Question

the total of individual weights of garbage discarded by 20 households in one week is normally distributed with a mean of 29.6 lbs with a sample standard deviation of 12.3 lbs. find the 90% confidence interval of the mean. < < do not round in between steps. round answers to at least 4 decimal places. check answer

Explanation:

Step1: Determine the critical value

Since the sample size \(n = 20\) (small sample, \(n<30\)) and the population standard deviation \(\sigma\) is unknown, we use the \(t -\)distribution. The degree of freedom \(df=n - 1=20 - 1 = 19\). For a \(90\%\) confidence interval, \(\alpha=1 - 0.90 = 0.10\), and \(\frac{\alpha}{2}=0.05\). Using the \(t -\)table or a calculator, \(t_{\frac{\alpha}{2},df}=t_{0.05,19}=1.7291\)

Step2: Calculate the margin of error \(E\)

The formula for the margin of error for a \(t -\)distribution is \(E = t_{\frac{\alpha}{2}}\frac{s}{\sqrt{n}}\), where \(s = 12.3\) (sample standard deviation) and \(n = 20\) (sample size).

$$
LATEXBLOCK0
$$

Step3: Calculate the confidence interval

The confidence interval formula for the population mean \(\mu\) (when using the \(t -\)distribution) is \(\bar{x}-E<\mu<\bar{x} + E\), where \(\bar{x}=29.6\) (sample mean)

$$
LATEXBLOCK1
$$

Answer:

\(24.8450<\mu<34.3550\)