Topic 7 - Sampling Distributions, Central Limit Theorem (CLT), Confidence Intervals and Sample Size

Contents

Gerald Keller (2011), Statistics for Management and Economics (Abbreviated), 9th Edition, pp. value.

Sampling Distribution of a Sample Mean

There are 2 methods to find the sampling distribution of the sample mean.

1. Draw samples of the same size from the population and calculate the statistic of interest
2. Rely on Probability, E(X) & V(X)

Bank Teller Example

In this example, the sample size is 2 (there are 2 bank tellers), and they can make either 0, 1 or 2 errors ('X' number of errors). In the following picture the probability distribution is given and the mean and variance is calculated.

Now, we create a new random variable, X bar - representing the sample mean. For example, if Teller 1 makes 2 mistakes and Teller 2 makes 0 mistakes, X bar = 1 with probability 1/5 * 3/5 (from the original probability distribution) = 3/25.

The full list of every combination of X bar is shown above.

Sampling from a Normal Distribution - Example

Let X = the amount of money customers owe a firm. 'Mu' = Average = 40. The standard deviation of 'X' is 10. In this example we are told that X ~ N(40,100), this means that with a sample size of n = 25, X bar ~ N(40,4). (100/25 = 4)

If an auditor took a a random sample of 25, what is the probability that the average amount outstanding is less than 36.

P(X bar < 36) = P(Z < (36-40)/2) = P(Z < -2) = P(Z > 2) [By Symmetry] = 0.5 - P(0 < Z < 2) = 0.0228

Central Limit Theorem

The sampling distribution of the mean of a random sample will be ‘normal’ for a large sample size (typically n > 30).‘Large’ is dependent on the abnormality of the population’s distribution. To fully grasp the concept of the central limit theorem, we highly recommend you click here

Claire Auditing Example (Lecture Notes) – Sampling From a Non-Normal Distribution [1]

Interval Estimation

Produce a range of values with a degree of confidence attached - Confidence Interval.

Interpretation: If we repeatedly draw samples of size ‘n’, (1-α)% of the values of ¯X will be such that μ would lie within the confidence interval ¯X± Z_(α/2)×σ/√n

Example – If 100 samples were drawn, we would expect 95 of them to include the population mean. The confidence Level determines the multiple of standard errors of the end points

Clare Example, Using the interval of +/- 4, 95% Confidence Level, Sample Size = 250

Sampling Distribution of a Sample Proportion & Estimation

• Sample Proportion = P ̂ = X/n
• If ‘n’ =10 & ‘p’ = 0.4, the probability of P ̂≤0.5=0.8338
• If we assume a normal distribution
• Estimator: A Statistic whose purpose is to estimate a parameter
• Point Estimator: A formula for combining Sample information to estimate a parameter (single value)

Requirements & Rules

• NP and N(1-P)≥5
• E(P^)= P,
• V(P^)= (σp^)^2 = P(1-P)/N,
• Standard Deviation/Error = σp^ = √(P(1-P)/N)
• P^= X/n (X = number of successes, n = sample size)

Textbook Example (Page 326)

• n=300 p=0.52
• SD = 8.65, E(X) = 156, X = 150, P(Z > (X-(E(X) / SD)), P(Z > -0.69) = 0.7549
• OR
• P (P^) > 0.50) = (P^- p)/√(p(1-p)/n)= (0.5-0.52)/0.0288= .7549

End

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References

Textbook refers to Gerald Keller (2011), Statistics for Management and Economics (Abbreviated), 9th Edition,.

1. ASB, UNSW