# Level 1 CFA® Exam:

Tests Concerning Population Mean

The table presents tests for distributions with unknown variance:

small sample (n < 30) | large sample (n ≥ 30) | |
---|---|---|

Normal Distribution | t-test | t-test (z-test is also acceptable) |

Nonnormal Distribution | - | t-test (z-test is also acceptable) |

If we don’t know the population variance and the sample is small and the population sampled is not normally distributed or approximately normally distributed, then there are no common tests for the population mean.

However, if we have the distribution with unknown variance but:

- the sample is large, or
- the sample is small, but the population sampled is normally distributed, or approximately normally distributed,

then we can apply the following test statistic for hypothesis tests concerning a single population mean:

\(t_{n-1} = \frac{\overline{X}-\mu_{0}}{s/\sqrt{n}}\)

- \(t_{n-1}\) - t-statistic with "n-1" degrees of freedom
- \(\overline{X}\) - sample mean
- \(\mu_{0}\) - hypothesized value of the population mean
- \(s\) - sample standard deviation
- \(n\) - sample size

If we have the distribution with unknown variance but:

1. the sample is large, or

2. the sample is small, but the population sampled is normally distributed or approximately normally distributed,

then we can apply the t-test for hypothesis tests concerning a single population mean.

The t-test is based on Student's t-distribution and we use it in most tests concerning the mean. However, there are some cases when we may use the z-test alternative which is based on the standardized normal distribution.

If the sample is drawn from a normally distributed population with known variance, then the z-test concerning a single population mean is as follows:

\(z = \frac{\overline{X}-\mu_{0}}{\sigma/\sqrt{n}}\ or\ z = \frac{\overline{X}- \mu_{0}}{s/\sqrt{n}}\)

- \(z\) - z-statistic
- \(\overline{X}\) - sample mean
- \(\mu_{0}\) - hypothesized value of the population mean
- \(\sigma\) - standard deviation from the population
- \(s\) - sample standard deviation
- \(n\) - sample size

If the sample is drawn from a normally distributed population with known variance, then the z-test concerning a single population mean can be used.

A fund manager claims that the average return on his portfolio is 15%. Some analysts at a brokerage house decided to verify the manager's statement. They sampled 20 stocks from his portfolio. They noticed that the returns on the stocks are approximately normally distributed and the average sample return is 13% with a standard deviation of 4%. Is the manager's statement true if the analysts want to be 95% certain that they will not reject the true null hypothesis?

(...)

When we assume that the populations are distributed normally, for tests concerning differences between population means we use a t-test.

Depending on data, we can assume two cases, namely that:

- unknown population variances are equal, or that
- unknown population variances are not equal.

(...)

### Unknown Population Variances Are Not Equal

If unknown variances are assumed unequal, then the formula for the t-test is very similar as in the first case:

\(t = \frac{(\overline{X_{1}} - \overline{X_{2}}) - (\mu_{1}-\mu_{2})}{(\frac{s_{1}^{2}}{n_{1}} + \frac{s^{2}_{2}}{n_{2}})^{\frac{1}{2}}}, \text{ where:} \\ df = \frac{(\frac{s^{2}_{1}}{n_{1}} + \frac{s^{2}_{2}}{n_{2}})^{2}}{\frac{(s^{2}_{1}/n_{1})^{2}}{n_{1}} + \frac{(s^{2}_{2}/n_{2})^{2}}{n_{2}}}\)

- \(t\) - t-statistic
- \(\overline{X}\) - sample mean
- \(\mu\) - population mean
- \(s^{2}\) - sample variance
- \(n\) - sample size
- \(df\) - number of degrees of freedom

However in this case, we don’t use the pooled estimator of the common variance in the denominator, but instead we use sample standard deviations of the two samples.

As you can see, when we compare these two cases it turns out that in the latter we simply need to calculate the degrees of freedom differently and there is no pooled estimator of the common variance.

Note one important thing. These tests can be used only, when two samples are independent of each other.

If samples are not independent of each other, we use the so-called paired comparisons test to examine paired observations.

A test statistic concerning the difference between population means assuming that populations are normally distributed, variances are unknown, and observations are paired is given by the following formulas:

\(t_{n-1}=\frac{\bar{d}-\mu_{d0}}{s_{\bar{d}}}\)

- \(t_{n-1}\) - t-statistic with "n-1" degrees of freedom
- \(\bar{d}\) - sample mean difference
- \(\mu_{d0}\) - value of the difference between means under null hypothesis
- \(s_{\bar{d}}\) - standard error of mean difference

We use the paired-comparisons test, when samples drawn from both populations are not independent of each other.

\(\overline{d} = \frac{1}{n}\times \Sigma_{i=1}^{n}d_{i}\)

- \(d_{i}\) - difference for
th pair of observations*i* - \(n\) - number of pairs of observations

\(s^{2}_{d} = \frac{\Sigma_{i=1}^{n}(d_{i}-\overline{d})^{2}}{n-1}\)

- \(\overline{d}\) - sample mean difference
- \(d_{i}\) - difference for
th pair of observations*i* - \(n\) - number of pairs of observations

\(s_{\overline{d}} = \frac{s_{d}}{\sqrt{n}}\)

- \(s_{d}\) - sample standard deviation
- \(n\) - sample size

Let’s assume that we want to check at the 0.01 significance level if marathon runners run at the same pace before and after they take an energy gel. We draw a sample of 20 marathon runners that take the energy gel at 10 km of the marathon race. We measure their pace throughout the kilometer before taking the energy gel and throughout the kilometer after taking the energy gel.

\(\overline{d}=4\)

\(s_d=9\)

The difference (d) is defined as the runner’s pace after 10 km and before 10 km.

(...)

- If we have the distribution with unknown variance but the sample is large or the sample is small, but the population sampled is normally distributed, or approximately normally distributed, then we can apply the t-test for hypothesis tests concerning a single population mean.
- If the sample is drawn from a normally distributed population with known variance, then the z-test concerning a single population mean is also suitable.
- When we assume that the populations are distributed normally, for tests concerning differences between population means we use a t-test.
- If samples are not independent of each other, we use the so-called paired comparisons test to examine paired observations.