Answer:
The p-value of the test is 0.049.
Step-by-step 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:
<em>H</em>₀: The experimental toothbrush was not effective, i.e. <em>d</em> = 0.
<em>Hₐ</em>: The experimental toothbrush was effective, i.e. <em>d</em> < 0.
The information provided is:
![\bar d=5.5\\S_{d}=11.6\\n=14](https://tex.z-dn.net/?f=%5Cbar%20d%3D5.5%5C%5CS_%7Bd%7D%3D11.6%5C%5Cn%3D14)
Compute the test statistic value as follows:
![t=\frac{\bar d}{S_{d}/\sqrt{n}}](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B%5Cbar%20d%7D%7BS_%7Bd%7D%2F%5Csqrt%7Bn%7D%7D)
![=\frac{5.5}{11.6/\sqrt{14}}\\\\=1.7740617\\\\\approx 1.774](https://tex.z-dn.net/?f=%3D%5Cfrac%7B5.5%7D%7B11.6%2F%5Csqrt%7B14%7D%7D%5C%5C%5C%5C%3D1.7740617%5C%5C%5C%5C%5Capprox%201.774)
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}](https://tex.z-dn.net/?f=p-value%3DP%28t_%7Bn-1%7D%3C1.774%29)
![=P(t_{13}](https://tex.z-dn.net/?f=%3DP%28t_%7B13%7D%3C1.774%29%5C%5C%3D0.049)
*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.