A decision rule refers to the procedure followed by analysts and researchers when deciding whether to reject or not to reject the null hypothesis. We use the phrase “not to reject” because it’s considered statistically incorrect to “accept” the null hypothesis. Rather, we can only assemble enough evidence to support it.

**Breaking down the Decision Rule**

The decision to reject or not to reject the null hypothesis is based on the distribution assumed by the test statistic. This means if the variable involved follows a normal distribution, we use the level of significance of the test to come up with critical values that lie along the standard normal distribution.

Note that prior to decision making, one must decide whether the test should be one-tailed or two-tailed as the number of tails determines the value of α (significance level). The following is a brief summary of the decision rules under different scenarios:

**Left One-tailed Test**

H_{1}: parameter < X

Decision rule: Reject H_{0} if the test statistic is less than the critical value. Otherwise, **do not reject** H_{0.}

**Right One-tailed Test**

H_{1}: parameter > X

Decision rule: Reject H_{0} if the test statistic is greater than the critical value. Otherwise, **do not reject** H_{0.}

**Two-tailed Test**

H_{1}: parameter *≠* X (not equal to X)

Decision rule: Reject H_{0} if the test statistic is greater than the upper critical value or less than the lower critical value.

**The Power of a Test**

The power of a test is the direct opposite of the level of significance. While the level of significance gives us the probability of rejecting the null hypothesis when it’s, in fact, true, the power of a test gives the probability of correctly discrediting and rejecting the null hypothesis when it is false. In other words, it gives the likelihood of rejecting H_{0 }when indeed it’s false. Expressed mathematically,

$$ \text{Power of a test} = 1 – P(\text{type } II \text{ error}) $$

When presented with a situation where there are multiple test results for the same purpose, it’s the test with the highest power that’s considered the best.

**The Link between Confidence Interval and Hypothesis Testing**

Confidence intervals and hypothesis tests are linked by critical values. For example, to construct a 95% confidence interval assuming a normal distribution, we would need to determine the critical values that correspond to 5% level of significance. Similarly, if we were to conduct a test of some given hypothesis at the 5% significance level, we would use the same critical values used for the confidence interval to subdivide the distribution space into the rejection and non-rejection regions.

**Example: Hypothesis testing**

A survey carried out using a sample of 50 CFA level I candidates reveals an average IQ of 105. Assuming that IQs are normally distributed, carry out a statistical test to determine whether the mean IQ is greater than 100. You are instructed to use 5% level of significance. (Previous studies give a standard deviation of IQs of approximately 20.)

**Solution**

First, state the hypothesis:

H_{0}: μ = 100 vs H_{1}: μ > 100

Since IQs follow a normal distribution, under \(H_0, \frac {(X’ – 100)}{\left( \frac {\mu}{\sqrt n} \right)} \sim N(0,1)\)

Next, we compute the test statistic, which is \(\frac {(105 – 100)}{\left(\frac {20}{\sqrt {50}} \right)} = 1.768\)

This is a right one-tailed test and IQs are normally distributed. Therefore, we should compare our test statistic to the upper 5% point of the normal distribution.

From the normal distribution table, this value is 1.6449. Since 1.768 is greater than 1.6449, we have **sufficient evidence** to reject H_{0} at the 5% significance level. Thus, it’s **reasonable **to conclude that the mean IQ of CFA candidates is greater than 100.

(*Note the choice of words used in the decision-making part and the conclusion.)*

QuestionUse the data from the previous example to carry out a test at 5% significance to determine whether the average IQ of candidates is greater than 102.

A. There is sufficient evidence to reject H

_{0 }and conclude that the average IQ is greater than 102.B. There is insufficient evidence to reject H

_{0 }and therefore it’s reasonable to conclude that the average IQ is not more than 102.C. There is sufficient evidence to reject H

_{0 }and therefore it’s reasonable to conclude that the average IQ is greater than 102.

SolutionThe correct answer is B.

Just like in the example above, start with stating the hypothesis;

H

_{0}: μ = 100 vs. H_{1}: μ > 102The test statistic is \(\frac {(105 – 102)}{\left( \frac {20}{\sqrt{50}} \right)} = 1.061\)

Again, this is a right one-tailed test. 1.061 is less than the upper 5% point of a standard normal distribution (1.6449). Therefore, we

do not have sufficient evidenceto reject H_{0}at the 5% level of significance. Thus, it’s reasonable to conclude that the average IQ of CFA candidatesis not more than102.

*Reading 11 LOS 11d:*

*Explain a decision rule, the power of a test, and the relation between confidence intervals and hypothesis tests*