You have given the first 90 patients to enter the student clinic a questionnaire

Question

You have given the first 90 patients to enter the student clinic a questionnaire regarding general demographics and oral hygiene habits. After participants finish the questionnaire they are provided a full clinical assessment and provide salivary samples. You are interested in the relationship between the following 3 variables. The questions and possible responses are:

- Do you have access to dental insurance [0-no, 1-yes]

- How may teeth have you lost in the last year? [0, 1, 2, >2]

- P. gingivalis bacterial count from plaque samples (bacterial/ml) [assume normal distribution]

Please answer the following questions: a) You are first interested in examining the relationship between access to dental insurance and P. gingivalis bacterial count. What type of statistical test would be appropriate and why? Options t-tests ,Whitney or Wilcoxon.

b) Suppose that you are interested in whether teeth lost in the past year is associated with P. gingivalis. Could you answer this question with the statistical tools we have covered thus far?Options t-tests ,Whitney or Wilcoxon

c) Suppose that instead of sampling 90 patients you are only able to sample 30, would that change the test you choose?

Explanation / Answer

a)

We would use the t-test because the data for P.gingivalis is normally distributed. And a t-test is better for normally distributed variable. We use Whitney or Wilcoxon test when the data is not normally distributed.

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b)

No, this question can not be answered with either t-test or wilcoxon test. As there are more than 2 groups now for the number of teeths lost last year, so using a t-test is not possible. A chi-square test of independence is required in this case.

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c)

No, that would not change anything because in first case, the variable P.gingivalis is already given to be normally distributed. So, even using a small sample size would be good to go for a t-test. And, in the second case decreasing the sample size wouldn't help as we need to do a chi-squre test of independence.

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