Question

A study of 25 graduates of four-year public colleges revealed the mean amount owed by a student in student loans was $55,051. The standard deviation of the sample was $7,568.
Required:
a. Construct a 90% confidence interval for the population mean.
b. Confidence interval for the population men between _______ up to_______________

Answer 1

Answer:

a

   The  90%  confidence interval is  $52561.13 < \mu < 57540.8$

b

Confidence interval for the population men between $52561.13  up to $57540.8

Step-by-step explanation:

From the question we are told that

   The sample size is  $n = 25$

     The  sample mean is  $\= x = \ 55,051$

     The standard deviation is  $\sigma = \ 7,568$

Given that the confidence level is  90% then the level of confidence is mathematically represented as

             $\alpha = 100 -90$

              $\alpha = 10\%$

             $\alpha = 0.10$

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

               $Z_{\frac{\alpha }{2} } = 1.645$

Generally the margin of error is mathematically  represented as

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

substituting values

               $E = 1.645 * \frac{ 7568}{ \sqrt{ 25} }$

             $E = 2489.9$

The 90% confidence interval is mathematically evaluated as

       $\= x -E < \mu < \= x +E$

substituting values

     $55051 - 2489.8 < \mu < 55051 + 2489.8$

     $52561.13 < \mu < 57540.8$