Question

a) Suppose that a random sample of size 80 has a sample mean of 1.69 ppb and a sample standard deviation of 0.25 ppb. Find a 95% confidence interval for the PCB concentration.

Answer 1

Answer: The confidence interval is 1.69ppb±0.055

Step-by-step explanation:

To find the confidence interval, first find the z* which is the appropriate z*-value related to the confidence interval of preference. As in this is 95%, z*-value = 1.96 (Note: The value is from the standard normal Z distribution).

With the z*-value, multiply by the division of the standard deviation and the square root of the sample size:

z*-value . σ/√n = 1.96 · $\frac{0.25}{\sqrt{80} }$ = 0.055

The 95% confidence interval of the sample mean of 1.69 ppb is 1.69ppb±0.055ppb