Complete Question
suppose a study estimated the population mean for a variable of interest using a 99% confidence interval. If the width of the estimated confidence interval( the difference between the upper limit and the lower limit) is 600 and the sample size used in estimating the mean is 1000, what is the population standard deviation?
Answer:
The  value is  [tex]\sigma = 3677[/tex]
Step-by-step explanation:
From the question we are told that
  The width of the 99% confidence interval is  w =  600
  The sample size is  n  =  1000
From the question we are told the confidence level is  99% , hence the level of significance is  Â
   [tex]\alpha = (100 - 99) \%[/tex]
=> Â [tex]\alpha = 0.01[/tex]
Generally from the normal distribution table the critical value  of  [tex]\frac{\alpha }{2}[/tex] is Â
  [tex]Z_{\frac{\alpha }{2} } =  2.58 [/tex]
Generally the margin of error is mathematically evaluated as
     [tex]E = \frac{w}{2}[/tex]
=> Â Â Â [tex]E = \frac{600}{2}[/tex]
=> Â Â Â [tex]E = 300[/tex]
Generally this margin of error can be mathematically represented as
    [tex]E = Z_{\frac{\alpha }{2} } * \frac{\sigma}{\sqrt{n} }[/tex]
=> Â Â [tex]300 = 2.58 * \frac{\sigma}{\sqrt{1000} }[/tex]
=> Â [tex]\sigma = 3677[/tex]
