# 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.

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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.

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## 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 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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## Normal Distribution Hypothesis Tests 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.

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