𝒙
- 11. Lecture (2) Sampling Distribution Sampling Distribution of the Mean? Two Dice (x=number of spots showing on any one throw) Freq. 1.0 1 1/36 1.5 2 2/36 2.0 3 3/36 2.5 4 4/36 3.0 5 5/36 3.5 6 6/36 4.0 5 5/36 4.5 4 4/36 5.0 3 3/36 5.5 2 2/36 6.0 1 1/36 Sum 36 1.0 Samples Samples Samples Samples Samples Samples 1,1 1.00 2,1 1.50 3,1 2.00 4,1 2.50 5,1 3.00 6,1 3.50 1,2 1.50 2,2 2.00 3,2 2.50 4,2 3.00 5,2 3.50 6,2 4.00 1,3 2.00 2,3 2.50 3,3 3.00 4,3 3.50 5,3 4.00 6,3 4.50 1,4 2.50 2,4 3.00 3,4 3.50 4,4 4.00 5,4 4.50 6,4 5.00 1,5 3.00 2,5 3.50 3,5 4.00 4,5 4.50 5,5 5.00 6,5 5.50 1,6 3.50 2,6 4.00 3,6 4.50 4,6 5.00 5,6 5.50 6,6 6.00 &𝒙 &𝒙 &𝒙 &𝒙 &𝒙 &𝒙 &𝒙 𝑃 &𝒙
- 12. Lecture (2) Sampling Distribution Sampling Distribution of the Mean? Two Dice (x=number of spots showing on any one throw) Freq. 1.0 1 1/36 = 0.03 1.5 2 2/36 = 0.06 2.0 3 3/36 = 0.08 2.5 4 4/36 = 0.11 3.0 5 5/36 = 0.14 3.5 6 6/36 = 0.17 4.0 5 5/36 = 0.14 4.5 4 4/36 = 0.11 5.0 3 3/36 = 0.08 5.5 2 2/36 = 0.06 6.0 1 1/36 = 0.03 Sum 36 1.0 1.0 Samples Samples Samples Samples Samples Samples 1,1 1.00 2,1 1.50 3,1 2.00 4,1 2.50 5,1 3.00 6,1 3.50 1,2 1.50 2,2 2.00 3,2 2.50 4,2 3.00 5,2 3.50 6,2 4.00 1,3 2.00 2,3 2.50 3,3 3.00 4,3 3.50 5,3 4.00 6,3 4.50 1,4 2.50 2,4 3.00 3,4 3.50 4,4 4.00 5,4 4.50 6,4 5.00 1,5 3.00 2,5 3.50 3,5 4.00 4,5 4.50 5,5 5.00 6,5 5.50 1,6 3.50 2,6 4.00 3,6 4.50 4,6 5.00 5,6 5.50 6,6 6.00 &𝒙 &𝒙 &𝒙 &𝒙 &𝒙 &𝒙 &𝒙 𝑃 &𝒙
- 13. Lecture (2) Sampling Distribution Sampling Distribution of the Mean? Two Dice (x=number of spots showing on any one throw) Freq. 1.0 1 0.03 1.5 2 0.06 2.0 3 0.08 2.5 4 0.11 3.0 5 0.14 3.5 6 0.17 4.0 5 0.14 4.5 4 0.11 5.0 3 0.08 5.5 2 0.06 6.0 1 0.03 Sum 36 1.0 𝜇 ̅# = - !"" # ̅𝑥𝑃 ̅𝑥 𝜎 ̅# ( = - !"" ̅# ̅𝑥( 𝑃 ̅𝑥 − 𝜇 ̅# ( Mean: Variance: 0.00 0.02 0.04 0.06 0.08 0.10 0.12 0.14 0.16 0.18 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0 &𝒙 𝑃 &𝒙
- 14. Lecture (2) Sampling Distribution Sampling Distribution of the Mean? Two Dice (x=number of spots showing on any one throw) Freq. 1.0 1 0.03 0.0 0.0 1.5 2 0.06 0.1 0.1 2.0 3 0.08 0.2 0.3 2.5 4 0.11 0.3 0.7 3.0 5 0.14 0.4 1.3 3.5 6 0.17 0.6 2.0 4.0 5 0.14 0.6 2.2 4.5 4 0.11 0.5 2.3 5.0 3 0.08 0.4 2.1 5.5 2 0.06 0.3 1.7 6.0 1 0.03 0.2 1.0 Sum 36 1.0 3.5 13.7 𝜇 ̅# = - !"" # ̅𝑥𝑃 ̅𝑥 = 3.5 𝜎 ̅# ( = - !"" ̅# ̅𝑥( 𝑃 ̅𝑥 − 𝜇 ̅# ( = 13.7 − 3.5( = 1.46 Mean: Variance: 𝜎 ̅# = 1.46 = 1.21 0.00 0.02 0.04 0.06 0.08 0.10 0.12 0.14 0.16 0.18 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0 &𝒙 𝑃 &𝒙 &𝒙 𝑃 &𝒙 &𝒙$ 𝑃 &𝒙
- 15. Lecture (2) Sampling Distribution Sampling Distribution of the Mean? (x=number of spots showing on any one throw) 𝜇 = 3.5 𝜎( = 2.92 Mean: Variance: 𝜇 ̅# = 3.5 𝜎 ̅# ( = 1.46 Population Distribution 𝑥 Variance: 𝜎 ̅# ( = 𝜎( 2 = 2.92 2 = 1.46 0.00 0.02 0.04 0.06 0.08 0.10 0.12 0.14 0.16 0.18 1.0 2.0 3.0 4.0 5.0 6.0 0.00 0.02 0.04 0.06 0.08 0.10 0.12 0.14 0.16 0.18 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0 𝜇 = 𝜇 ̅# Sampling Distribution ̅𝑥Mean: Population Distribution 𝑥 Sampling Distribution ̅𝑥
