Inferential Statistics - Hypothesis Testing

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Components of a statistical test

1. Hypotheses: Null (H₀) and Alternative (H₁)
2. Level of significance (α)
3. Test statistic
4. Decision rule
5. Conclusion

Before observing the data, the null and alternative hypotheses should be stated, a significance level (α) should be chosen (often equal to 0.05), and the test statistic that will summarize the information in the sample should be chosen as well. Based on the hypotheses, test statistic, and sampling distribution of the test statistic, we can find the critical region of the test statistic which is the set of values for the statistical test that show evidence in favor of the alternative hypothesis and against the null hypothesis. This region is chosen such that the probability of the test statistic falling in the critical region when the null hypothesis is correct (Type I error) is equal to the previously chosen level of significance (α).

The test statistic is then calculated:

• if the value of the test statistic falls inside the critical region, then the null hypothesis is rejected at the chosen significance level.
• if the value of the test statistic falls outside the critical region, then there is not enough evidence to reject the null hypothesis at the chosen significance level.

The p-value, the probability of a test result at least as extreme as the one observed if the null hypothesis was true, can also be calculated.

Example - Paired t-test of change in cholesterol from 1952 to 1962

d is the difference in cholesterol for each individual from 1952 to 1962

 

Hypotheses:

H₀: There is no change, on average, in cholesterol level from 1952 to 1962

(H₀: μ_d = 0)

H₁: There is an average non-zero change in cholesterol level from 1952 to 1962

(H₁: μ_d ≠ 0)

Test statistic:

Decision rule: Reject H₀ at α=0.05 if |t| > 2.093

The decision rule is constructed from the sampling distribution for the test statistic t. For this example, the sampling distribution of the test statistic, t, is a student t-distribution with 19 degrees of freedom. The critical value 2.093 can be read from a table for the t-distribution.

Results:

.

Conclusion:

Cholesterol levels decreased, on average, 69.8 units from 1952 to 1962. For a significance level of 0.05 and 19 degrees of freedom, the critical value for the t-test is 2.093. Since the absolute value of our test statistic (6.70) is greater than the critical value (2.093) we reject the null hypothesis and conclude that there is on average a non-zero change in cholesterol from 1952 to 1962.

Note that this summary includes:

• The test being performed. (The paired t-test in this example)
• A statement of the null hypothesis and alternative hypothesis in terms of the population parameter of interest. (The mean difference in the previous example)
• The magnitude, direction, and units of the effect (observed mean difference).
• a 69.8 unit mean decrease from 1952 to 1962
• Note, this should be reported regardless of whether or not it is statistically significant!
• The test statistic and degrees of freedom
• A statement of whether the effect (observed difference) is statistically significant and the significance level (α)
• Note that here we compared the test statistic to the critical value. Using the p-value also satisfies this criteria.

 

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