# 8.10: Different Significance Level

Finally, let’s take a look at an example phrased in generic terms, rather than in the context of a specific research question, to see the individual pieces one more time. This time, however, we will use a stricter significance level, $$α$$ = 0.01, to test the hypothesis.

Step 1: State the Hypotheses

We will use 60 as an arbitrary null hypothesis value:

$$H_0$$: The average score does not differ from the population

$$H_0: \mu = 60$$

We will assume a two-tailed test:

$$H_A$$: The average score does differ

$$H_A: μ ≠ 60$$

Step 2: Find the Critical Values

We have seen the critical values for $$z$$-tests at $$α$$ = 0.05 levels of significance several times. To find the values for $$α$$ = 0.01, we will go to the standard normal table and find the $$z$$-score cutting of 0.005 (0.01 divided by 2 for a two-tailed test) of the area in the tail, which is $$z*$$ = ±2.575. Notice that this cutoff is much higher than it was for $$α$$ = 0.05. This is because we need much less of the area in the tail, so we need to go very far out to find the cutoff. As a result, this will require a much larger effect or much larger sample size in order to reject the null hypothesis.

Step 3: Calculate the Test Statistic

We can now calculate our test statistic. We will use $$σ$$ = 10 as our known population standard deviation and the following data to calculate our sample mean:

 61 62 65 61 58 59 54 61 60 63

The average of these scores is $$M$$= 60.40. From this we calculate our $$z$$-statistic as:

$z=\dfrac{60.40-60.00}{10.00 / \sqrt{10}}=\dfrac{0.40}{3.16}=0.13 \nonumber$

Step 4: Make the Decision

Our obtained $$z$$-statistic, $$z$$ = 0.13, is very small. It is much less than our critical value of 2.575. Thus, this time, we fail to reject the null hypothesis. Our conclusion would look something like:

Based on the sample of 10 scores, we cannot conclude that there is no effect causing the mean ($$M$$= 60.40) to be statistically significantly different from 60.00, $$z$$ = 0.13, $$p$$ > 0.01.

Notice two things about the end of the conclusion. First, we wrote that $$p$$ is greater than instead of p is less than, like we did in the previous two examples. This is because we failed to reject the null hypothesis. We don’t know exactly what the $$p$$-value is, but we know it must be larger than the $$α$$ level we used to test our hypothesis. Second, we used 0.01 instead of the usual 0.05, because this time we tested at a different level. The number you compare to the $$p$$-value should always be the significance level you test at.

Finally, because we did not detect a statistically significant effect, we do not need to calculate an effect size.