- 16. Lecture (2) Sampling Distribution Sampling Distribution of the Mean? (x=number of spots showing on any one throw) 𝜇 = 3.5 𝜎( = 2.92 Mean: Variance: 𝜇 ̅# = 3.5 𝜎 ̅# ( = 1.46 Mean: Variance: 𝜇 = 𝜇 ̅# (n) Samples 𝜎 ̅# ( = 𝜎( 𝑛 𝜇 ̅# = 𝜇 The sampling distribution of ̅𝑥 𝜎 ̅# ( = 𝜎( 2 = 2.92 2 = 1.46 mean of the sampling distribution of ̅𝑥 variance of the sampling distribution of ̅𝑥 Population Distribution 𝑥 Sampling Distribution ̅𝑥 𝜎 ̅# = 𝜎 𝑛 standard error of ̅𝑥
- 17. Lecture (2) Sampling Distribution Central Limit Theorem (n) Samples 𝜎 ̅# ( = 𝜎( 𝑛 𝜇 ̅# = 𝜇 𝜎 ̅# = 𝜎 𝑛 The sampling distribution of ̅𝑥 mean of the sampling distribution of ̅𝑥 variance of the sampling distribution of ̅𝑥 standard error of ̅𝑥 The sampling distribution of the drawn from any population is approximately normal for a sufficiently large sample size. The larger the sample size, the more closely the sampling distribution of will resemble a normal distribution.
- 18. Lecture (2) Sampling Distribution Central Limit Theorem The sampling distribution of the drawn from any population is approximately normal for a sufficiently large sample size. The larger the sample size, the more closely the sampling distribution of will resemble a normal distribution. Population Distribution 𝑥 Sampling Distribution ̅𝑥 For all values of (n) For only Large values of (n)
- 19. Lecture (2) Sampling Distribution Sampling Distribution of the Mean? if the the is: Standard Error of ̅𝑥 𝜎 ̅# = 𝜎 𝑛 Infinite population 𝜎 ̅# = 𝜎 𝑛 𝑁 − 𝑛 𝑁 − 1 Finite population Correction Factor
- 20. Lecture (2) Sampling Distribution Contents of a 2-Litre Bottle The foreman of a bottling plant has observed that the amount of sparkling water in each 2 litre bottle is actually a normally distributed random variable, with a mean of 2.2 litres and a standard deviation of 0.3 litres. (A) If a customer buys one bottle, what is the probability that the bottle will contain more than 2 litres? (B) If a customer buys a carton of four bottles, what is the probability that the mean amount of the four bottles will be greater than 2 litres? Sampling Distribution of the Mean?Random variable ... Amount of Sparkling Water in each bottle 𝜇 = 2.2 litres 𝜎 = 0.3 litres -A- Because we will use one bottle, then we will work on the random variable 𝑥 𝑃 𝑥 > 2 = 𝑃 𝑥 − 𝜇 𝜎 > 2 − 𝜇 𝜎 = 𝑃 𝑥 − 𝜇 𝜎 > 2 − 2.2 0.3 = 𝑃 𝑧 > −0.67
- 21. 𝒛 Lecture (2) Sampling Distribution Contents of a 2-Litre Bottle The foreman of a bottling plant has observed that the amount of sparkling water in each 2 litre bottle is actually a normally distributed random variable, with a mean of 2.2 litres and a standard deviation of 0.3 litres. (A) If a customer buys one bottle, what is the probability that the bottle will contain more than 2 litres? (B) If a customer buys a carton of four bottles, what is the probability that the mean amount of the four bottles will be greater than 2 litres? Sampling Distribution of the Mean? 𝜇 = 2.2 litres -A- Because we will use one bottle, then we will work on the random variable 𝑥 Random variable ... Amount of Sparkling Water in each bottle 𝑃 𝑥 > 2 = 𝑃 𝑥 − 𝜇 𝜎 > 2 − 𝜇 𝜎 = 𝑃 𝑥 − 𝜇 𝜎 > 2 − 2.2 0.3 = 𝑃 𝑧 > −0.67 𝜎 = 0.3 litres 𝑃 𝑧 > −0.67 𝒇𝒙
