Sample: is a subset of population used as a representation of the entire population
Central Limit Theorem
If (X1, X2, …, Xn) is a random sample of size n, drawn from any population with mean μ and variance σ², then the sample has:
Expected mean, μ
Expected variance, σ²/n
It forms a normal distribution:
X ~ N (μ, σ²/n)
Examples
The weights of the trout at a trout farm are normally distributed with mean 1 kg and standard deviation 0.25 kg
Find, to 4 decimal places, the probability that a trout chosen at random weighs more than 1.25 kg
Write down the distribution: X ~ N (1, 0.25²)
Write down the desired probability:
P(X > 1.25) = 1 – P(X < 1.25)
Standardize and evaluate:
1 – P(Z < 1.25 – 1/0.25) = 0.1587
If X̄ kg represents the mean weight of a sample of 10 trouts chosen at random, state the distribution of Ȳ, evaluate the mean and variance. Find the probability that the mean weight of a sample of 10 trouts will be less than 0.9 kg
Write down the distribution: X ~ N (1, 0.25²)
For a sample, the mean remains equal but the variance changes. Find the new variance:
Variance of sample = σ²/n = 0.25²/10 = 0.00625
Write the distribution of the sample: Y ~ N (1, 0.00625)
Write the desired probability: P(Y < 0.9)
Standardize and evaluate:
P(Z < 0.9 – 1/0.00625)
= 1 – P(Z < 0.1/0.00625)
= 0.103
Point Estimate and Confidence Interval
A point estimate is a numerical value calculated from a set of data (sample) which is used as an estimate of an unknown parameter in a population
Examples of point estimates are:
Sample mean, x̄ estimates the population mean, μ
Sample proportion, r/n estimates the population proportion, p
Sample variance, s² estimates the population variance, σ²
Point estimates are close to the population value but are not the value
We can determine a confidence interval where the population value is likely to lie in (x̄ – δ, x̄ + δ)
The Variance
Variance can be calculated/given for either a sample or a population and there is a difference between them
Using the divisor, n: This is appropriate when:
the data is given for the whole population and you are interested in the variance of the whole
the data is given for the sample and you are interested in the variance of just the sample
Using the divisor (n – 1):
Appropriate to use when the data is given for a sample and you are estimating the variance of the whole population
The quantity of calculated, s², is known as the unbiased estimate of the population variance
Percentage Points for a Normal Distribution
The percentage points are determined by finding the z-value of specific percentages
For example, to find the z-value of a 95% confidence level, we can see that the 5% would be removed equally from both sides (2.5% on each side), so the z-value we would actually be finding would be of 100% – 2.5% = 97.5%
Confidence Interval for a Population Mean
A Sample Taken from a Normal Population Distribution with Known Population Variance:
z is the value corresponding to the confidence level required and n is the sample size
The confidence interval calculated is exact
Large Sample Taken from an Unknown Population Distribution with Known Population Variance:
By the Central Limit Theorem, the distribution of x̄ will be approximately normal, so the same method as above
The confidence interval calculated is an approximate
Large Sample Taken from an Unknown Population Distribution with Unknown Population Variance:
As the population variance is unknown, you must first estimate the population variance, s, using the sample data
The confidence interval is an approximate
Examples
Heights of a certain species of animal are normally distributed with σ = 0.17 m. Obtain a 99% confidence interval for the population mean, with the total width less than 0.2 m. Find the smallest sample size required
For a 99% confidence interval, find z where Φ(z) = 0.995 (think of the 1% cut from both sides):
z = 2.576
Subtract the limits of the interval and equate to 0.2:
Substitute the information given and find n:
√n = 0.2/2 × 2.576 × 0.17
n = 4126.53 ≈ 4130
Confidence Interval for a Population Proportion
Calculating the confidence interval from a random sample of n observations from a population in which the proportion of successes is p and the proportion if failures is q
The observed proportion of success p is r/n, where r represents the number of successes
Examples
A random sample of n people were questioned about their internet use. 87 of them had a high-speed internet connection. A confidence interval for the population proportion having high-speed internet connection is 0.1129 < p < 0.1771.
Write the mid-point of this confidence interval and hence find the value of n
Find the midpoint of the limits, finding p:
0.1129 + 0.1771 – 0.1129/2 = 0.145
The mid-point is equal to the proportion of people with high-speed internet use, so:
87/n = 0.145
∴ n = 600
This interval is an α% confidence interval. Find α
Using the upper limit, this was calculated by:
0.1771 = 0.145 + z[√(87/600 × 513/600) ÷ 600]
∴ z = 2.233
Use normal tables and find the corresponding probability
Φ(z) = 0.9872
The same area is chopped off from both sides of the graph, so: