Question

In a random sample of 8 residents of the state of Montana, the mean waste recycled per person per day was 3.0 pounds with a standard deviation of 0.23 pounds. Determine the 99% confidence interval for the mean waste recycled per person per day for the population of Montana. Assume the population is approximately normal. Step 1 of 2: Find the critical value that should be used in constructing the confidence interval. Round your answer to three decimal places.

Answer 1

at 99%, Confidence interval is ( 2.716, 3.285 )

( 2.716 < μ < 3.285 )

Explanation:

Given the data in the question;

sample size n = 8

mean x' = 3.0

standard deviation σ = 0.23

To calculate the confidence interval, we use the following expression;

x' ±
`t_{\alpha /2, df`( σ/√n) ------- let this be equation

now, at 99% confidence interval;

∝ = 1 - 99% = 1 - 0.99 = 0.01

degree of freedom df = n - 1 = 8 - 1 = 7

hence t-critical value = 3.499

in equation 1, we substitute in our values;

⇒ 3.0 ± 3.499( 0.23/√8 )

⇒ 3.0 ± 0.2845

⇒ 3.0 - 0.2845, 3.0 + 0.2845

⇒ ( 2.7155, 3.2845 )

Therefore at 99%, Confidence interval is ( 2.716, 3.285 )

( 2.716 < μ < 3.285 )