- 22. 𝒛 𝒇𝒙 Lecture (2) Sampling Distribution Contents of a 2-Litre Bottle The foreman of a bottling plant has observed that the amount of sparkling water in each 2 litre bottle is actually a normally distributed random variable, with a mean of 2.2 litres and a standard deviation of 0.3 litres. (A) If a customer buys one bottle, what is the probability that the bottle will contain more than 2 litres? (B) If a customer buys a carton of four bottles, what is the probability that the mean amount of the four bottles will be greater than 2 litres? Sampling Distribution of the Mean? 𝜇 = 2.2 litres -A- Because we will use one bottle, then we will work on the random variable 𝑥 Random variable ... Amount of Sparkling Water in each bottle 𝑃 𝑥 > 2 = 𝑃 𝑧 > −0.67 = 𝑃 0 < 𝑧 < 0.67 + 𝑃 𝑧 > 0 =? ? ? +0.5 𝜎 = 0.3 litres 𝑃 𝑧 > 0 = 0.5𝑃 −0.67 < 𝑧 < 0 𝑃 𝑧 > −0.67
- 23. Lecture (2) Sampling Distribution Sampling Distribution of the Mean? 𝜇 = 2.2 litres -A- Because we will use one bottle, then we will work on the random variable 𝑥 Random variable ... Amount of Sparkling Water in each bottle 𝑃 𝑥 > 2 = 𝑃 𝑧 > 0.67 = 𝑃 0 < 𝑧 < 0.67 + 𝑃 𝑧 > 0 = 0.2486 + 0.5 = 0.7486 ≃ 0.75 𝜎 = 0.3 litres 𝒛 𝒇𝒙 𝑃 𝑧 > 0 = 0.5𝑃 −0.67 < 𝑧 < 0 𝑃 𝑧 > −0.67
- 24. Lecture (2) Sampling Distribution Contents of a 2-Litre Bottle The foreman of a bottling plant has observed that the amount of sparkling water in each 2 litre bottle is actually a normally distributed random variable, with a mean of 2.2 litres and a standard deviation of 0.3 litres. (A) If a customer buys one bottle, what is the probability that the bottle will contain more than 2 litres? (B) If a customer buys a carton of four bottles, what is the probability that the mean amount of the four bottles will be greater than 2 litres? Sampling Distribution of the Mean?Random variable ... Amount of Sparkling Water in each bottle 𝜇 = 2.2 litres -B- Because we will use four bottles, then we will work on the random variable ̅𝑥 𝑃 ̅𝑥 > 2 = 𝑃 ̅𝑥 − 𝜇 ̅# 𝜎 ̅# > 2 − 𝜇 ̅# 𝜎 ̅# 𝜇 ̅#=𝜇 = 2.2 𝜎 ̅# = 𝜎 𝑛 = 0.3 4 = 0.15 𝑃 ̅𝑥 > 2 = 𝑃 𝑧 > 2 − 2.2 0.15 = 𝑃(𝑧 > −1.33) 𝜎 = 0.3 litres
- 25. Lecture (2) Sampling Distribution Contents of a 2-Litre Bottle The foreman of a bottling plant has observed that the amount of sparkling water in each 2 litre bottle is actually a normally distributed random variable, with a mean of 2.2 litres and a standard deviation of 0.3 litres. (A) If a customer buys one bottle, what is the probability that the bottle will contain more than 2 litres? (B) If a customer buys a carton of four bottles, what is the probability that the mean amount of the four bottles will be greater than 2 litres? Sampling Distribution of the Mean?Random variable ... Amount of Sparkling Water in each bottle 𝜇 = 2.2 litres -B- Because we will use four bottles, then we will work on the random variable ̅𝑥 𝑃 ̅𝑥 > 2 = 𝑃 𝑧 > −1.33 𝒛 𝒇𝒙 𝑃 𝑧 > −1.33 𝑃 −1.33 < 𝑧 < 0 𝑃 𝑧 > 0 = 0.5 𝜎 = 0.3 litres = 𝑃 −1.33 < 𝑧 < 0 + 𝑃(𝑧 > 0) = 𝑃 0 < 𝑧 < 1.33 + 0.5 =? ? ? +0.5
