# Normal Distribution Hypothesis Tests

## Normal Distribution Hypothesis Tests Revision

**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.

**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 \mu which is**always the mean**of something. - Write down the
**null hypothesis**and the**alternate hypothesis**. - Define the
**test statistic**X in the**context of the question**. - Write down the
**distribution**of X under the**null hypothesis**. - State the
**significance level**\alpha – 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 H_{0} to the**context**.

**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.

**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}<0.05Reject 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 X\sim 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}: \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>0.005Do not reject H_{0}. Insufficient evidence to suggest H_{1}.

**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 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<0.05Reject H_{0}. Sufficient evidence to suggest journey times have decreased.

**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: 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<0.05Reject H_{0}. Sufficient evidence to suggest the manufacturer is lying.