#### Sampling and Estimation

__Sample and Population__

__Sample and Population__

- Population: is a collection of all items
- Sample: is a subset of population used as a representation of the entire population

__Central Limit Theorem__

__Central Limit Theorem__

- If (X
_{1}, X_{2}, …, X_{n}) 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})

- X ~ N (μ,

__Examples__

__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

- 1 – P(Z <

- 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

- Variance of sample =
- 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

- P(Z <

- Find, to 4 decimal places, the probability that a trout chosen at random weighs more than 1.25 kg

__Point Estimate and Confidence Interval__

__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__

__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__

__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__

__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

- By the Central Limit Theorem, the distribution of x̄ will be approximately normal, so the same method as above
**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

- As the population variance is unknown, you must first estimate the population variance, s, using the sample data

__Examples__

__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

- √n =

- For a 99% confidence interval, find z where Φ(z) = 0.995 (think of the 1% cut from both sides):

__Confidence Interval for a Population Proportion__

__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__

__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

- 0.1129 +
- The mid-point is equal to the proportion of people with high-speed internet use, so:
^{87}/_{n}= 0.145- ∴ n = 600

- Find the midpoint of the limits, finding p:
- 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:
- 1 – 2(1 – 0.9872) = 0.9744
- Hence α% confidence is 97.44%

- Using the upper limit, this was calculated by:

- Write the mid-point of this confidence interval and hence find the value of n