Question

Genes Samples of DNA are collected, and the four DNA bases of A, G, C, and T are codedas 1, 2, 3, and 4, respectively. The results are listed below. Construct a 95% confidence intervalestimate of the mean. What is the practical use of the confidence interval?

2 2 1 4 3 3 3 3 4 1

Short answer

The 95% confidence interval of mean is $1.8<\mu <3.4$.

The genes samples of DNA is a nominal level data, therefore the confidence interval does not have any practical use.

Step-by-step solution

Given information

The genes samples of DNA are collected, where DNA bases are coded as 1,2,3,4 for A, G, C, T respectively.

Results are given as 2,2,1,4,3,3,3,3,4,1.

Check the requirements

Assume that the samples are randomly selected and the observations are normally distributed.

As the standard deviation of population is unknown, t-distribution would be used.

The sample size of genes sample of DNA (n) =10

Describe the formula for confidence interval

The formula for $\left(1-\alpha \right)\%$confidence interval is $\overline{x}-E<\mu <\overline{x}+E$.

Here, E is margin of error which is given by,$E=t_{\frac{\alpha }{2}}\times \frac{s}{\sqrt{n}}$

Where,$t_{\frac{\alpha }{2}}$ is the critical value with $\alpha$for Student’s t distribution.

Here,$\overline{x}$represents the sample mean ofgenes samples of DNAand$\mu$represents the population mean ofgenes samples of DNA.

Calculate the critical value

For 95% confidence level,$\alpha =0.05$.

The degree of freedom is,

$$\begin{gathered} \text{df}=n-1 \\ =9 \end{gathered}$$

In the t-distribution table, find the value corresponding to the row value of degree of freedom 15 and column value of area in one tail 0.025 is 2.262 which is critical value ${\text{t}}_{\text{0.025}}$.

Therefore, the critical value${\text{t}}_{\text{0.01}}$is 2.262.

Calculate sample mean and standard deviation

From the given sample, the mean is computed as,

$$\begin{gathered} \overline{x}=\frac{\sum X_i}{n} \\ =\frac{2+2+...+1}{10} \\ =2.6 \end{gathered}$$

The standard deviation is given by,

$$\begin{gathered} s=\sqrt{\frac{\sum {\left(X_i-\overline{X}\right)}^2}{n-1}} \\ =\sqrt{\frac{{\left(2-2.6\right)}^2+{\left(2-2.6\right)}^2+...+{\left(1-2.6\right)}^2}{10-1}} \\ =1.0749 \end{gathered}$$

Calculate margin of error

Margin of error is given by,

$$\begin{gathered} E=t_{\frac{\alpha }{2}}\times \frac{s}{\sqrt{n}} \\ =2.262\times \frac{1.0749}{\sqrt{10}} \\ =0.7690 \end{gathered}$$

Construct the confidence interval.

The 95% confidence interval for codes is,

$$\begin{gathered} \overline{x}-E<\mu <\overline{x}+E=2.6-0.7690<\mu <2.6+0.7690 \\ =1.83<\mu <3.37 \\ \approx 1.8<\mu <3.4 \end{gathered}$$

Therefore, 95% confidence interval is $1.8<\mu <3.4$.

Interpret the result.

The 95% confidence interval of the mean of genes sample of DNA is $1.8<\mu <3.4$.

Here, the numbers are just a code name for four DNA bases which are specific count or measures. The genes sample of DNA is a nominal level data. Therefore, there is no practical use of this confidence interval.