Questions: Use the following sample to estimate a population mean μ. 13.3 16.2 16.6 36.6 57.8 22.2 24.3 Assuming the population is normally distributed, find the 80% confidence interval about the population mean. Enter your answer as an open-interval (i.e., parentheses) accurate to two decimal places. 80% C.I. =

Use the following sample to estimate a population mean μ.
13.3
16.2
16.6
36.6
57.8
22.2
24.3

Assuming the population is normally distributed, find the 80% confidence interval about the population mean. Enter your answer as an open-interval (i.e., parentheses) accurate to two decimal places.
80% C.I. =

Transcript text: Use the following sample to estimate a population mean $\mu$. \begin{tabular}{|l|} \hline 13.3 \\ \hline 16.2 \\ \hline 16.6 \\ \hline 36.6 \\ \hline 57.8 \\ \hline 22.2 \\ \hline 24.3 \\ \hline \end{tabular} Assuming the population is normally distributed, find the $80 \%$ confidence interval about the population mean. Enter your answer as an open-interval (i.e., parentheses) accurate to two decimal places. 80\% C.I. = $\square$

Solution

Solution Steps

Step 1: Calculate the Sample Mean

The sample mean $ \mu $ is calculated using the formula:

\[ \mu = \frac{\sum_{i=1}^N x_i}{N} = \frac{187.0}{7} = 26.71 \]

Step 2: Calculate the Sample Standard Deviation

The variance $ \sigma^2 $ is calculated as follows:

\[ \sigma^2 = \frac{\sum (x_i - \mu)^2}{n-1} = 247.17 \]

The standard deviation $ \sigma $ is then:

\[ \sigma = \sqrt{247.17} = 15.72 \]

Step 3: Calculate the 80% Confidence Interval

For the confidence interval, we use the formula:

\[ \bar{x} \pm t \frac{s}{\sqrt{n}} \]

Where:

$ \bar{x} = 26.71 $ (sample mean)
$ t $ is the t-value for 80% confidence level with $ n-1 = 6 $ degrees of freedom, which is approximately $ 1.44 $
$ s = 15.72 $ (sample standard deviation)
$ n = 7 $ (sample size)

Substituting the values:

\[ 26.71 \pm 1.44 \cdot \frac{15.72}{\sqrt{7}} = (18.16, 35.26) \]