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Hypothesis Testing (Normal Distribution) (AQA A Level Maths: Statistics): Flashcards

Exam code: 7357

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FrontSample Mean Distribution
Why is the population mean usually estimated from a sample rather than measured directly?
BackSample Mean Distribution
Because taking a census is often impractical or impossible: the population may simply be too large to collect data from every member.
Or collecting the data may destroy or compromise what is being measured, so that testing every item would leave nothing behind: the lifetime of a light bulb can only be found by running it until it fails.
So the mean of a sample is used as an estimate of $μ$ , and the point of this subtopic is working out how good an estimate that is.
BackSample Mean Distribution
The mean is unchanged because a sample is as likely to come out above $μ$ as below it, so sample means centre on the same value the population does.
The variance is reduced because averaging dilutes extreme values. One unusually large value in a sample of 20 is divided by 20 before it reaches the sample mean, so it moves the mean far less than it would move a single observation.
Sample means are therefore clustered more tightly around $μ$ than individual values are, which is exactly what makes a sample mean a useful estimate.

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Why is the population mean usually estimated from a sample rather than measured directly?
Because taking a census is often impractical or impossible: the population may simply be too large to collect data from every member.
Or collecting the data may destroy or compromise what is being measured, so that testing every item would leave nothing behind: the lifetime of a light bulb can only be found by running it until it fails.
So the mean of a sample is used as an estimate of $μ$ , and the point of this subtopic is working out how good an estimate that is.
A sample of size $n$ is taken from a population $X ~ N (μ, σ^{2})$ . Complete the distribution of the sample mean:
$\bar{X} ~ N (______,______)$
The completed distribution is:
$\bar{X} ~ N (μ, \frac{σ^{2}}{n})$
The mean is unchanged and the variance is divided by the sample size.
$\bar{X}$ is the distribution of all the values the sample mean could take, over every sample of size $n$ that might have been drawn. It is a normal distribution in its own right, narrower than the population it came from.
Why does taking a sample mean leave the mean unchanged but reduce the variance?
The mean is unchanged because a sample is as likely to come out above $μ$ as below it, so sample means centre on the same value the population does.
The variance is reduced because averaging dilutes extreme values. One unusually large value in a sample of 20 is divided by 20 before it reaches the sample mean, so it moves the mean far less than it would move a single observation.
Sample means are therefore clustered more tightly around $μ$ than individual values are, which is exactly what makes a sample mean a useful estimate.
What is the standard deviation of the distribution of the sample mean, and how does it depend on the sample size?
It is
$\frac{σ}{\sqrt{n}}$
which comes from square rooting the variance $\frac{σ^{2}}{n}$ .
It is inversely proportional to the square root of the sample size, not to the sample size itself. So the larger the sample, the narrower the distribution of the sample means and the more reliable the estimate, but the improvement slows down as $n$ grows.
True or False?
Doubling the sample size halves the standard deviation of the sample mean.
False.
The standard deviation is $\frac{σ}{\sqrt{n}}$ , so doubling $n$ divides it by $\sqrt{2}$ , which is about 1.41.
To halve it you have to quadruple the sample size, since $\sqrt{4} = 2$ .
That is the practical cost of precision: each further halving of the spread needs four times as much data as the last one.
A random sample of 10 observations is taken from $X ~ N (30, 25)$ . What is the distribution of the sample mean?
Divide the variance by the sample size:
$\bar{X} ~ N (30, \frac{25}{10}) = N (30, 2.5)$
The 25 in the original bracket is the variance, so the population standard deviation is 5 and the standard deviation of the sample mean is $\frac{5}{\sqrt{10}} = 1.58$ to 3 significant figures.
Check which one a question has given you before dividing: a distribution written as $N (30, 5^{2})$ means the same thing, but $N (30, 5)$ would not.

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