Question

samples of dna are collected, and the four dna bases of a, g, c, and t are coded as 1, 2, 3, and 4, respectively. the results are listed below: construct a 90% confidence interval estimate of the mean. what is the practical use of the confidence interval? 1, 1, 2, 3, 4, 3, 4, 3, 3, 2 what is the confidence interval for the population mean μ? □ < μ < □ (round to one decimal place as needed.)

Explanation:

Step1: Calculate the sample mean $\bar{x}$

The data set is \(1,1,2,3,4,3,4,3,3,2\).
The formula for the sample mean is \(\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}\).
Here, \(n = 10\) and \(\sum_{i=1}^{n}x_{i}=1 + 1+2+3+4+3+4+3+3+2=26\).
So, \(\bar{x}=\frac{26}{10}=2.6\).

Step2: Calculate the sample standard deviation \(s\)

The formula for the sample standard deviation is \(s=\sqrt{\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}}\).
\((x_{1}-\bar{x})^{2}=(1 - 2.6)^{2}=(-1.6)^{2}=2.56\), \((x_{2}-\bar{x})^{2}=(1 - 2.6)^{2}=2.56\), \((x_{3}-\bar{x})^{2}=(2 - 2.6)^{2}=0.36\), \((x_{4}-\bar{x})^{2}=(3 - 2.6)^{2}=0.16\), \((x_{5}-\bar{x})^{2}=(4 - 2.6)^{2}=1.96\), \((x_{6}-\bar{x})^{2}=(3 - 2.6)^{2}=0.16\), \((x_{7}-\bar{x})^{2}=(4 - 2.6)^{2}=1.96\), \((x_{8}-\bar{x})^{2}=(3 - 2.6)^{2}=0.16\), \((x_{9}-\bar{x})^{2}=(3 - 2.6)^{2}=0.16\), \((x_{10}-\bar{x})^{2}=(2 - 2.6)^{2}=0.36\).
\(\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}=2.56+2.56 + 0.36+0.16+1.96+0.16+1.96+0.16+0.16+0.36=10.4\).
\(s=\sqrt{\frac{10.4}{9}}\approx1.1\).

Step3: Determine the critical value \(t_{\alpha/2}\)

The confidence level is \(90\%\), so \(\alpha=1 - 0.90 = 0.10\) and \(\alpha/2=0.05\).
The degrees of freedom \(df=n - 1=10 - 1 = 9\).
From the \(t\)-distribution table, \(t_{0.05,9}=1.833\).

Step4: Calculate the margin of error \(E\)

The formula for the margin of error for a \(t\)-distribution is \(E=t_{\alpha/2}\frac{s}{\sqrt{n}}\).
Substitute \(t_{\alpha/2}=1.833\), \(s = 1.1\), and \(n = 10\) into the formula:
\(E=1.833\times\frac{1.1}{\sqrt{10}}\approx1.833\times0.348\approx0.6\).

Step5: Calculate the confidence interval

The confidence interval for the population mean \(\mu\) is \(\bar{x}-E<\mu<\bar{x}+E\).
Substitute \(\bar{x}=2.6\) and \(E = 0.6\) into the formula:
\(2.6-0.6<\mu<2.6 + 0.6\).

Answer:

\(2.0<\mu<3.2\)