# Hypothesis Test Concerning the Equality of the Population Means

Analysts are often interested in establishing whether there exists a significant difference between the means of two different populations. For instance, they might want to know whether the average returns for two subsidiaries of a given company exhibit a significant variance. Such a test may then be used to make decisions regarding resource allocation or the rewarding of the directors. Before embarking on such an exercise, it is paramount to ensure that the samples taken are not only independent but also sourced from normally distributed populations. It can either be assumed that the population variances are equal or unequal. In this reading, we will assume that the population variances are equal.

Assume that $$\mu_1$$ is the mean of the first population while $$\mu_2$$ is the mean of the second population. In testing the equality of two population means, we wish to determine if they are equal or not. As such, the hypotheses can be any of the following:

I. Two-sided

$$H_0: µ_1 – µ_2 = 0\ \text{vs.}\ H_a∶ µ_1 – µ_2 ≠ 0,$$

This can be stated as:

$$H_0: µ_1 = µ_2\ \text{vs.}\ H_a ∶ µ_1 ≠ µ_2$$

II. One-sided (right)

$$H_0: µ_1 – µ_2 \leq 0\ \text{vs.}\ H_a∶ µ_1 – µ_2 > 0,$$

This can be stated as:

$$H_0: µ_1 \leq µ_2\ \text{vs.}\ H_a ∶ µ_1 > µ_2$$

III. One-sided (Left)

$$H_0: µ_1 – µ_2 \geq 0\ \text{vs.}\ H_a∶ µ_1 – µ_2 < 0,$$

This can be stated as:

$$H_0: µ_1 \geq µ_2\ \text{vs.}\ H_a ∶ µ_1 < µ_2$$

However, note that tests such as $$H_0: µ_1 – µ_2 =3\ vs. H_a∶ µ_1 – µ_2 ≠ 3$$. The process is similar in both cases.

### Test Statistic for Testing Differences in Population Mean

When testing for the difference between two population means, we assume that the two populations are distributed normally. Further, we assume that they have equal and unknown variances. We always make use of the student’s t-distribution where the test statistic is given by:

$$t=\frac{\left(\bar{X}_{1}-\bar{X}_{2}\right)-\left(\mu_{1}-\mu_{2}\right)}{\sqrt{\frac{s_{p}^{2}}{n_{1}}+\frac{s_{p}^{2}}{n_{2}}}}=\frac{\left(\bar{X}_{1}-\bar{X}_{2}\right)-\left(\mu_{1}-\mu_{2}\right)}{s_p\left(\sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}}}\right)}$$

Where $$s_p^2$$ is pooled estimator of the common variance and is given by:

$$s_{p}^{2}=\frac{\left(n_{1}-1\right) s_{1}^{2}+\left(n_{2}-1\right) s_{2}^{2}}{n_{1}+n_{2}-2}$$

Also, the variables are defined as follows:

$$\bar{X}_1$$ = Mean of the first sample.

$$\bar{X}_2$$ = Mean of the second sample.

$$s_1^2$$ = Variance of the first sample.

$$s_2^2$$ = Variance of the second sample.

$$n_1$$ = Sample size of the first sample.

$$n_2$$ = Sample size of the second sample.

#### Example 1: Hypothesis Test Concerning the Equality of the Population Means

Nutritionists want to establish whether obese patients on a new special diet have a lower weight than the control group. After six weeks, the average weight of 10 patients (group A) on the special diet is 75kg, while that of 10 more patients of the control group (B) is 72kg. Carry out a 5% test to determine if the patients on the special diet have a lower weight.

Additional information: $$\sum A^2 = 59520$$ and $$\sum B^2 =56430$$.

