Questions: Use the t-distribution to find a confidence interval for a mean μ given the relevant sample results. Give the best point estimate for μ, the margin of error; and the confidence interval. Assume the results come from a random sample from a population that is approximately normally distributed. A 95% confidence interval for μ using the sample results x̄=87.0, s=7.2, and n=42 Round your answer for the point estimate to one decimal place, and your answers for the margin of error and the confidence interval to two decimal places. point estimate = margin of error = The 95% confidence interval is to .

Solution Steps

The best point estimate for the population mean \( \mu \) is given by the sample mean \( \bar{x} \): \[ \text{Point estimate} = \bar{x} = 87.0 \]

To calculate the margin of error, we use the formula: \[ \text{Margin of Error} = Z \cdot \frac{s}{\sqrt{n}} \] where \( Z \) is the Z-score corresponding to the 95% confidence level, \( s \) is the sample standard deviation, and \( n \) is the sample size. For a 95% confidence level, \( Z = 1.96 \).

Substituting the values: \[ \text{Margin of Error} = 1.96 \cdot \frac{7.2}{\sqrt{42}} \approx 2.18 \]

The confidence interval for the mean is calculated as: \[ \bar{x} \pm \text{Margin of Error} \] Thus, we have: \[ 87.0 \pm 2.18 \] Calculating the lower and upper bounds: \[ \text{Lower Bound} = 87.0 - 2.18 = 84.82 \] \[ \text{Upper Bound} = 87.0 + 2.18 = 89.18 \] Therefore, the 95% confidence interval is: \[ (84.82, 89.18) \]

Final Answer