Questions: Find the necessary confidence interval for a population mean μ for the following values. (Round your answers to two decimal places.) 0.95% confidence interval, n=81, x̄=2.93, s²=0.1017

Solution Steps

Given the variance \( s^2 = 0.1017 \), the sample standard deviation \( s \) is calculated as follows:

\[ s = \sqrt{s^2} = \sqrt{0.1017} \approx 0.3189 \]

Using the formula for the margin of error \( E \) at a confidence level of \( 95\% \) (where the Z-score \( Z \) is approximately \( 1.96 \)), we have:

\[ E = Z \cdot \frac{s}{\sqrt{n}} = 1.96 \cdot \frac{0.3189}{\sqrt{81}} = 1.96 \cdot \frac{0.3189}{9} \approx 0.0694 \]

The confidence interval for the population mean \( \mu \) is given by:

\[ \bar{x} \pm E = 2.93 \pm 0.0694 \]

Calculating the lower and upper bounds:

\[ \text{Lower Bound} = 2.93 - 0.0694 \approx 2.8606 \quad \text{(rounded to 2.86)} \] \[ \text{Upper Bound} = 2.93 + 0.0694 \approx 3.0004 \quad \text{(rounded to 3.00)} \]

Thus, the confidence interval is:

\[ (2.86, 3.00) \]

Final Answer