Question

Data were collected on random specimens of Yellowfin tuna from 1991 and 2010. For the sample of 231 specimens of Yellowfin tuna, the average mercury level was found to be 0.354 ppm and the standard deviation of mercury level to be 0.231 ppm. The data are not strongly skewed. Find the 95% confidence interval for the average mercury level. Round to three decimal places.

Answer 1

Answer:

95% confidence interval: (0.325 ,0.383)

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 231

Sample mean = 0.354 ppm

Sample standard deviation = 0.231 ppm

95% confidence interval:

$\bar{x} \pm t_{critical}\displaystyle\frac{s}{\sqrt{n}}$  

Putting the values, we get,  

$t_{critical}\text{ at degree of freedom 4 and}~\alpha_{0.01} = \pm 1.97$  

$0.354 \pm 1.97(\dfrac{0.231}{\sqrt{231}} ) = 0.354 \pm 0.029 = (0.325 ,0.383)$