Answer:
The 80% confidence interval is  [tex]0.0328 <  \mu <  0.0392[/tex]
Step-by-step explanation:
From the question we are told that
 The sample size is  n = 23
 The sample mean is  [tex]\= x = 0.036 \ cc/m^3[/tex]
  The standard deviation is  [tex]\sigma = 0.012[/tex]
From the question we are told the confidence level is  80% , hence the level of significance is  Â
   [tex]\alpha = (100 - 80 ) \%[/tex]
=> Â [tex]\alpha = 0.20[/tex]
Generally from the normal distribution table the critical value  of  [tex]\frac{\alpha }{2}[/tex] is Â
  [tex]Z_{\frac{\alpha }{2} } =  1.282[/tex]
Generally the margin of error is mathematically represented as Â
   [tex]E = 1.282  *  \frac{0.012 }{\sqrt{23} }[/tex]
=> Â Â [tex]E = 0.00321 [/tex]
Generally 80% confidence interval is mathematically represented as Â
   [tex]0.036  -0.00321 <  \mu <  0.036  + 0.00321[/tex]
=> Â [tex]0.0328 < Â \mu < Â 0.0392[/tex]
 Â
