91Ó°ÊÓ

Jan Lindhe conducted a study \(^{\star}\) on the effect of an oral antiplaque rinse on plaque buildup on teeth. Fourteen subjects, 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 \(y,\) a measure of plaque buildup, was recorded at \(4,7,\) and 14 days. The mean and standard deviation for the 14-day plaque measurements for the two groups are given in the following table: $$\begin{array}{lcc} \hline & \text { Control Group } & \text { Antiplaque Group } \\ \hline \text { Sample size } & 7 & 7 \\ \text { Mean } & 1.26 & .78 \\ \text { Standard deviation } & 32 & .32 \\ \hline \end{array}$$ 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. Bound or find the \(p\) -value for the test.

Step by step solution

01

Define Hypotheses

To test the effectiveness of the antiplaque rinse, define the null and alternative hypotheses. - **Null Hypothesis (H_0):** There is no difference in plaque buildup between the control group and the antiplaque group at 14 days, i.e., \( \mu_1 = \mu_2 \). - **Alternative Hypothesis (H_a):** The antiplaque rinse reduces plaque buildup compared to the control, i.e., \( \mu_1 > \mu_2 \).

02

Calculate the Test Statistic

Using the means, standard deviations, and sample sizes, calculate the test statistic for a two-sample t-test. The formula for the test statistic is: \[ t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \]Substitute the values: \( \bar{x}_1 = 1.26 \), \( \bar{x}_2 = 0.78 \), \( s_1^2 = 0.32^2 \), \( s_2^2 = 0.32^2 \), \( n_1 = n_2 = 7 \).\[ t = \frac{1.26 - 0.78}{\sqrt{\frac{0.32^2}{7} + \frac{0.32^2}{7}}} \]

03

Calculate the Test Statistic Numerically

Compute the value of the test statistic using the obtained formula:1. Calculate the numerator: \( 1.26 - 0.78 = 0.48 \).2. Calculate the denominator: \( \sqrt{\frac{0.32^2}{7} + \frac{0.32^2}{7}} = \sqrt{\frac{0.1024}{7} + \frac{0.1024}{7}} = \sqrt{0.0293} = 0.171 \).3. Calculate the t-value: \( t = \frac{0.48}{0.171} \approx 2.81 \).

04

Determine the Critical Value or p-value

Determine the critical value from the t-distribution table for \( df = n_1 + n_2 - 2 = 12 \) degrees of freedom at \( \alpha = 0.05 \).For a one-tailed test, the critical value of t at \( df = 12 \) is approximately 1.782. Since \( t = 2.81 \) is greater than 1.782, the null hypothesis can be rejected.

05

Find the p-value

Find the p-value corresponding to the calculated t-value.For \( t = 2.81 \) and \( df = 12 \), use statistical software or tables to find that \( p < 0.05 \).

06

Conclusion

Because the calculated t-value exceeds the critical value and \( p < 0.05 \), we reject the null hypothesis. Thus, there is sufficient evidence to conclude that the antiplaque rinse is effective.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Null Hypothesis

In hypothesis testing, the **null hypothesis** is a crucial concept. It's essentially a statement of no effect or no difference. It's what we assume to be true until we have enough evidence to conclude otherwise. In our exercise, the null hypothesis (H_0) claims that there is no difference in plaque buildup between the control group and the antiplaque group after 14 days. Mathematically, this is expressed as \( \mu_1 = \mu_2 \), where \( \mu_1 \) and \( \mu_2 \) are the population means of the control and antiplaque groups, respectively.
To ensure a correct interpretation, remember that failing to reject the null hypothesis doesn't prove that it's true. It just means the data didn't provide strong enough evidence to conclude it's false.

• It's tested indirectly. We look for evidence against it rather than for it.
• Failing to reject it doesn't mean it's true, just that we don't have enough evidence to reject it.
• It's often a statement of "no effect" or "no difference."

Alternative Hypothesis

The **alternative hypothesis** is what you want to prove true. It's the statement that there is an effect or a difference. In this study of plaque buildup, the alternative hypothesis (H_a) suggests that the antiplaque rinse indeed reduces plaque compared to the control solution. In formal terms, it's expressed as \( \mu_1 > \mu_2 \), indicating that the average plaque index for the antiplaque rinse group is lower.
When we conduct a test, our statistical calculations target rejecting the null hypothesis in favor of the alternative hypothesis. This inquiry leads to scientific discoveries or actionable results, like a new oral hygiene product being proven effective.

• It directly challenges the null hypothesis.
• It's what the researcher aims to support.
• Its verification means the treatment or change has an effect.

t-test

The **t-test** is a statistical test used to compare the means of two groups. In the context of our exercise, we applied a two-sample t-test to determine if there is a significant difference between the plaque buildup of the control and antiplaque groups. The t-test formula for our study is: \[ t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \]Here, \( \bar{x}_1 \) and \( \bar{x}_2 \) are the sample means, \( s_1 \) and \( s_2 \) are the standard deviations, and \( n_1 \) and \( n_2 \) are the sample sizes of the two groups respectively.
In our case, after plugging in the values, we found a t-value of approximately 2.81. This figure helps decide if the mean difference is statistically significant.

• It is crucial for comparing two group means.
• It helps determine if the observed difference could have happened by random chance.
• The computed t-value is compared to a critical value to make a decision.

p-value

The **p-value** is a probability that measures the evidence against the null hypothesis. A low p-value indicates strong evidence against the null hypothesis, so you reject it. In our exercise, for the calculated t-value of 2.81 and 12 degrees of freedom, the p-value was found to be less than 0.05.
In simpler terms, this means that there was less than a 5% probability that the difference in means between our treatment and control groups happened by random chance alone. Hence, it provides enough evidence to reject the null hypothesis and accept that the antiplaque rinse is indeed effective.

• It's used to determine the significance of results.
• A small p-value (<0.05) usually leads to rejecting the null hypothesis.
• Helps to communicate how unlikely the observed data is under the null hypothesis.