BUS204: Business Statistics

The comparison of two independent population means is very common and provides a way to test the hypothesis that the two groups differ from each other. Is the night shift less productive than the day shift, are the rates of return from fixed asset investments different from those from common stock investments, and so on? An observed difference between two sample means depends on both the means and the sample standard deviations. Very different means can occur by chance if there is great variation among the individual samples. The test statistic will have to account for this fact. The test comparing two independent population means with unknown and possibly unequal population standard deviations is called the Aspin-Welch t-test. The degrees of freedom formula we will see later was developed by Aspin-Welch.

When we developed the hypothesis test for the mean and proportions we began with the Central Limit Theorem. We recognized that a sample mean came from a distribution of sample means, and sample proportions came from the sampling distribution of sample proportions. This made our sample parameters, the sample means and sample proportions, into random variables. It was important for us to know the distribution that these random variables came from. The Central Limit Theorem gave us the answer: the normal distribution. Our Z and t statistics came from this theorem. This provided us with the solution to our question of how to measure the probability that a sample mean came from a distribution with a particular hypothesized value of the mean or proportion. In both cases that was the question: what is the probability that the mean (or proportion) from our sample data came from a population distribution with the hypothesized value we are interested in?

Now we are interested in whether or not two samples have the same mean. Our question has not changed: Do these two samples come from the same population distribution? To approach this problem we create a new random variable. We recognize that we have two sample means, one from each set of data, and thus we have two random variables coming from two unknown distributions. To solve the problem we create a new random variable, the difference between the sample means. This new random variable also has a distribution and, again, the Central Limit Theorem tells us that this new distribution is normally distributed, regardless of the underlying distributions of the original data. A graph may help to understand this concept.

Figure 10.2

Pictured are two distributions of data, X₁ and X₂, with unknown means and standard deviations. The second panel shows the sampling distribution of the newly created random variable $(\overline X_1−\overline X_2)$. This distribution is the theoretical distribution of many sample means from population 1 minus sample means from population 2. The Central Limit Theorem tells us that this theoretical sampling distribution of differences in sample means is normally distributed, regardless of the distribution of the actual population data shown in the top panel. Because the sampling distribution is normally distributed, we can develop a standardizing formula and calculate probabilities from the standard normal distribution in the bottom panel, the Z distribution.

The Central Limit Theorem, as before, provides us with the standard deviation of the sampling distribution, and further, that the expected value of the mean of the distribution of differences in sample means is equal to the differences in the population means. Mathematically this can be stated:

$E(µ\overline x_1− µ \overline x_2)=µ_1−µ_2$

Because we do not know the population standard deviations, we estimate them using the two sample standard deviations from our independent samples. For the hypothesis test, we calculate the estimated standard deviation, or standard error, of the difference in sample means, $\overline X_1−\overline X_2$.

The standard error is:

$\sqrt{\dfrac{(s_1)^2}{n_1}+\dfrac{(s_2)^2}{n_2}}$

We remember that substituting the sample variance for the population variance when we did not have the population variance was the technique we used when building the confidence interval and the test statistic for the test of hypothesis for a single mean back in Confidence Intervals and Hypothesis Testing with One Sample. The test statistic (t-score) is calculated as follows:

$t_c=\dfrac{(\overline x_1–\overline x_2)–δ_0}{\sqrt{\dfrac{(s_1)^2}{n_1}+\dfrac{(s_2)^2}{n_2}}}$

where:

• s₁ and s₂, the sample standard deviations, are estimates of σ₁ and σ₂, respectively and
• σ₁ and σ₂ are the unknown population standard deviations.
• $\overline x_1$ and $\overline x_2$ are the sample means. μ₁ and μ₂ are the unknown population means.

The number of degrees of freedom (df) requires a somewhat complicated calculation. The df are not always a whole number. The test statistic above is approximated by the Student's t-distribution with df as follows:

The standard error is:

$df=\dfrac{(\dfrac{(s_1)^2}{n_1}+\dfrac{(s_2)^2}{n_2})^2}{(\dfrac{1}{n_1–1})(\dfrac{(s_1)^2}{n_1})^2+(\dfrac{1}{n_2–1})(\dfrac{(s_2)^2}{n_2})^2}$

When both sample sizes n1 and n2 are 30 or larger, the Student's t approximation is very good. If each sample has more than 30 observations then the degrees of freedom can be calculated as n1 + n2 - 2.

The format of the sampling distribution, differences in sample means, specifies that the format of the null and alternative hypothesis is:

$H_0:µ_1−µ_2=δ_0$

$H_a:µ_1−µ_2≠δ_0$

where $δ_0$ is the hypothesized difference between the two means. If the question is simply “is there any difference between the means?” then $δ_0 = 0$ and the null and alternative hypotheses becomes:

$H_0:µ_1=µ_2$

$H_a:µ_1≠µ_2$

An example of when $δ_0$ might not be zero is when the comparison of the two groups requires a specific difference for the decision to be meaningful. Imagine that you are making a capital investment. You are considering changing from your current model machine to another. You measure the productivity of your machines by the speed they produce the product. It may be that a contender to replace the old model is faster in terms of product throughput, but is also more expensive. The second machine may also have more maintenance costs, setup costs, etc. The null hypothesis would be set up so that the new machine would have to be better than the old one by enough to cover these extra costs in terms of speed and cost of production. This form of the null and alternative hypothesis shows how valuable this particular hypothesis test can be. For most of our work we will be testing simple hypotheses asking if there is any difference between the two distribution means.

Example 10.1

Independent groups
The Kona Iki Corporation produces coconut milk. They take coconuts and extract the milk inside by drilling a hole and pouring the milk into a vat for processing. They have both a day shift (called the B shift) and a night shift (called the G shift) to do this part of the process. They would like to know if the day shift and the night shift are equally efficient in processing the coconuts. A study is done sampling 9 shifts of the G shift and 16 shifts of the B shift. The results of the number of hours required to process 100 pounds of coconuts is presented in Table 10.1. A study is done and data are collected, resulting in the data in Table 10.1.

NOTE

Example 10.2

• $H_o:μ_A≤μ_B$
• $H_a:μ_A>μ_B$

Example 10.3