Normal Distribution Hypothesis Tests

Normal Distribution Hypothesis Tests

A LevelAQAEdexcelOCRAQA 2022OCR 2022

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.

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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.

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How to do a Normal Hypothesis Test

  1. Define the parameter in the context of the question – for a normal hypothesis test the parameter is \mu which is always the mean of something.
  2. Write down the null hypothesis and the alternate hypothesis.
  3. Define the test statistic X in the context of the question.
  4. Write down the distribution of X under the null hypothesis.
  5. State the significance level \alpha – even though you are likely given it in the question, not stating it risks losing a mark.
  6. Test for significance or find the critical region.
  7. Write a concluding sentence, linking the acceptance or rejection of H_{0} 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 X\sim N(\mu,\sigma^{2}), the average of n observations of X has the distribution \bar{X}\sim N\left(\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.

A LevelAQAEdexcelOCR
A Level AQA Edexcel OCR

Example: Multiple Observations

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

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

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 N\left(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)=3.872\times 10^{-6}<0.05

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

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Example Questions

\mu is the mean of the distribution.

H_{0}: \mu=300

H_{1}: \mu\neq 300

Test statistic X is the observed value.

X\sim 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}(X\leq 288)=0.3156>0.005

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

\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 N\left(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.

\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: X\sim N(0.5,0.3^{2})

\begin{aligned}X&\sim N(0.5,0.09)\\[1.2em]\bar{X}&\sim N\left(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.

Related Topics


Hypothesis Testing

A Level

Additional Resources


Exam Tips Cheat Sheet

A Level

Formula Booklet

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