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

Answer:

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

Step-by-step 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  $\frac{\alpha }{2}$ is  

   $Z_{\frac{\alpha }{2} } =  1.282$

Generally the margin of error is mathematically represented as  

      $E = 1.282  *  \frac{0.012 }{\sqrt{23} }$

=>    $E = 0.00321$

Generally 80% confidence interval is mathematically represented as  

      $0.036  -0.00321 <  \mu <  0.036  + 0.00321$

=>  $0.0328 <  \mu <  0.0392$