Question

An economist is interested in studying the spending habits of consumers in a particular region. The population standard deviation is known to be $1,000. A random sample of 50 individuals resulted in an average expense of $15,000. What is the width of the 99% confidence interval for the mean of expense? a. 364.28 b. 728.55 c. 329.00 d. 657.99

Answer 1

Answer:

The  width is  $w = \ 729.7$

Step-by-step explanation:

From the question we are told that

   The population standard deviation is  $\sigma = \% 1,000$

    The  sample size is  $n = 50$

    The sample mean  is  $\= x = \ 15,000$

Given that the confidence level is  99% then the level of significance is mathematically represented as  

               $\alpha = 100 - 99$

=>            $\alpha = 1\%$

=>             $\alpha = 0.01$

Next we obtain the critical value of  $\frac{\alpha }{2}$ from the normal distribution table, the value is

             $Z_{\frac{\alpha }{2} } = Z_{\frac{0.01 }{2} } = 2.58$

Generally margin of error is mathematically represented as

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

substituting values

              $E = 2.58 * \frac{1000 }{\sqrt{50} }$

              $E = 2.58 * \frac{1000 }{\sqrt{50} }$

               $E = 364.9$

The width of the 99% confidence interval is mathematically evaluated as

         $w = 2 * E$

substituting values

          $w = 2 * 364.9$

          $w = \ 729.7$