Question

In a random sample of 5 residents of the state of Texas, the mean waste recycled per person per day was 1.9 pounds with a standard deviation of 0.89 pounds. Determine the 99% confidence interval for the mean waste recycled per person per day for the population of Texas. 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

Answer with explanation:

The confidence interval for population mean is given by :-

$\overline{x} \pm\ t_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}$

Given : Sample size : n= 5, since n<30 , so the test we use here is t-test.

Sample mean : $\overline{x}=1.9\text{ pounds}$

Standard deviation: $\sigma=0.89\text{ pounds}$

Significance level : $1-\alpha:1-0.99=0.01$

By using the standard normal distribution table , the critical value corresponds to the given significance level will be :-

$t_{n-1,\alpha/2}=t_{5-1,0.01/2}=t_{4,0.005}=4.604$

Now, the 99% confidence interval for the mean waste recycled per person per day for the population of Texas will be :-

$1.9\pm\ (4.604)\dfrac{0.89}{\sqrt{5}}\\\\\approx1.9\pm1.832\\\\=(1.9-1.832,1.9+1.832)=(0.068,\ 3.732)$

Hence, the 99% confidence interval for the mean waste recycled per person per day for the population of Texas  = (0.068, 3.732)