Question

In a random sample of 16 residents of the state of Washington, the mean waste recycled per person per day was 2.8 pounds with a standard deviation of 0.24 pounds. Determine the 80% confidence interval for the mean waste recycled per person per day for the population of Washington. Assume the population is approximately normal.

Answer 1

Answer: The required confidence interval would be (2.72,2.89)

Step-by-step explanation:

Since we have given that

Mean = 2.8 pounds

Standard deviation = 0.24 pounds

n = sample size = 16

We need to find the 80% confidence interval for the mean waste.

z=1.341

So, the confidence interval will be

$\bar{x}\pm z\dfrac{\sigma}{\sqrt{n}}\\\\=(2.8\pm 1.341\times \dfrac{0.24}{\sqrt{16}})\\\\=(2.8\pm 1.314\times 0.06)\\\\=(2.8-0.07884,2.8+0.07884)\\\\=(2.72116,2.87884)$

Hence, the required confidence interval would be (2.72,2.89)