Respuesta :
Answer:
The 95% confidence interval for the mean calcium concentration in the river is (96.2482mg/L, 100.5518mg/L).
Step-by-step 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
[tex]df = 24-1 = 23[/tex].
Then, we need to subtract one by the confidence level [tex]\alpha[/tex] and divide by 2. So:
[tex]\frac{1-0.95}{2} = \frac{0.05}{2} = 0.025[/tex]
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 [tex]T = 2.069[/tex]
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
[tex]s = \frac{5.1}{\sqrt{24}} = 1.04[/tex]
Now, we multiply T and s
[tex]M = 2.069*1.04 = 2.1518[/tex]
Then
The lower end of the confidence interval is the mean subtracted by M. So:
[tex]L = 98.4 - 2.1518 = 96.2482[/tex]
The upper end of the confidence interval is the mean added to M. So:
[tex]LCL = 98.4 + 2.1518 = 100.5518[/tex]
The 95% confidence interval for the mean calcium concentration in the river is (96.2482mg/L, 100.5518mg/L).