- 26. Lecture (2) Sampling Distribution Sampling Distribution of the Mean?Random variable ... Amount of Sparkling Water in each bottle 𝜇 = 2.2 litres -B- Because we will use four bottles, then we will work on the random variable ̅𝑥 𝑃 ̅𝑥 > 2 = 𝑃 𝑧 > −1.33 = 𝑃 0 < 𝑧 < 1.33 + 0.5 = 0.4082 + 0.5 = 0.9082 ≃ 0.91 𝒛 𝒇𝒙 𝑃 𝑧 > −1.33 𝑃 −1.33 < 𝑧 < 0 𝑃 𝑧 > 0 = 0.5 𝜎 = 0.3 litres
- 27. Lecture (2) Sampling Distribution Contents of a 2-Litre Bottle The foreman of a bottling plant has observed that the amount of sparkling water in each 2 litre bottle is actually a normally distributed random variable, with a mean of 2.2 litres and a standard deviation of 0.3 litres. (A) If a customer buys one bottle, what is the probability that the bottle will contain more than 2 litres? (B) If a customer buys a carton of four bottles, what is the probability that the mean amount of the four bottles will be greater than 2 litres? Sampling Distribution of the Mean? 0.75 → 𝟕𝟓% 0.91 → 𝟗𝟏%
- 28. Lecture (2) Sampling Distribution Contents of a 2-Litre Bottle The foreman of a bottling plant has observed that the amount of sparkling water in each 2 litre bottle is actually a normally distributed random variable, with a mean of 2.2 litres and a standard deviation of 0.3 litres. (C) What is the expected sampling mean amount of the four bottles, under a probability of 95%? Sampling Distribution of the Mean?Random variable ... Amount of Sparkling Water in each bottle 𝜇 = 2.2 litres 𝜎 = 0.3 litres -C- Because we know the probability, we need to inference the random variable ̅𝑥 𝒛 𝒇𝒙 𝑃 𝑎 < 𝑧 < 𝑏 = 0.95 𝑎 𝑏 0.0250.025 0.4750.475 𝑃 0 < 𝑧 < 𝑏 = 𝑃 𝑎 < 𝑧 < 0 = 0.475
- 29. Lecture (2) Sampling Distribution Sampling Distribution of the Mean?Random variable ... Amount of Sparkling Water in each bottle 𝜇 = 2.2 litres 𝜎 = 0.3 litres -C- Because we know the probability, we need to inference the random variable ̅𝑥 𝒛 𝒇𝒙 𝑃 𝑎 < 𝑧 < 𝑏 = 0.95 𝑎 𝑏 0.0250.025 0.4750.475 𝑃 0 < 𝑧 < 𝑏 = 𝑃 𝑎 < 𝑧 < 0 = 0.475 𝑎 = −1.96 𝑃 −1.96 < 𝑧 < 1.96 = 0.95 𝑏 = 1.96
- 30. Lecture (2) Sampling Distribution Sampling Distribution of the Mean?Random variable ... Amount of Sparkling Water in each bottle 𝜇 = 2.2 litres 𝜎 = 0.3 litres -C- Because we know the probability, we need to inference the random variable ̅𝑥 𝑃 1.96 < 𝑧 < 1.96 = 0.95 Contents of a 2-Litre Bottle The foreman of a bottling plant has observed that the amount of sparkling water in each 2 litre bottle is actually a normally distributed random variable, with a mean of 2.2 litres and a standard deviation of 0.3 litres. (C) What is the expected sampling mean amount of the four bottles, under a probability of 95%? 𝜇 ̅#=𝜇 = 2.2 𝜎 ̅# = 𝜎 𝑛 = 0.3 4 = 0.15 𝑃 −1.96 < ̅𝑥 − 𝜇 ̅# 𝜎 ̅# < 1.96 = 0.95 𝑃 −1.96 < ̅𝑥 − 2.2 0.15 < 1.96 = 0.95 𝑃 −1.96×0.15 < ̅𝑥 − 2.2 < 1.96×0.15 = 0.95 𝑃 −0.294 < ̅𝑥 − 2.2 < 0.294 = 0.95 𝑃 −0.294 + 2.2 < ̅𝑥 < 0.294 + 2.2 = 0.95 𝑃 1.906 < ̅𝑥 < 2.494 = 0.95