Normal Distribution Hypothesis Tests

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Normal Distribution Hypothesis Tests

We have done a few normal hypothesis tests on an earlier page, Hypothesis Testing, but on this page we shall dive deeper. You need to know when to do a normal hypothesis test and also how to do a normal hypothesis test, as well as other skills such as handling multiple observations.

Make sure you are happy with the following topics before continuing.

Hypothesis Testing

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When to do a Normal Hypothesis Test

There are two types of hypothesis tests you need to know about: binomial distribution hypothesis tests and normal distribution hypothesis tests. In binomial hypothesis tests, you are testing the probability parameter $p$ . In normal hypothesis tests, you are testing the mean parameter $mu$ . This gives us a key difference that we can use to determine what test to do and when.

Since normal hypothesis tests test a mean parameter, words like mean, average and overall are all clues that you should use a normal hypothesis test. The situation and the context should also help – if you would model it with a normal distribution, you want to do a normal hypothesis test.

How to do a Normal Hypothesis Test
Define the parameter in the context of the question – for a normal hypothesis test the parameter is μmuμ which is always the mean of something.
Write down the null hypothesis and the alternate hypothesis.
Define the test statistic XXX in the context of the question.
Write down the distribution of XXX under the null hypothesis.
State the significance level αalphaα – even though you are likely given it in the question, not stating it risks losing a mark.
Test for significance or find the critical region.
Write a concluding sentence, linking the acceptance or rejection of H0H_{0}H0​ to the context.


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Handling Multiple Observations

If we are given multiple observations on which to base our hypothesis test, we can produce a more accurate result and have a larger critical region.

If we have a normally distributed variable $Xsim N(mu,sigma^{2})$ , the average of $n$ observations of $X$ has the distribution $bar{X}sim Nleft(mu,dfrac{sigma^{2}}{n}right)$ .

This means that if we are given multiple observations we can do the hypothesis test with the average of the observations rather than a single observation. Indeed, this is much better than using a single observation because it has a larger critical region.

Example: Multiple Observations

An office water cooler is said by the manufacturers to dispense an average of $250$ ml of water each use with a standard deviation of $5$ ml. Jordan suspects the manufacturer’s claim is too high, and observes the following amounts dispensed over five uses during a day:

$240$ ml, $235$ ml, $250$ ml, $245$ ml, $230$ ml

Test the manufacturer’s claim at the $5%$ level.

[7 marks]

$mu$ is the mean amount the water cooler dispenses in each use.

$H_{0}: mu=250$

$H_{1}: mu<250$

Test statistic $bar{X}$ is the mean of five observations of the water cooler.

Distribution is:

$begin{aligned}X&sim N(250,5^{2})[1.2em]&sim N(250,25)[1.2em]bar{X}&sim Nleft(250,dfrac{25}{5}right)[1.2em]&sim N(250,5)end{aligned}$

Significance level: $alpha=0.05$

Test statistic is $bar{X}=dfrac{1}{5}(240+235+250+245+230)=240$

$mathbb{P}(bar{X}leq 240)=3.872times 10^{-6}<0.05$

Reject $H_{0}$ . Sufficient evidence to suggest the manufacturer is not correct.

Normal Distribution Hypothesis Tests Example Questions

Question 1: Perform a two tail hypothesis test on $Xsim N(300,25)$ with significance level $1%$ , having observed $288$ .

[5 marks]

$mu$ is the mean of the distribution.

$H_{0}: mu=300$

$H_{1}: muneq 300$

Test statistic $X$ is the observed value.

$Xsim N(300,25)$

Significance level is $1%$ , however since this is a two tail test we are looking for probabilities of less than $0.005$ .

$mathbb{P}(Xleq 288)=0.3156>0.005$

Do not reject $H_{0}$ . Insufficient evidence to suggest $H_{1}$ .

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Question 2: The journey times by car to Brighton from London had an average of $63$ minutes in 2019, with a standard deviation of $2$ minutes. Journey times are thought to have decreased in the early 2020s. Ten drivers are asked to report their journey times upon arrival in Brighton. Their times are listed below. Test, at the $5%$ significance level, if journey times really are decreasing.

$59,61,58,62,56,56,61,64,69,68$

[6 marks]

$mu$ is the mean travel time to Brighton.

$H_{0}: mu=63$

$H_{1}: mu<63$

Test statistic $bar{X}$ is the mean travel time of $10$ observed journeys to Brighton.

Distribution is:

$begin{aligned}X&sim N(63,2^{2})[1.2em]&sim N(63,4)[1.2em]bar{X}&sim Nleft(63,dfrac{4}{10}right)[1.2em]&sim N(63,0.4)end{aligned}$

Significance level: $alpha=0.05$

Test statistic is $bar{X}=dfrac{1}{10}(59+61+58+62+56+56+61+64+69+68)=61.4$

$mathbb{P}(bar{X}leq 61.4)=0.0057<0.05$

Reject $H_{0}$ . Sufficient evidence to suggest journey times have decreased.

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Question 3: A biscuit company promises that their Prime biscuits are covered in a layer of chocolate at least $0.5$ cm thick. A newsagent buys thirty chocolate bars. He finds a standard deviation of $0.3$ cm and his chocolate bars have a mean thickness of $0.4$ cm. Is the manufacturer lying?

[8 marks]

$mu$ is the average thickness of the chocolate layer on the biscuit.

$H_{0}: mu=0.5$

$H_{0}: mu<0.5$

Test statistic $bar{X}$ is the mean of the newsagent’s observations.

Distribution: $Xsim N(0.5,0.3^{2})$

$begin{aligned}X&sim N(0.5,0.09)[1.2em]bar{X}&sim Nleft(0.5,dfrac{0.09}{30}right)[1.2em]&sim N(0.5,0.003)end{aligned}$

Significance level is not given in the question. Choose a sensible significance level: $alpha=5%$

$mathbb{P}(bar{X}leq 0.4)=0.0339<0.05$

Reject $H_{0}$ . Sufficient evidence to suggest the manufacturer is lying.

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Additional Resources

Exam Tips Cheat Sheet

A Level

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Formula Booklet

A Level

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Specification Points Covered

N2 – Understand and use the Normal distribution as a model; find probabilities using the Normal distribution Link to histograms, mean, standard deviation, points of inflection and the binomial distribution
O3 – Conduct a statistical hypothesis test for the mean of a Normal distribution with known, given or assumed variance and interpret the results in context