Question

the following data represent the concentration of dissolved organic carbon (mg/l) collected from 20 samples of organic soil. assume that the population is normally distributed. complete parts (a) through (c) on the right. 22.49 29.80 27.10 16.51 15.72 8.81 5.20 20.46 14.90 33.67 30.91 14.86 15.42 15.35 9.72 19.80 14.86 8.09 11.40 18.30 (a) find the sample mean. the sample mean is 17.67. (round to two decimal places as needed.) (b) find the sample standard deviation. the sample standard deviation is 7.85. (round to two decimal places as needed.) (c) construct a 80% confidence interval for the population mean μ. the 80% confidence interval for the population mean μ is (□,□,). (round to two decimal places as needed.)

Explanation:

Step1: Determine the critical value

Since the population standard deviation is unknown and the population is normally distributed, we use the \(t -\)distribution. The sample size \(n = 20\), so the degrees of freedom \(df=n - 1=19\). For an \(80\%\) confidence interval, \(\alpha=1 - 0.80 = 0.20\), and \(\frac{\alpha}{2}=0.10\).
Using the \(t -\)table or a calculator, \(t_{\frac{\alpha}{2},df}=t_{0.10,19}=1.328\)

Step2: Calculate the margin of error

The formula for the margin of error \(E\) is \(E = t_{\frac{\alpha}{2}}\frac{s}{\sqrt{n}}\), where \(s = 7.85\) (sample standard deviation) and \(n = 20\) (sample size)

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Step3: Construct the confidence interval

The formula for the confidence interval for the population mean \(\mu\) when \(\sigma\) is unknown is \(\bar{x}-E<\mu<\bar{x} + E\), where \(\bar{x}=17.67\) (sample mean)

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Answer:

\((15.34,20.00)\)