​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

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:

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