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`

Explanation:

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

`\overline{x}\pm t_c(s)/(√(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)(0.0036)/(√(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`