Question

Five students were asked how many months had passed since their last visit to a dentist. They said 6, 16, 12, 22, 29. How do you construct a $95$ confidence interval for the mean number of months elapsed since their last visit?

Answer 1

We are required to know and understand the basic concepts of statistics in order to solve such problems. Using basic formulae, we shall solve this question. We first calculate the mean and standard deviation for the given data. Then we need to tune our answer according to the $95$ confidence interval and obtain the result.

Complete step-by-step solution:

To solve this question, let us first find the mean and the standard deviation for the given data. To calculate the mean, we use the formula
$\Rightarrow \overline{X}=\dfrac{\sum\limits_{i=1}^{n}{{{X}_{i}}}}{n}$
Here, ${{X}_{i}}$ represents the all-individual data in the set that is, 6, 16, 12, 22 and 29 in this case. n represents the total amount of data. Substituting these values,
$\Rightarrow \overline{X}=\dfrac{6+16+12+22+29}{5}$
Adding all the terms in the numerator and dividing by 5,
$\Rightarrow \overline{X}=\dfrac{85}{5}=17$
Hence, the mean is 17. Now we calculate the standard deviation by using the formula,
$\Rightarrow s=\sqrt{\dfrac{1}{n-1}\left( \sum\limits_{i=1}^{n}{{{X}_{i}}^{2}}-n{{\overline{X}}^{2}} \right)}$
Substituting the values,
$\Rightarrow s=\sqrt{\dfrac{1}{5-1}\left( \left( {{6}^{2}}+{{16}^{2}}+{{12}^{2}}+{{22}^{2}}+{{29}^{2}} \right)-5\times {{17}^{2}} \right)}$
Squaring and adding the terms in the brackets,
$\Rightarrow s=\sqrt{\dfrac{1}{4}\left( 1761-1445 \right)}$
Subtracting the terms and dividing by 4,
$\Rightarrow s=\sqrt{\dfrac{316}{4}}=\sqrt{79}$
Taking the square root of 79,
$\Rightarrow s=8.8881$
We can now calculate the confidence upper and lower levels by using the t-critical value. The significance level $\alpha$ for the $95$ confidence interval can be calculated as,
$\Rightarrow \alpha =1-0.95=0.05$
The degrees of freedom for the t-distribution is
$\Rightarrow \text{Degrees of freedom=}n-1$
Substituting n as 5,
$\Rightarrow \text{Degrees of freedom=5}-1=4$
For the given significance level $\alpha$ and degree of freedom, the t-critical value ${{t}_{\dfrac{\alpha }{2},n-1}}$ can be obtained from the t-table as,
$\Rightarrow 2P\left( {{t}_{4}}>{{t}_{\dfrac{\alpha }{2},n-1}} \right)=0.05$
$\Rightarrow 2P\left( {{t}_{4}}>2.77 \right)=0.05$
To calculate the interval, we use the formula
$\Rightarrow \left( \overline{X}-{{t}_{\dfrac{\alpha }{2},n-1}}.\dfrac{s}{\sqrt{n}}<\mu <\overline{X}+{{t}_{\dfrac{\alpha }{2},n-1}}.\dfrac{s}{\sqrt{n}} \right)$
Substituting the values,
$\Rightarrow \left( 17-2.776.\dfrac{8.8881}{\sqrt{5}}<\mu <17+2.776.\dfrac{8.8881}{\sqrt{5}} \right)$
Simplifying the above expression,
$\Rightarrow \left( 17-2.776\times 3.9749<\mu <17+2.776\times 3.9749 \right)$
Multiplying the terms,
$\Rightarrow \left( 17-11.0343<\mu <17+11.0343 \right)$
Adding and subtracting,
$\Rightarrow 5.966<\mu <28.034$
**Hence, the $95$ confidence interval for the mean number of months elapsed since their last visit $5.966 <\mu <28.034.$ **

Note: It is important to know the basics of statistics and we need to know the concept of confidence level calculation in order to answer such questions. We need to remember the formulae for all the important parameters such as significance level, mean, standard deviation, degrees of freedom, etc in order to calculate to solve such questions.