Question

An industrial plant dumps its waste into a nearby river, but claims that it is not impacting the native fish that live in the river. You’ve measure the calcium concentration from a random sample of 24 locations in the river and calculated a mean of 98.4 mg/L, with a standard deviation of 5.1 mg/L. The fish are able to tolerate calcium concentrations up to 95 mg/L. Assuming a normal distribution, calculate a 95% confidence interval for the mean calcium concentration in the river.

Answer 1

The 95% confidence interval for the mean calcium concentration in the river is (96.2482mg/L, 100.5518mg/L).

Explanation:

Our sample size is 24.

The first step to solve this problem is finding our degrees of freedom, that is, the sample size subtracted by 1. So

`df = 24-1 = 23`.

Then, we need to subtract one by the confidence level
`\alpha` and divide by 2. So:

`(1-0.95)/(2) = (0.05)/(2) = 0.025`

Now, we need our answers from both steps above to find a value T in the t-distribution table. So, with 23 and 0.025 in the two-sided t-distribution table, we have
`T = 2.069`

Now, we find the standard deviation of the sample. This is the division of the standard deviation by the square root of the sample size. So

`s = (5.1)/(√(24)) = 1.04`

Now, we multiply T and s

`M = 2.069*1.04 = 2.1518`

Then

The lower end of the confidence interval is the mean subtracted by M. So:

`L = 98.4 - 2.1518 = 96.2482`

The upper end of the confidence interval is the mean added to M. So:

`LCL = 98.4 + 2.1518 = 100.5518`

The 95% confidence interval for the mean calcium concentration in the river is (96.2482mg/L, 100.5518mg/L).