Question

A physicist examines 28 sedimentary samples for mercury concentration. The mean mercury concentration for the sample data is 0.863 cc/cubic meter with a standard deviation of 0.0036. Determine the 98% confidence interval for the population mean mercury concentration. Assume the population is approximately normal. Step 2 of 2: Construct the 98% confidence interval. Round your answer to three decimal places.

Answer 1

Answer: $0.861<\mu< 0.865$

Step-by-step explanation:

The confidence interval for population mean $(\mu)$ is given by :-

$\overline{x}\pm t_c\dfrac{s}{\sqrt{n}}$ , where n=sample size

$\overline{x}$= sample mean

$s$=sample standard deviation

$t_c$= critical t-value (for two tailed )

Let  $\mu$ be the confidence interval for the population mean mercury concentration.

As per given , we have

Sample size : n= 28

degree of freedom = 27   [df=n-1]

Sample mean : $\overline{x}=0.863$ cc/cubic meter

Sample standard deviation : s= 0.0036

Significance level : $\alpha=1-0.98=0.02$

Critical two-tailed test value :

$t_c=t_{\alpha/2,df}=t_{0.01,\ 27}= 2.473$  (Using t-distribution table.)

We assume the population is approximately normal.

Now, the 98% confidence interval for the population mean mercury concentration will be :-

$0.863\pm (2.473)\dfrac{0.0036}{\sqrt{28}}\\\\=0.863\pm(0.002)\\\\=(0.863-0.002,\ 0.863+0.002)=(0.861,\ 0.865)$

Required confidence interval : $0.861<\mu< 0.865$