98  The One-Sided Hypothesis Test

98.1 Introduction

Assume that the general population has an average length of \(\mu_0\) = 170 cm and a standard deviation \(\sigma = 10\). A simple random sample is drawn from the population of statistics students at a local university with \(N = 100\) and \(m = 171.6\) cm.

It does not come as a surprise that \(m \neq \mu\), after all \(m\) is only based on sample measurements. On the other hand, one may wonder whether the positive difference \(m - \mu_0 = 171.6 - 170 = 1.6\) can be attributed to sampling uncertainty under H\(_0\).

98.2 Definition of the One-Sided Hypothesis

A hypothesis is a claim about one or more population parameters. Typically, one Hypothesis is compared to another -- in practice one often formulates a Null Hypothesis H\(_0\) versus an Alternative Hypothesis H\(_A\).

98.2.1 The Null Hypothesis

H\(_0\): the length of statistics students (at our university) is normally distributed with \(\mu = \mu_0 = 170\) and \(\sigma = 10\) which implies that the probability density function is

\[Y = \frac{1}{10 \sqrt{2 \pi}} e^{-\frac{1}{2} \left( \frac{l - 170}{10} \right)^2}\]

98.2.2 The Alternative Hypothesis

H\(_A\): the length of statistics students (at our university) is normally distributed with \(\mu > \mu_0 = 170\) and \(\sigma = 10\) which implies that the probability density function is

\[Y = \frac{1}{10 \sqrt{2 \pi}} e^{-\frac{1}{2} \left( \frac{l - \mu }{10} \right)^2}\]

where \(\mu > \mu_0 = 170\).

Unlike H\(_0\), H\(_A\) is composite -- the formula above holds for any specific \(\mu > 170\), but H\(_A\) does not identify a single distribution.

98.3 H\(_0\) versus H\(_A\)

The Null Hypothesis H\(_0\) provides a complete description of the population (there are no unknown parameters). Therefore, we define H\(_0\) as the default position. In other words, we assume H\(_0\) to be true and investigate the consequences of this assumption.

There are two possible scenarios we need to consider:

• the consequences of H\(_0\) are improbable
• the consequences of H\(_0\) are not particularly improbable

98.3.1 Consequences of H\(_0\) are improbable

In this scenario, the assumption made by the Null Hypothesis is unlikely to be true. In other words, the Null Hypothesis is rejected and the Alternative Hypothesis is supported. Is it possible that we make errors when rejecting a Null Hypothesis?

Of course this is possible because the consequences of H\(_0\) are only improbable (not impossible). We define the type I error as the incorrect rejection of a true Null Hypothesis (a “false positive”). The symbol for the type I error is \(\alpha\) and is chosen by the researcher (many authors use \(\alpha = 5\%\) or \(\alpha = 1\%\) but these are by no means “holy” numbers).

ImportantDecision Threshold Choice

In this chapter, \(\alpha\) is introduced as a researcher-chosen threshold for controlling the probability of a Type I error under H\(_0\).

The value of \(\alpha\) is not a universal constant. In confirmatory settings, stricter thresholds (often 1% to 5%) are common, while diagnostic/screening tests may justify higher thresholds (e.g. 10% to 20%) to reduce false reassurance.

Choose \(\alpha\) based on the role of the test and the decision context before interpreting the result. For the general framework, see Chapter 112.

98.3.2 Consequences of H\(_0\) are not improbable

In this scenario, there is no reason to reject the assumption made by the Null Hypothesis. In this case, we fail to reject H\(_0\).

98.3.3 What are the consequences of H\(_0\)?

We investigate the consequences regarding the Arithmetic Mean of a simple random sample which is drawn from the population.

The Null Hypothesis describes the population

\[Y = \frac{1}{\sigma \sqrt{2 \pi}} e^{-\frac{1}{2} \left( \frac{X - \mu}{\sigma} \right)^2} = \frac{1}{10 \sqrt{2 \pi}} e^{-\frac{1}{2} \left( \frac{X - 170}{10} \right)^2}\]

and the variable \(\bar{X}\) (which represents all possible values of Arithmetic Means from simple random samples) is known to be normally distributed

\[Y = \frac{1}{\frac{\sigma}{\sqrt{N}} \sqrt{2 \pi}} e^{-\frac{1}{2} \left( \frac{\bar{X} - \mu}{\frac{\sigma}{\sqrt{N}}} \right)^2} = \frac{1}{1 \sqrt{2 \pi}} e^{-\frac{1}{2} \left( \frac{\bar{X} - 170}{1} \right)^2}\]

We divide the \(\bar{X}\)-axis into two regions:

• the non-rejection region, which -- if H\(_0\) is true -- contains the sample mean \(\bar{X}\) with probability 95%
• the region of rejection, which -- if H\(_0\) is true -- contains the sample mean \(\bar{X}\) with probability 5%

Both regions are separated by the value \(\mu + k\) which can be computed by finding \(k\) in P\(\left( \bar{X} < \mu + k \right) = 95\%\).

