Question

Jan Lindhe conducted a study on the effect of an oral antiplaque rinse on plaque buildup on teeth. \(^{6}\) Fourteen people whose teeth were thoroughly cleaned and polished were randomly assigned to two groups of seven subjects each. Both groups were assigned to use oral rinses (no brushing) for a 2 -week period. Group 1 used a rinse that contained an antiplaque agent. Group \(2,\) the control group, received a similar rinse except that, unknown to the subjects, the rinse contained no antiplaque agent. A plaque index \(x\), a measure of plaque buildup, was recorded at 14 days with means and standard deviations for the two groups shown in the table. $$ \begin{array}{lll} & \text { Control Group } & \text { Antiplaque Group } \\ \hline \text { Sample Size } & 7 & 7 \\ \text { Mean } & 1.26 & .78 \\ \text { Standard Deviation } & 32 & 32 \end{array} $$ a. State the null and alternative hypotheses that should be used to test the effectiveness of the antiplaque oral rinse b. Do the data provide sufficient evidence to indicate that the oral antiplaque rinse is effective? Test using \(\alpha=.05 .\) c. Find the approximate \(p\) -value for the test.

Short answer

Answer: No, there isn't sufficient evidence to support the effectiveness of the antiplaque oral rinse in reducing plaque buildup based on the given data and an alpha level of 0.05.

Step-by-step solution

State the null and alternative hypothesis

The null hypothesis (\(H_0\)) states that there is no difference in the plaque index means between the two groups, while the alternative hypothesis (\(H_A\)) claims that there is a difference and that the antiplaque rinse is effective in reducing plaque buildup. Mathematically, the hypotheses are: \(H_0: \mu_1 = \mu_2\) \(H_A: \mu_1 > \mu_2\) where \(\mu_1\) is the mean plaque index of the control group, and \(\mu_2\) is the mean plaque index of the antiplaque group.

Conduct a two-sample t-test

To perform a two-sample t-test, we will use the following formula for the test statistic: \(t = \frac{(\bar{x}_1 - \bar{x}_2) - 0}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}\) where \(\bar{x}_1\) and \(\bar{x}_2\) are the sample means, \(s_1\) and \(s_2\) are the sample standard deviations, and \(n_1\) and \(n_2\) are the sample sizes for the control and antiplaque groups, respectively. We are given: - Control group: \(n_1 = 7\), \(\bar{x}_1 = 1.26\), \(s_1 = 32\) - Antiplaque group: \(n_2 = 7\), \(\bar{x}_2 = 0.78\), \(s_2 = 32\) Plugging these values into the formula, we get: \(t = \frac{(1.26 - 0.78) - 0}{\sqrt{\frac{32^2}{7} + \frac{32^2}{7}}} \approx 0.48\) Since we are conducting a one-tailed test, we will compare this value to the critical t-value at an alpha level of 0.05. Using a t-distribution table or calculator, we find that the critical t-value for a one-tailed test with 12 degrees of freedom (2(7) - 2) is approximately 1.782. Because \(0.48 < 1.782\), we fail to reject the null hypothesis.

Find the approximate p-value

Using a t-distribution calculator or table, we can find the p-value for the test statistic obtained above. For a t-value of 0.48 with 12 degrees of freedom, we find that the p-value is approximately 0.32. Since \(p \approx 0.32 > \alpha = 0.05\), we fail to reject the null hypothesis. In conclusion, there isn't sufficient evidence to support the claim that the antiplaque oral rinse is effective in reducing plaque buildup, based on the given data and an alpha level of 0.05.

Key concepts

Null Hypothesis

In hypothesis testing, the null hypothesis is a crucial concept. It is a statement that suggests no effect or no difference regarding the population parameter of interest. Think of it as the default position where nothing new or surprising happens.
In our exercise, the null hypothesis (\(H_0\)) is that there is no difference in the mean plaque index between the control group and the antiplaque group. In other words, the rinse with the antiplaque agent does not work any better than the rinse without it.
We express this mathematically as: \(H_0: \mu_1 = \mu_2\). Here, \(\mu_1\) is the mean plaque index for the control group, and \(\mu_2\) is the mean for the antiplaque group.
When testing a hypothesis, we initially assume the null hypothesis is true. The aim is to determine if there's enough statistical evidence from the data to reject this assumption.

Alternative Hypothesis

The alternative hypothesis serves as the opposite of the null hypothesis. It represents the new effect or difference we aim to detect. In many experiments, researchers hope to find grounds to support the alternative hypothesis.
For the oral rinse study, the alternative hypothesis (\(H_A\)) is that the antiplaque rinse does indeed lead to a lower mean plaque index compared to the control rinse. This suggests that the rinse is effective.
Mathematically, the alternative hypothesis is written as: \(H_A: \mu_1 > \mu_2\), which implies the mean plaque index of the control group is greater than that of the antiplaque group. In hypothesis testing, a successful test outcome is typically the rejection of the null hypothesis in favor of the alternative one.
This is why the alternative hypothesis is sometimes regarded as the research hypothesis, as it aligns with the researcher's expectations or predictions.

t-test

A t-test is a statistical method used to compare the means of two groups, and it is useful when determining if the differences between them are significant. For this exercise, a two-sample t-test is appropriate because we have two independent groups: the control group and the antiplaque group.
The t-test uses a calculated t-value to assess whether the observed difference in means could have arisen under the null hypothesis.
The formula used in the calculation is: \[ t = \frac{(\bar{x}_1 - \bar{x}_2) - 0}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \] where:

• \(\bar{x}_1\) and \(\bar{x}_2\) are the sample means,
• \(s_1\) and \(s_2\) are the sample standard deviations,
• \(n_1\) and \(n_2\) are the sample sizes for each group.

In our example, the t-value calculated was approximately 0.48.
A critical component in the t-test is deciding whether this t-value is significant or not, which will depend on the p-value and the chosen significance level (in the exercise, \(\alpha = 0.05\)).

p-value

The p-value is a probability measure used in hypothesis testing to quantify the strength of the evidence against the null hypothesis. It represents the likelihood of obtaining test results at least as extreme as the observed data, assuming the null hypothesis is true.
A small p-value indicates strong evidence against the null hypothesis, suggesting the alternative might be true.
In our study, the p-value for the calculated t-value (0.48) was approximately 0.32. This is much greater than the significance level (\(\alpha = 0.05\)), meaning there is insufficient evidence to reject the null hypothesis.
The higher the p-value compared to the alpha level, the more support there is for maintaining the null hypothesis.
While a p-value less than 0.05 often leads researchers to reject the null hypothesis, in this exercise, the high p-value suggests that the antiplaque rinse is not significantly more effective than the control, based on the data.
Hence, the conclusion is to "fail to reject" the null hypothesis.