Question

In a sample of eight whitefish caught in Yellowknife Bay, the mean arsenic concentration in the liver was 0.32 mg/kg, with a standard deviation of 0.06 mg/kg. Find a 95% confidence interval for the concentration in whitefish found in Yellowknife Bay. (Round the final answers to four decimal places.)

Answer 1

Answer:

The 95% confidence interval for the concentration in whitefish found in Yellowknife Bay is (0.2698 mg/kg, 0.3702 mg/kg).

Step-by-step explanation:

We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 8 - 1 = 7

95% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 7 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.95}{2} = 0.975$. So we have T = 2.3246

The margin of error is:

$M = T\frac{s}{\sqrt{n}} = 2.3646\frac{0.06}{\sqrt{8}} = 0.0502$

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 0.32 - 0.0502 = 0.2698 mg/kg

The upper end of the interval is the sample mean added to M. So it is 0.32 + 0.0502 = 0.3702 mg/kg

The 95% confidence interval for the concentration in whitefish found in Yellowknife Bay is (0.2698 mg/kg, 0.3702 mg/kg).