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Hypothesis testing with the normal distribution

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Summary

Hypothesis testing with the normal distribution

In a nutshell

It is possible to use hypothesis testing to make a conclusion on whether or not the mean of a sample is equal to that of the population.

Sample mean distribution

Given a random variable $X \sim N(\mu ,\sigma^2)$ , the sample mean of $X$ is denoted by $\overline X$ , and is given by:

$\boxed{\overline X \sim N \left(\mu, \dfrac{\sigma^2}{n} \right)}$

Conducting a hypothesis test

To conduct a hypothesis test with a normal distribution, follow this procedure:

procedure

1.	Define the random variable.
2.	Write down the null and alternative hypotheses.
3.	Assume the null hypothesis to be true.
4.	Define the sample mean.
5.	Calculate the relevant probability.
6.	Make a conclusion on whether or not the null hypothesis is to be accepted by comparing the relevant probability against the significance level.

Example 1

A certain company sells coffee beans. The company claims each packet has $500g$ worth of coffee beans, with a standard deviation of $5g$ . A regular customer claims that the company is overstating the mass of coffee beans per packet. The customer takes a sample of $20$ packets, and finds that the mean mass of coffee is $497g$ . Is there evidence, at the $5\%$ level, that the customer's claim is correct?

First, define the variable:

Let $X$ be the mass of coffee in a packet.

$X \sim N(\mu, 5^2)$ 

State the null and alternative hypotheses:

$H_0: \mu =500, H_1: \mu \lt 500$ 

Assume the null hypothesis to be true:

$\mu = 500 \rightarrow X \sim N(500,5^2)$ 

Calculate the sample mean distribution:

$\begin{aligned}\overline X &\sim N\left(\mu, \dfrac{\sigma^2}{n}\right) \\& \sim N\left(500, \dfrac{5^2}{20}\right)\\ &\sim N\left(500, \left(\dfrac{\sqrt{5}}{2}\right)^2\right) \end{aligned}$ 

Calculate the relevant probability:

$P(\overline X \lt 497) = 0.0036$ 

Compare this to the significance level, and decide whether or not to reject $H_0$ :

$0.0036 \lt 0.05$ , so reject $H_0$ .

Conclude in context:

There is sufficient evidence, at the $5\%$ level, that the company is overstating the mass of coffee in their packets.

Finding critical values

To find critical values and critical regions, use the inverse normal distribution.

Example 2

A hypothesis test has the following information:

$X \sim N(\mu, 20^2)\\H_0: \mu = 50, H_1: \mu \neq 50$ 

Find the critical region(s) for this test, assuming a $10\%$ level of significance and a sample value of $25$ .

Because $H_1$ is in the form $\mu \neq ...$ , this is a two-tailed test. Therefore, this will have two critical values: call these $c_1$ and $c_2$ with critical regions $\overline X \lt c_1$ and $\overline X \gt c_2$ respectively.

Assume the null hypothesis to be true:

$\mu =50 \rightarrow X\sim N(50,20^2)$ 

Calculate the sample mean distribution:

$\begin{aligned}\overline X &\sim N\left(50, \dfrac{20^2}{25}\right)\\&\sim N(50, 4^2) \end{aligned}$ 

Use the inverse normal distribution to calculate $c_1$ and $c_2$ , remembering to halve the significance level at each end:

Want $c_1$ such that $P(\overline X \lt c_1 )= 0.05$ .

$c_1 = 43.4206$

Want $c_2$ such that $P(\overline X \gt c_2 )= 0.05$ .

Rearranging to get in the form $P(\overline X\lt ...)$ :

$\begin{aligned} P(\overline X \gt c_2) &= 0.05\\1 - P(\overline X \lt c_2)&=0.05\\P(\overline X \lt c_2)&=0.95 \end{aligned}$ 

$c_2 = 56.5794$ 

The critical region is therefore $\underline{\overline X \lt 43.4206}$ or $\underline{\overline X \gt 56.5794}$ .

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