Questions: Question 8 Use the following sample to estimate a population mean μ. 3.3 1.4 34.4 -4.6 30.9 18.5 15.9 49 30 45.4 -12.4 44.7 -1.5 25 14.8 34.8 19.6 45.6 32.9 -7.4 20.1 35.1 18.3 19.5 23.3 21.9 21.4 34.9 35.4 12.8 37.5 25.4 5.1 -3.7 20.5 46.9 11.1 42.7 24 30.7 Find the 99% confidence interval about the population mean. Enter your answer as a tri-linear inequality accurate to two decimal place. < μ <

Question 8 Use the following sample to estimate a population mean μ. 3.3 1.4 34.4 -4.6 30.9 18.5 15.9 49 30 45.4 -12.4 44.7 -1.5 25 14.8 34.8 19.6 45.6 32.9 -7.4 20.1 35.1 18.3 19.5 23.3 21.9 21.4 34.9 35.4 12.8 37.5 25.4 5.1 -3.7 20.5 46.9 11.1 42.7 24 30.7 Find the 99% confidence interval about the population mean. Enter your answer as a tri-linear inequality accurate to two decimal place. < μ <
Transcript text: Question 8 Use the following sample to estimate a population mean $\mu$. \begin{tabular}{|r|r|r|r|} \hline 3.3 & 1.4 & 34.4 & -4.6 \\ \hline 30.9 & 18.5 & 15.9 & 49 \\ \hline 30 & 45.4 & -12.4 & 44.7 \\ \hline-1.5 & 25 & 14.8 & 34.8 \\ \hline 19.6 & 45.6 & 32.9 & -7.4 \\ \hline 20.1 & 35.1 & 18.3 & 19.5 \\ \hline 23.3 & 21.9 & 21.4 & 34.9 \\ \hline 35.4 & 12.8 & 37.5 & 25.4 \\ \hline 5.1 & -3.7 & 20.5 & 46.9 \\ \hline 11.1 & 42.7 & 24 & 30.7 \\ \hline \end{tabular} Find the $99 \%$ confidence interval about the population mean. Enter your answer as a tri-linear inequality accurate to two decimal place. $\square$ $<\mu<$ $\square$
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Solution

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Solution Steps

Step 1: Calculate the Sample Mean

The sample mean \( \bar{x} \) is calculated using the formula:

\[ \bar{x} = \frac{\sum_{i=1}^N x_i}{N} = \frac{903.2}{40} = 22.58 \]

Step 2: Calculate the Sample Standard Deviation

The variance \( \sigma^2 \) is calculated as follows:

\[ \sigma^2 = \frac{\sum (x_i - \mu)^2}{n-1} = 261.22 \]

The sample standard deviation \( s \) is then:

\[ s = \sqrt{261.22} = 16.16 \]

Step 3: Calculate the 99% Confidence Interval

For a 99% confidence level, we use the formula for the confidence interval:

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

Where \( z \) for 99% confidence is approximately \( 2.58 \). Thus, the confidence interval is:

\[ 22.58 \pm 2.58 \cdot \frac{16.16}{\sqrt{40}} = (16.0, 29.16) \]

Final Answer

The 99% confidence interval for the population mean \( \mu \) is:

\[ \boxed{16.0 < \mu < 29.16} \]

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