Question

A similar follow-up study is done on a sample of 25 meat-lovers who never eat vegetables, again randomly selected from the same general population (population mean life expectancy = 75, population standard deviation = 5). This new sample of meat-eaters live to an average age of 77. What is the lower limit and upper limit of the 95% confidence interval for the life expectancy of this sample of meat-lovers?

Answer 1

Answer:

The lower limit is  75.04

The upper  limit is  78.96

Step-by-step explanation:

From the question we are told that

   The sample size is  $n = 25$

   The sample mean is  $\= x = 77$

    The standard deviation is  $\sigma = 5$

Given that the confidence level is 95%  then the level of significance is mathematically represented as

        $\alpha = (100 - 95)\%$

         $\alpha = 0.05$

The critical value for  $\frac{\alpha }{2}$ obtained from the normal distribution table is  

     $Z_{\frac{\alpha }{2} } = 1.96$

Generally the margin of error is mathematically represented as

       $E = Z_{\frac{\alpha }{2} } * \frac{\sigma }{\sqrt{n} }$

=>   $E = 1.96* \frac{5 }{\sqrt{25} }$

=>    $E = 1.96$

The 95% confidence interval is mathematically represented as

        $\= x -E < \mu < \= x +E$

=>   $77 - 1.96 < \mu < 77 + 1.96$

=>   $75.04 < \mu < 78.96$