Chapter 9, Problem 1P

In a study, 28 adults with mild periodontal disease are assessed before and 6 months after implementation of a dental-education program intended to promote better oral hygiene. After 6 months, periodontal status improved in 15 patients, declined in 8, and remained the same in 5.

Assess the impact of the program statistically (use a two-sided test)

Suppose patients are graded on the degree of change in periodontal status on a 7-point scale, with +3 indicating the greatest improvement, 0 indicating no change, and −3 indicating the greatest decline. The data are given in Table 9.9.

Table 9.9 Degree of change in periodontal status

To determine

Illustrate the impact of the program statistically.

Explanation of Solution

From the given information, the degree of change in periodontal status before and after the oral dental- education program is given. The scores are given on a 7 point scale with +3 indicate greatest improvement, 0 indicate no improvement and -3 indicate the greatest decline.

It is appropriate to use sign test.

Null hypothesis:

H₀: The impact of program is not statistically significant.

Alternative hypothesis:

H₁: The impact pf program is statistically significant.

Thus, C = 15 and n =15+8 = 23.

Calculation:

There are 15 positive and 8 negative scores. Since 23 > 20, the normal theory test can be used. The rejection region is given as $C>c_2\text{ or }C<c_1$.

Where,

$c_2=\frac{n}{2}+\frac{1}{2}+z_{1-\frac{\alpha }{2}}\sqrt{\frac{n}{4}}\text{ and }{c}_1=\frac{n}{2}-\frac{1}{2}-z_{1-\frac{\alpha }{2}}\sqrt{\frac{n}{4}}$

Here n = 23 and

 $\begin{array}{l}{z}_{1-\frac{\alpha }{2}}=z_{1-\frac{0.05}{2}}\\ =z_{1-0.025}\\ =z_{0.975}\\ =1.96\left(\text{using Excel formula "=NORM}\text{.S}\text{.DIST}\left(0.975\right)"\right)\end{array}$

Substitute the values in the above formula

$\begin{array}{c}{c}_2=\frac{23}{2}+\frac{1}{2}{z}_{0.975}\sqrt{\frac{23}{4}}\\ =12+1.96\left(2.398\right)\\ =12+4.7\\ =16.7\end{array}$

$\begin{array}{c}{c}_1=\frac{23}{2}-\frac{1}{2}-z_{0.975}\sqrt{\frac{23}{4}}\\ =11-1.96\left(2.398\right)\\ =11-4.7\\ =6.3\end{array}$

Thus, the rejection region is $c_2=16.7\text{ and }{c}_1=6.3$. Here $C=15$, $6.3\le 15\le 16.7$. Thus, fail to reject the null hypothesis at 5% level of significance.

Thus, it can be concluded that patients does not significantly changes on the program. It can be assumed that the periodontal status of patients will remain unchanged in the absence of program, which is a questionable assumption. It is appropriate to use a study design of comparing two groups with a control group over 6 months who did not receive the education program.