Question

Construct 90%, 95%, and 99% confidence intervals to estimate μ from the following data. State the point estimate. Assume the data come from a normally distributed population. 12.9 11.6 11.9 12.2 12.5 11.4 12.0 11.7 11.8 12.9

(Round the intermediate values to 4 decimal places. Round your answers to 2 decimal places.)

90% confidence interval: ?≤ μ ?≤

95% confidence interval: ?≤ μ? ≤

99% confidence interval:? ≤ μ ?≤

The point estimate is?

Answer 1

The point estimate is = (12.9 + 11.6 + 11.9 + 12.2 + 12.5 + 11.4 + 12 + 11.7 + 11.8 + 12.9)/10 = 12.09

s = sqrt(((12.9 - 12.09)^2 + (11.6 - 12.09)^2 + (11.9 - 12.09)^2 + (12.2 - 12.09)^2 + (12.5 - 12.09)^2 + (11.4 - 12.09)^2 + (12 - 12.09)^2 + (11.7 - 12.09)^2 + (11.8 - 12.09)^2 + (12.9 - 12.09)^2)/9) = 0.526

At 90% confidence level, the critical value is t^* = 1.833

The 90% confidence interval is

= 11.785, 12.395

= 11.79, 12.40

The 90% confidence interval: 11.79 < < 12.40

At 95% confidence level, the critical value is t^* = 2.262

The 95% confidence interval is

= 11.714, 12.466

= 11.71, 12.47

The 95% confidence interval: 11.71 < < 12.47

At 99% confidence level, the critical value is t^* = 3.250

The 99% confidence interval is

= 11.549, 12.631

= 11.55, 12.63

The 99% confidence interval: 11.55 < < 12.63