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Quantitative Methods Varsha Varde.

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Presentation on theme: "Quantitative Methods Varsha Varde."— Presentation transcript:

1 Quantitative Methods Varsha Varde

2 Sampling Distributions
Concept The Central Limit Theorem The Sampling Distribution of the Sample Mean The Sampling Distribution of the Sample Proportion The Sampling Distribution of the Difference Between Two Sample Means The Sampling Distribution of the Difference Between Two Sample Proportions Varsha Varde

3 Sample Population Varsha Varde

4 A number describing a population
Parameter A number describing a population Varsha Varde

5 A number describing a sample
Statistic A number describing a sample Varsha Varde

6 Random Sample Every unit in the population has an equal probability of being included in the sample

7 Common Sense Thing #1 A random sample should represent the population well, so sample statistics from a random sample should provide reasonable estimates of population parameters

8 Common Sense Thing #2 All sample statistics have some error in estimating population parameters

9 Common Sense Thing #3 If repeated samples are taken from a population and the same statistic (e.g. mean) is calculated from each sample, the statistics will vary, that is, they will have a distribution

10 sampling distribution.
The probability distribution of a statistic is called its sampling distribution. Varsha Varde

11 Mean and Standard Deviation of

12 Distribution of when sampling from a normal distribution
has a normal distribution with mean = and standard deviation =

13 The Central Limit Theorem (CLT
Whatever be the probability distribution of the population from which sample is drawn the probability distribution of sample mean will approximately normal if size of sample is large ;generally larger than 30 Strictly speaking CLT was derived only for sample means but it applies also for sample totals and sample proportions Varsha Varde

14 Central Limit Theorem If the sample size (n) is large enough, has a normal distribution with mean = and standard deviation = regardless of the population distribution

15 What is Large Enough? Varsha Varde

16 Does have a normal distribution?
Is the population normal? Yes No is normal Is ? Yes No is considered to be normal may or may not be considered normal (We need more info) Varsha Varde

17 The Sampling Distribution of the Sample Mean
Suppose X is a random variable with mean µ and standard deviation  . (i) What is the the mean (expected value) and standard deviation of sample mean x ¯ ? Answer: μx ¯ = E(x ¯ ) = µ σ x ¯ = S.D.(x¯ ) =  /√n Varsha Varde

18 (ii)What is the sampling distribution of the sample mean x ¯?
Answer: x¯ has a normal distribution with mean µ and standard deviation  /√n , Equivalently Z =(x ¯ − μx¯ )/σ x ¯ Z = (x¯ − µ ) /  /√n provided n is large (i.e. n ≥ 30) Varsha Varde

19 Example. Consider a population, X, with mean μ = 4 and standard deviation σ = 3. A sample of size 36 is to be selected. (i) What is the mean and standard deviation of X¯? (ii) Find P(4 <X ¯ <5), (iii) Find P(X ¯ >3.5), (exercise) (iv) Find P(3.5 ≤ X ¯ ≤ 4.5). (exercise) Varsha Varde

20 The Sampling Distribution of the Sample Proportion
Suppose the distribution of X is binomial with parameters n and p. (ii) What is the the mean (expected value) and standard deviation of sample proportion p ¯ ? Answer:μP ¯ ˆ = E(p ¯ ) = p σ P ¯ = S.E.(p ¯ ) = √pq/n Varsha Varde

21 What is the sampling distribution of the sample proportion p ¯?
Answer: p ¯ has a normal distribution with mean p and standard deviation √pq /n , Equivalently Z =(p ¯ − μ P ¯ )/σ P ¯ Z = (p ¯ − p) / √pq/n provided n is large (i.e. np ≥ 5, and nq ≥ 5) Varsha Varde

22 Example. It is claimed that at least 30% of all adults favour brand A versus brand B. To test this theory a sample n = 400 is selected. Suppose 130 individuals indicated preference for brand A. DATA SUMMARY: n = 400, x = 130, p = .30, p ¯ = 130/400 = .325 (i) Find the mean and standard deviation of the sample proportion p ¯. Answer: μp ¯ = p = .30 σp ¯ =√pq/n = .023 (ii) Find Prob( p ¯ > 0.30) Varsha Varde

23 Comparing Two Sample Means
E(X ¯ 1 − X ¯ 2) = μ1 − μ2 σX ¯ 1−X ¯ 2 =√(σ21/n1 +σ22/n2) ( X ¯ 1 − X ¯ 2) − (μ1 − μ2) Z = √ σ12/n1+ σ22/n2 provided n1, n2 ≥ 30. Varsha Varde

24 Comparing Two Sample Proportions
E( P ¯ 1 - P ¯ 2) = p1 - p2 σ2P ¯ 1- P ¯ 2 = p1q1/n1+p2q2/n2 (P ¯ 1 - P ¯ 2) - (p1 - p2) Z= √ p1q1/n1+ p2q2/n2 provided n1 and n2 are large. Varsha Varde

25 Varsha Varde

26 Varsha Varde


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