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: [tex]0.861<\mu< 0.865[/tex]Step-by-step explanation:The confidence interval for population mean [tex](\mu)[/tex] is given by :-[tex]\overline{x}\pm t_c\dfrac{s}{\sqrt{n}}[/tex] , where n=sample size[tex]\overline{x}[/tex]= sample mean[tex]s[/tex]=sample standard deviation[tex]t_c[/tex]= critical t-value (for two tailed )Let  [tex]\mu[/tex] be the confidence interval for the population mean mercury concentration.As per given , we haveSample size : n= 28degree of freedom = 27   [df=n-1]Sample mean : [tex]\overline{x}=0.863[/tex] cc/cubic meterSample standard deviation : s= 0.0036Significance level : [tex]\alpha=1-0.98=0.02[/tex]Critical two-tailed test value : [tex]t_c=t_{\alpha/2,df}=t_{0.01,\ 27}= 2.473[/tex]  (Using t-distribution table.)We assume the population is approximately normal. Now, the 98% confidence interval for the population mean mercury concentration will be :-[tex]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)[/tex] Required confidence interval : [tex]0.861<\mu< 0.865[/tex]