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dem82 [27]
3 years ago
10

A physicist examines 23 sedimentary samples for mercury concentration. The mean mercury concentration for the sample data is 0.0

36 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.
Mathematics
1 answer:
amm18123 years ago
6 0

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

   

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