Mean of a Sample & Confidence Intervals (AQA Level 3 Mathematical Studies (Core Maths))

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Author

Naomi C

Expertise

Maths

Mean of a Sample

What is the sample mean?

For any given population it can often be difficult or impractical to find the true value of the population mean, $μ$
- The population could be too large to collect data using a census
- Collecting the data could compromise the individual data values and therefore taking a census could destroy the population
The population mean can be estimated by taking the mean from a sample from within the population
- The mean of a random sample is known as the point estimate for the mean of the population
- A point estimate is a 'best guess' of the mean of the population
A bigger sample size will most likely lead to a more accurate point estimate
- A sample size of 1, for example, will not give a very good estimate of the mean of the population

What is the distribution of the sample means?

If a number of samples are taken from a population with a normal distribution, then the means of these samples will themselves have a normal distribution
For a random variable, $X$ , whose population has a distribution $N (μ, σ^{2})$ , the normal distribution of the sample means, $X$ , will be $N (μ, \frac{σ^{2}}{n})$
- The distribution of the population and the sample means have the same mean, $μ$
- The distribution of the sample means has variance $\frac{σ^{2}}{n}$ , where $n$ is the sample size
- Therefore, the standard deviation of the sample means is $\frac{σ}{\sqrt{n}}$ , which is the standard deviation of the population divided by the square root of the sample size
The standard deviation of the sample means is also known as the standard error of the mean

Worked Example

The weights of bags of sugar are normally distributed with mean 500 g and variance 25 g.

(a) What is the probability that the weight of a bag of sugar chosen at random is less than 496 g?

Find $P (X < 492)$ , given $N (500, 25)$ using your calculator

Lower bound $= - 99999999$
Upper bound $= 496$
$μ = 500$
$σ = \sqrt{25} = 5$

$P (X < 496) = 0.211855 . . .$

(Alternatively you can find the z-value of 496 g and use the statistical tables)

Convert to a percentage and round to 3 significant figures

$21.2 %$ (3 s.f.)

Bags of sugar are shipped to supermarkets in boxes. Each box contains 8 bags of sugar.

(b) What is the probability that for a randomly selected box, the mean weight of a bag of sugar is less than 496 g?

The distribution of the sample means will be $N (μ, \frac{σ^{2}}{n})$

$N (500, \frac{25}{8})$

So using the calculator, find $P (X < 496)$

Lower bound $= - 99999999$
Upper bound $= 496$
$μ = 500$
$\sqrt{\frac{σ^{2}}{n}} = \sqrt{\frac{25}{8}}$

$P (X < 492) = 0.011825 . . .$

Convert to a percentage and round to 3 significant figures

$1.18 %$ (3 s.f.)

Confidence Intervals

A range of possibilities is often given rather than a single value, this is because we cannot always be exact when dealing with probabilities
The range of possibilities around a measurement is called the confidence interval
- A confidence interval describes how precise the measurement is
- A confidence interval of 95% means that 95% of the data falls within the given range

Exam Tip

When tackling an exam question, think about the context and the impact of a result being outside a given range.

For example, when we need to ensure that something is accurate, such as the safety of a piece of equipment, we would expect a high confidence interval (99%), but for something less vital, such as the results of a class test, a lower confidence interval (90%) would be acceptable.

A confidence interval is described using the notation $[a, b]$ , where $a$ is the lowest values in the range and $b$ is the greatest value in the range
If a percentage of the data falls within a number of standard deviations of the mean for a normal distribution
- Then for samples from that distribution, the same percentage of the data falls within the same number of standard errors of the mean
You can work out the confidence interval for a population with a normal distribution
- Find the z-value associated with the given percentage for the confidence interval
- Add/subtract the product of the z-value and the standard deviation to/from the mean
The confidence interval for a sample can be found in a similar way
- Find the z-value associated with the given percentage for the confidence interval
- Add/subtract the product of the z-value and the standard error to/from the mean
The confidence interval is always assumed to be symmetrical about the mean for a normal distribution
The confidence level required (%) and the sample size will always be given to you in an exam

Exam Tip

It can be useful to remember the key confidence intervals for speed in an exam:

Confidence interval for 99% of the distribution, $[- 2.58, 2.58]$
Confidence interval for 95% of the distribution, $[- 1.96, 1.96]$
Confidence interval for 90% of the distribution, $[- 1.64, 1.64]$

However, you can use the statistical tables in the data booklet to help you work them out if you can't remember them.

Worked Example

The reaction times of a population is normally distributed with mean $μ$ milliseconds and standard deviation 75 milliseconds.

A sample of 60 people had a total reaction time of 24360 milliseconds.

(a) What is the point estimate for the sample?

Divide the total reaction time by the number of people in the sample to find the point estimate

$\frac{24360}{60} = 406$

406 milliseconds

(b) Construct a 98% confidence interval for the mean reaction time for $μ$ .

Using the statistical tables find z for $Φ (z) = 0.98$

$Φ (2.0537) = 0.98$

Remember, the point estimate for the sample is the same as the mean for the population

Find the confidence interval for the population by multiplying the z-value by the standard deviation and adding to/subtracting from the mean

$406 - 2.0537 \times 75 = 251.9725 406 + 2.0537 \times 75 = 560.0275$

Round appropriately

$[560, 252]$

Find the standard error of the sample, $\frac{σ}{\sqrt{n}}$

$\frac{75}{\sqrt{60}}$

Find the confidence interval for the mean reaction time in the sample using the same z-value

$406 - 2.0537 \times \frac{75}{\sqrt{60}} = 386.11513 . . . 406 + 2.0537 \times \frac{75}{\sqrt{60}} = 425.88486 . . .$

Round appropriately

$[386, 426]$

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Mean of a Sample & Confidence Intervals (AQA Level 3 Mathematical Studies (Core Maths))

Revision Note

Mean of a Sample

What is the sample mean?

What is the distribution of the sample means?

Worked Example

Confidence Intervals

Exam Tip

Exam Tip

Worked Example

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Author: Naomi C