Question

A physicist examines 23 sedimentary samples for mercury concentration. The mean mercury concentration for the sample data is 0.036 cc/cubic meter with a standard deviation of 0.012. Determine the 80% confidence interval for the population mean mercury concentration. Assume the population is approximately normal. Step 1 of 2: Find the critical value that should be used in constructing the confidence interval. Round your answer to three decimal places.

Answer 1

The 80% confidence interval is
`0.0328 <  \mu <  0.0392`

Explanation:

From the question we are told that

The sample size is n = 23

The sample mean is
`\= x = 0.036 \ cc/m^3`

The standard deviation is
`\sigma = 0.012`

From the question we are told the confidence level is 80% , hence the level of significance is

`\alpha = (100 - 80 ) \%`

=>
`\alpha = 0.20`

Generally from the normal distribution table the critical value of
`(\alpha )/(2)` is

`Z_{(\alpha )/(2) } =  1.282`

Generally the margin of error is mathematically represented as

`E = 1.282  *  (0.012 )/(√(23) )`

=>
`E = 0.00321`

Generally 80% confidence interval is mathematically represented as

`0.036  -0.00321 <  \mu <  0.036  + 0.00321`

=>
`0.0328 <  \mu <  0.0392`