91Ó°ÊÓ

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.

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

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 ofgenes samples of DNAand\({\rm{\mu }}\)represents the population mean ofgenes samples of DNA.

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.

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,

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

Margin of error is given by,

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

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 \(1.8 < \mu < 3.4\).

The 95% confidence intervalof the mean ofgenes sample of DNAis\(1.8 < \mu < 3.4\).

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.