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

Exam code: 9MA0

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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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Cards in this collection (13)

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.
In a hypothesis test on the mean of a normal distribution, what is the parameter being tested and what is the test statistic?
The parameter is the population mean $μ$ , so both hypotheses are statements about $μ$ , and $H_{0}$ always has the form $H_{0} : μ = \dots$
The test statistic is the sample mean $\bar{x}$ , and assuming $H_{0}$ is true its distribution is
$\bar{X} ~ N (μ, \frac{σ^{2}}{n})$
Define $μ$ in words first if the question has not, exactly as you would define $p$ in a binomial test.
Why must the test use $\frac{σ^{2}}{n}$ rather than the population variance $σ^{2}$ ?
Because the evidence is a sample mean, not a single observation, and sample means vary far less than individual values do.
Using $σ^{2}$ would test it against the spread of the whole population, which is much too wide: with $σ = 9$ and $n = 12$ the sample mean's standard deviation is only $\frac{9}{\sqrt{12}} = 2.60$ .
Getting this wrong makes a significant result look insignificant, and it is the characteristic mistake in this topic.
Complete the standardised test statistic for a hypothesis test on the mean of a normal distribution:
$z = \frac{\bar{x} - μ}{______}$
The completed formula is:
$z = \frac{\bar{x} - μ}{\frac{σ}{\sqrt{n}}}$
The denominator is the standard deviation of the sample mean, not of the population, and writing $σ$ there instead is the commonest slip.
It is the ordinary standardising formula applied to $\bar{X}$ rather than to $X$ , so it measures how many sample-mean standard deviations the observed mean is away from the mean claimed by $H_{0}$ .
How do you find the critical value for a test on the mean of a normal distribution?
Write down the distribution of $\bar{X}$ assuming $H_{0}$ is true, then use the inverse normal function with an area equal to the significance level in the tail being tested, remembering $\frac{σ}{\sqrt{n}}$ as the standard deviation, not $σ$ .
For a two-tailed test there are two critical values, one in each tail, each found from $\frac{α}{2} %$ .
Unlike the binomial case the two are symmetrical, so they should be the same distance either side of the mean, and a sketch with the critical region shaded makes the comparison with the observed value much harder to get backwards.
Puzzle times are $X ~ N (204, 81)$ minutes and a sample of 12 gives a mean of 201, against a claim that the time has decreased. How is the test carried out at the 5% level?
With $H_{0} : μ = 204$ and $H_{1} : μ < 204$ , assuming $H_{0}$ gives
$\bar{X} ~ N (204, \frac{9^{2}}{12})$
since the 81 is the variance and so $σ = 9$ ; then, testing for a decrease, find the probability of a value that low or lower:
$P (\bar{X} \leq 201) = 0.1241$
Since $0.1241 > 0.05$ , do not reject $H_{0}$ : there is insufficient evidence at the 5% level that the time taken has decreased.
Never write "accept $H_{0}$ ": failing to find evidence of a decrease is not evidence that there was none.
A question about a change in the mean adds that you may assume the variance is unchanged. Why does that assumption have to be stated?
Because the test needs a known value of $σ^{2}$ to build the distribution of $\bar{X}$ under $H_{0}$ .
Whatever caused the mean to change might have changed the spread as well, and if the variance were also unknown then $\bar{X} ~ N (μ, \frac{σ^{2}}{n})$ would have two unknowns in it and the test could not be run at all.
So the assumption is not a formality: it is what makes the null hypothesis a complete enough statement to calculate with.
True or False?
The same sample mean can be significant with one sample size and not significant with another.
True.
The test compares the sample mean against a distribution whose standard deviation is $\frac{σ}{\sqrt{n}}$ , and that gets smaller as $n$ grows.
So a fixed distance below the claimed mean sits further out in the tail of a narrower distribution: the same observed mean becomes more extreme, and the probability of it or anything more extreme falls.
A difference too small to be convincing from a sample of 12 can be convincing from a sample of 200, which is why the sample size has to be reported alongside any conclusion.

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