Question

The Bureau of Alcohol, Tobacco, and Firearms (BATF) has been concerned about lead levels in California wines. In a previous testing of wine specimens, lead levels ranging from 51 to 710 parts per billion were recorded. How many wine specimens should be tested if the BATF wishes to estimate the true mean lead level for California wines to within 10 parts per billion with 95% confidence

Answer 1

Answer:

The sample size is  $n = 1043$

Step-by-step explanation:

From the question we are told that

  The lower bound of the range is a = 51

  The upper bound of the range is b =  710

    The margin of error is $E = 10$

Generally the standard deviation is mathematically represented as

       $\sigma = \frac{b- a}{4}$

=>    $\sigma = \frac{710 - 51 }{4}$

=>    $\sigma = 164.75$

From the question we are told the confidence level is  95% , hence the level of significance is    

      $\alpha = (100 - 95 ) \%$

=>   $\alpha = 0.05$

Generally from the normal distribution table the critical value  of  $\frac{\alpha }{2}$ is  

   $Z_{\frac{\alpha }{2} } = 1.96$

Generally the sample size is mathematically represented as  

    $n = [\frac{Z_{\frac{\alpha }{2} } * \sigma }{E} ] ^2$

=> $n = [ 1.96 * 164.75 }{10} ] ^2$

=> $n = 1043$