Q2-15BSC - Elementary Statistics

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

Step-by-Step Solution

Answer

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

 

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

Step 1: Given information

The genes samplesof 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.

Step 2: 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.

 

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

Step 3: Describe the formula for confidence interval

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

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,\({\rm{\bar x}}\) represents the sample mean of genes samples of DNA and \({\rm{\mu }}\) represents the population mean of genes samples of DNA.

Step 4: Calculate the critical value

For 95% confidence level,\(\alpha  = 0.05\).

 

The degree of freedom is,

\(\begin{array}{c}{\rm{df}} = n - 1\\{\rm{ }} = 9\end{array}\)

 

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 \({{\rm{t}}_{{\rm{0}}{\rm{.025}}}}\).

Therefore, the critical value \({{\rm{t}}_{0.01}}\)is 2.262.

Step 5: Calculate sample mean and standard deviation

From the given sample, the mean is computed as,

 

\(\begin{array}{c}\bar x = \frac{{\sum {{X_i}} }}{n}\\ = \frac{{2 + 2 + ... + 1}}{{10}}\\ = 2.6\end{array}\)

 

The standard deviation is given by,

 

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

Step 6: Calculate margin of error

Margin of error is given by,

\(\begin{array}{c}E = {t_{\frac{\alpha }{2}}} \times \frac{s}{{\sqrt n }}\\ = 2.262 \times \frac{{1.0749}}{{\sqrt {10} }}\\ = 0.7690\end{array}\)

Step 7: Construct the confidence interval.

The 95% confidence interval for codes is,

\(\begin{array}{l}\bar x - E < \mu  < \bar x + E = 2.6 - 0.7690 < \mu  < 2.6 + 0.7690\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; = 1.83 < \mu  < 3.37\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \approx 1.8 < \mu  < 3.4\end{array}\)

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

Step 8: Interpret the result.

The 95% confidence interval of the mean ofgenes 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.