Solution

As is the norm, start by stating the hypothesis:

$$H_0: µ_1 – µ_2 = 0\ \text{Vs}\ H_a∶ µ_1 – µ_2 ≠ 0,$$

We assume that the two samples have equal variances, are independent, and distributed normally. Then, under H0,

$$\frac { \bar { B } -\bar { A } }{ S\sqrt { \frac { 1 }{ m } +\frac { 1 }{ n } } } \sim { t }_{ m+n-2 }$$

Note that the sample variance is given by:

$$s^2=\frac{\sum X^2 -n\bar{X}^2}{n-1}$$

So,

\begin{align*} { S }_{ A }^{ 2 } & =\frac { \left\{ 59520-{ \left( 10\ast { 75 }^{ 2 } \right) } \right\} }{ 9 } =363.33 \\ { S }_{ B }^{ 2 } & =\frac { \left\{ 56430-{ \left( 10\ast { 72}^{ 2 } \right) } \right\} }{ 9 } =510 \\ \end{align*}

$$Therefore,$$ S^p_2 =\cfrac {(9 × 363.33 + 9 × 510)}{(10 + 10 -2)} = 436.665 $$And$$ \text{Test statistic} =\cfrac {(75 -72)}{ \left\{ \sqrt{439.665} × \sqrt{ \left(\frac {1}{10} + \frac {1}{10}\right)} \right\} }= 0.3210 

Our test statistic (0.3210) is less than the upper 5% point (1.734) of the t-distribution with 18 degrees of freedom.

Therefore, we do not have sufficient evidence to reject the H0 at 5% significance. As such, it is reasonable to conclude that the special diet has the same effect on body weight as the placebo.

Note to candidates: You could choose to work with the p-value and determine P(t18 > 0.937) and then establish whether this probability is less than 0.05. Working out the problem this way would lead to the same conclusion as above.

#### Example 2: Hypothesis Test Concerning the Equality of the Population Means

Suppose we replace ‘>’ with ‘≠’ in H1 in the example above, would the decision rule change?

Replacing ‘>’ with ‘≠’ in H1 would change the test from a one-tailed one to a two-tailed test. We would compute the test statistic just as demonstrated above. However, we would have to divide the level of significance by two and compare the test statistic to both the lower and upper 2.5% points of the t18-distribution (±2.101).

Since our test statistic lies within these limits (non-rejection region), the decision rule would, therefore, remain unchanged.

Shop CFA® Exam Prep

Offered by AnalystPrep

Featured Shop FRM® Exam Prep Learn with Us

Subscribe to our newsletter and keep up with the latest and greatest tips for success
Shop Actuarial Exams Prep Shop Graduate Admission Exam Prep

Sergio Torrico
2021-07-23
Excelente para el FRM 2 Escribo esta revisión en español para los hispanohablantes, soy de Bolivia, y utilicé AnalystPrep para dudas y consultas sobre mi preparación para el FRM nivel 2 (lo tomé una sola vez y aprobé muy bien), siempre tuve un soporte claro, directo y rápido, el material sale rápido cuando hay cambios en el temario de GARP, y los ejercicios y exámenes son muy útiles para practicar.
diana
2021-07-17
So helpful. I have been using the videos to prepare for the CFA Level II exam. The videos signpost the reading contents, explain the concepts and provide additional context for specific concepts. The fun light-hearted analogies are also a welcome break to some very dry content. I usually watch the videos before going into more in-depth reading and they are a good way to avoid being overwhelmed by the sheer volume of content when you look at the readings.
Kriti Dhawan
2021-07-16
A great curriculum provider. James sir explains the concept so well that rather than memorising it, you tend to intuitively understand and absorb them. Thank you ! Grateful I saw this at the right time for my CFA prep.
nikhil kumar
2021-06-28
Very well explained and gives a great insight about topics in a very short time. Glad to have found Professor Forjan's lectures.
Marwan
2021-06-22
Great support throughout the course by the team, did not feel neglected
Benjamin anonymous
2021-05-10
I loved using AnalystPrep for FRM. QBank is huge, videos are great. Would recommend to a friend
Daniel Glyn
2021-03-24
I have finished my FRM1 thanks to AnalystPrep. And now using AnalystPrep for my FRM2 preparation. Professor Forjan is brilliant. He gives such good explanations and analogies. And more than anything makes learning fun. A big thank you to Analystprep and Professor Forjan. 5 stars all the way!
michael walshe
2021-03-18
Professor James' videos are excellent for understanding the underlying theories behind financial engineering / financial analysis. The AnalystPrep videos were better than any of the others that I searched through on YouTube for providing a clear explanation of some concepts, such as Portfolio theory, CAPM, and Arbitrage Pricing theory. Watching these cleared up many of the unclarities I had in my head. Highly recommended.