# 10.3: Two Population Means with Known Standard Deviations

Even though this situation is not likely (knowing the population standard deviations is not likely), the following example illustrates hypothesis testing for independent means, known population standard deviations. The sampling distribution for the difference between the means is normal and both populations must be normal. The random variable is $$\bar{X}_{1} - \bar{X}_{2}$$. The normal distribution has the following format:

Normal distribution is:

The standard deviation is:

$\sqrt{\dfrac{(\sigma_{1})^{2}}{n_{1}} + \dfrac{(\sigma_{2})^{2}}{n_{2}}}\label{eq2}$

The test statistic (z-score) is:

Example $$\PageIndex{1}$$

Independent groups, population standard deviations known: The mean lasting time of two competing floor waxes is to be compared. Twenty floors are randomly assigned to test each wax. Both populations have a normal distributions. The data are recorded in Table.

Wax Sample Mean Number of Months Floor Wax Lasts Population Standard Deviation
1 3 0.33
2 2.9 0.36

Does the data indicate that wax 1 is more effective than wax 2? Test at a 5% level of significance.

This is a test of two independent groups, two population means, population standard deviations known.

Random Variable: $$\bar{X}_{1} - \bar{X}_{2} =$$ difference in the mean number of months the competing floor waxes last.

• $$H_{0}: \mu_{1} \leq \mu_{2}$$
• $$H_{a}: \mu_{1} > \mu_{2}$$

The words "is more effective" says that wax 1 lasts longer than wax 2, on average. "Longer" is a “>” symbol and goes into $$H_{a}$$. Therefore, this is a right-tailed test.

Distribution for the test: The population standard deviations are known so the distribution is normal. Using Equation \ref{eq1}, the distribution is:

Since $$\mu_{1} \leq \mu_{2}$$ then $$\mu_{1} - \mu_{2} \leq 0$$ and the mean for the normal distribution is zero.

Calculate the $$p\text{-value}$$ using the normal distribution: $$p\text{-value} = 0.1799$$

Graph: