Question

in a random sample of 6 residents of the state of maine, the mean waste recycled per person per day was 1.1 pounds with a standard deviation of 0.6 pounds. determine the 99 % confidence interval for the mean waste recycled per person per day for the population of maine. assume the population is approximately normal.
step 2 of 2: construct the 99 % confidence interval. round your answer to one decimal place.

Explanation:

Step1: Recall t - interval formula

For a confidence interval for the population mean when the population standard deviation \(\sigma\) is unknown (we use sample standard deviation \(s\)) and the population is normal, the formula is \(\bar{x}\pm t_{\alpha/2,n - 1}\frac{s}{\sqrt{n}}\), where \(\bar{x}\) is the sample mean, \(t_{\alpha/2,n - 1}\) is the t - critical value, \(s\) is the sample standard deviation, and \(n\) is the sample size.

We know that \(\bar{x}=1.1\), \(s = 0.6\), \(n=6\), and the confidence level is \(99\%\). So, \(\alpha=1 - 0.99 = 0.01\), \(\alpha/2=0.005\), and the degrees of freedom \(df=n - 1=6 - 1 = 5\).

Looking up the t - critical value in the t - distribution table, \(t_{0.005,5}\approx4.032\).

Step2: Calculate the margin of error (E)

The margin of error \(E=t_{\alpha/2,n - 1}\frac{s}{\sqrt{n}}\).

Substitute the values: \(E = 4.032\times\frac{0.6}{\sqrt{6}}\).

First, calculate \(\sqrt{6}\approx2.449\). Then \(\frac{0.6}{2.449}\approx0.245\). Then \(E = 4.032\times0.245\approx0.98\).

Step3: Construct the confidence interval

The confidence interval is \(\bar{x}-E<\mu<\bar{x}+E\).

Substitute \(\bar{x} = 1.1\) and \(E\approx0.98\):

Lower bound: \(1.1-0.98 = 0.12\approx0.1\) (rounded to one decimal place)

Upper bound: \(1.1 + 0.98=2.08\approx2.1\) (rounded to one decimal place)

Answer:

The \(99\%\) confidence interval is \(0.1<\mu<2.1\)