Question

The proper choice of toothbrush is important in dental care. On 14 patients, measures of Gingivitis were taken at the beginning of the experiment. After this, each was given an experimental toothbrush to be used for the next 45 days. Afterwards, dental exam measures were given again of these same individuals. The mean of the differences in scores was 5.5 with a sample standard deviation of 11.6. The mean score before the experiment was 56.4 with a sample standard deviation of 6.4 The mean score after the experiment was 60.6 with a sample standard deviation of 4.3 Let alpha = 0.05

What is the p value from the test of the previous hypothesis?

Answer 1

The p-value of the test is 0.049.

Explanation:

The dependent t-test (also known as the paired t-test or paired samples t-test) compares the two means associated groups to conclude if there is a statistically significant difference amid these two means.

In this case a paired t-test is used to determine whether the experimental toothbrush was effective or not.

The hypothesis for the test can be defined as follows:

H₀: The experimental toothbrush was not effective, i.e. d = 0.

Hₐ: The experimental toothbrush was effective, i.e. d < 0.

The information provided is:

`\bar d=5.5\\S_(d)=11.6\\n=14`

Compute the test statistic value as follows:

`t=(\bar d)/(S_(d)/√(n))`

`=(5.5)/(11.6/√(14))\\\\=1.7740617\\\\\approx 1.774`

The test statistic value is 1.774.

Decision rule:

If the p-value of the test is less than the significance level then the null hypothesis will be rejected and vice-versa.

Compute the p-value of the test as follows:

`p-value=P(t_(n-1)<1.774)`

`=P(t_(13)<1.774)\\=0.049`

*Use a t-table.

The p-value of the test is 0.049.

p-value= 0.049 > α = 0.05

The null hypothesis will be rejected.

Thus, it can be concluded that experimental toothbrush was effective.