###### Point Estimate and Confidence Interval ...

Point Estimate A point estimate gives statisticians a single value as the estimate... **Read More**

It is common for analysts to establish whether there is 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** difference. The results of such a test may then inform decisions regarding resource allocation or the rewarding of directors. Before embarking on such an exercise, it is paramount to ensure that the samples taken are independent and sourced from normally distributed populations.

When testing for the difference between two population means, we always use the student’s t-distribution. Thus, we can subdivide the tests for the difference between means into two distinctive scenarios.

Basic situation: two independent random samples of sizes n_{1} and n_{2}, means X’_{1} and X’_{2}, and variances \(\sigma_1^2\) and \(\sigma_1^2\) respectively. We assume that \(\sigma_1^2 = \sigma_1^2 = \sigma^2\)

**H _{0}: μ_{1} – μ_{2} = 0**

We estimate the common variance for the two samples by \(S_p^2\) where,

$$ { S }_{ p }^{ 2 }=\frac { \left( { n }_{ 1 }-1 \right) { S }_{ 1 }^{ 2 }+\left( { n }_{ 2 }-1 \right) { S }_{ 2 }^{ 2 } }{ { n }_{ 1 }+{ n }_{ 2 }-2 } $$

We use the t-statistic with (n_{1} + n_{2} – 2) degrees of freedom, under the null hypothesis that μ_{1} – μ_{2} = 0. Therefore,

$$ { t }_{ { n }_{ 1 }+{ n }_{ 2 }-2 }=\frac { { \bar { x } }_{ 1 }-{ \bar { x } }_{ 2 } }{ { S }_{ p }\sqrt { \left( \frac { 1 }{ { n }_{ 1 } } +\frac { 1 }{ { n }_{ 2 } } \right) } } $$

**Basic situation: two independent random samples of sizes n _{1} and n_{2}, means X’_{1} and X’_{2}, and Unknown variances \(\sigma_1^2\) and \(\sigma_1^2\) respectively.**

**H _{0}: μ_{1} – μ_{2} = 0**

When dealing with large samples, we can use S^{2} to estimate σ^{2}. This is made possible by the central limit theorem. As such, the requirement to draw a sample from a normally distributed population is not necessary. The test statistic used is:

$$ Z=\frac { { \bar { x } }_{ 1 }-{ \bar { x } }_{ 2 } }{ \sqrt { \left( \frac { { \sigma }_{ 1 }^{ 2 } }{ { n }_{ 1 } } +\frac { { \sigma }_{ 2 }^{ 2 } }{ { n }_{ 2 } } \right) } } $$

We then compare the test statistic with the relevant percentage point of the normal distribution. The test statistic is also applicable when the variances are known.

Nutritional experts want to establish whether obese patients on a new special diet have a lower weight than the control group. After 6 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_{B} – μ_{A} = 0

H_{1: }μ_{B} > μ_{A}

We assume that the two samples have equal variance, are independent and distributed normally. Then, under the H_{0},

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

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

Therefore,

$$ S^p_2 =\cfrac {(9 * 363.33 + 9 * 510)}{(10 + 10 -2)} = 436.665 $$

And

$$ \text{the test statistic} =\cfrac {(75 -72)}{ \left\{ \sqrt{439.665} * \sqrt{ \left(\frac {1}{10} + \frac {1}{10}\right)} \right\} }= 0.3210 $$

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 H_{0} 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: You could choose to work with the p-value and determine P(t _{18} > 0.937) and then establish whether this probability is less than 0.05. However, working out the problem correctly would lead to the same conclusion as above*.

Suppose we replace ‘>’ with ‘≠’ in H_{1} in the example above, would the decision rule change?

Replacing ‘>’ with ‘*≠’ *in H_{1} 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 2 and compare the test statistic to both the lower and upper 2.5% points of the t_{18 }-distribution (±2.101). Our test statistic lies within these limits (non-rejection region). The decision rule would, therefore, remain unchanged.