In accordance with previous sections:

\[\text{P} \left( \bar{X} < \mu + k \right) = \text{P} \left( \frac{\bar{X} - \mu}{\frac{\sigma}{\sqrt{N}}} \leq \frac{\mu + k - \mu}{\frac{\sigma}{\sqrt{N}}} \right) = \text{P} \left( Z \leq \frac{k}{\frac{\sigma}{\sqrt{N}}} \right) = 95\%\]

From the Gaussian Table (Appendix E) it follows that

\[\frac{k}{\frac{\sigma}{\sqrt{N}}} = 1.645 \Rightarrow k = 1.645 \frac{\sigma}{\sqrt{N}}\]

which can be used to find the regions

\[\bar{X} \leq \mu + 1.645 \frac{\sigma}{\sqrt{N}}\]

or

\[\bar{X} \leq 170 + 1.645 \frac{10}{\sqrt{100}} = 171.645\]

From this it can be concluded that the Null Hypothesis implies the following consequences:

• 95% of simple random sample means are smaller than (or equal to) 171.645
• 5% of simple random sample means are larger than 171.645

The sample mean (\(m = 171.6 < 171.645\)) is contained in the region of non-rejection. This implies that we fail to reject H\(_0\). Hence, we do not conclude that statistics students from our university are taller than individuals from the general population.

Given a significance level of 5%, we cannot reject H\(_0\) because the sample mean \(m = 171.6\) is not significantly larger than the population mean \(\mu = 170\).

Note that the region of non-rejection and rejection are both dependent on the sample size \(N\). From the discussion above it is easy to see that

\[\lim\limits_{N \rightarrow \infty} \left( 170 + 1.645 \frac{10}{\sqrt{N}} \right) = 170\]

which implies that any non-zero difference between the sample mean and \(\mu\) would be declared “significantly different” when \(N\) becomes large enough. Hence, when we reject H\(_0\) (for any chosen value of \(\alpha\)) this does not necessarily imply that we have found an “important” difference.

98.3.4 Procedure

To test the one-sided hypothesis we performed the following steps:

1. Define H\(_0\) and H\(_A\). In our example:

\[ \begin{align*} \text{H}_0: \mu = \mu_0 = 170 \\ \text{H}_A: \mu > \mu_0 = 170 \end{align*} \]

2. Specify the significance level \(\alpha\). In our example we chose \(\alpha = 0.05\)
3. Draw a simple random sample of size \(N\) from the population and compute the Arithmetic Sample Mean \(m\)
4. Determine the regions of non-rejection and rejection for H\(_0\)
5. If \(m\) is contained in the region of rejection we reject H\(_0\) and support H\(_A\). Otherwise, we fail to reject H\(_0\).

98.4 Corollary

We define the type II error (labeled \(\beta\)) as the failure to reject a false Null Hypothesis (a “false negative”). Sometimes the type II error is referred to as the failure to detect an effect that is present.

To illustrate this we suppose that the true population mean \(\mu = 170.5\) (instead of \(170\)). In the previous sections, we concluded that the one-sided Null Hypothesis (\(\mu_0 = 170\)) should not be rejected -- but if the true parameter \(\mu = 170.5\) then we have come to the wrong conclusion (we should have rejected H\(_0\) and supported H\(_A\): \(\mu > \mu_0 = 170\)).

So the problem we need to solve, is to find the probability of making a type II error. In order to do that we need to realize that the population is not adequately described by

\[Y = \frac{1}{10 \sqrt{2 \pi}} e^{-\frac{1}{2} \left( \frac{l - 170}{10} \right)^2}\]

but, instead by

\[Y = \frac{1}{10 \sqrt{2 \pi}} e^{-\frac{1}{2} \left( \frac{l - 170.5}{10} \right)^2}\]

This implies that the mean of simple random samples of size \(N = 100\) are described by

\[Y = \frac{1}{\sqrt{2 \pi}} e^{-\frac{1}{2} \left( \frac{\bar{X} - 170.5}{1} \right)^2}\]

From the previous discussions we know that the region of non-rejection is \(\bar{X} \leq 171.645\). Hence, the type II error can be computed as follows

\[\beta = \text{P} \left( \bar{X} \leq 171.645 \right) = \int_{-\infty}^{171.645} \frac{1}{\sqrt{2 \pi}} e^{-\frac{1}{2} \left( \frac{\bar{X} - 170.5}{1} \right)^2} \text{d} \bar{X} \simeq 87.39\%\]

This computation shows that the type II error is rather large. The power of the test is \(1 - \beta\). In the first example, power \(\simeq 12.6\%\), indicating very low sensitivity to a true shift of only 0.5 cm. Some would argue that there is no need to worry about this error because the Null Hypothesis (\(\mu = 170\)) is still very close to the true value \(\mu = 170.5\)). If the true population mean \(\mu = 173\) then we would obtain a type II error of

\[\beta = \text{P} \left( \bar{X} \leq 171.645 \right) = \int_{-\infty}^{171.645} \frac{1}{\sqrt{2 \pi}} e^{-\frac{1}{2} \left( \frac{\bar{X} - 173}{1} \right)^2} \text{d} \bar{X} \simeq 8.77\%\]

This illustrates that the type II error is related to the difference between the Null value and the true population value. In other words, the assessment of making a type II error depends on more than just \(\beta\).