Question

In a survey, 29 people were asked how much they spent on their child's last birthday gift. The results were roughly bell-shaped with a mean of $41 and standard deviation of $8. Construct a confidence interval at a 99% confidence level.
Give your answers to one decimal place.

Answer 1

Answer:

The  99%  confidence interval is

                     $37.167< \= x < 44.833$

Step-by-step explanation:

From the question we are told that

  The sample size is  $n = 29$

  The  sample mean is  $\= x =$$41

  The  sample standard deviation is  $\sigma =$$8

   The  level of confidence is $C =$99%

Given that the confidence level id  99% the level of confidence is evaluated as

        $\alpha = 100 - 99$

        $\alpha = 1$%

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

      $Z_{\frac{\alpha }{2} } = 2.58$

The reason we are obtaining values for  is because  is the area under the normal distribution curve for both the left and right tail where the 99% interval did not cover while   is the area under the normal distribution curve for just one tail and we need the  value for one tail in order to calculate the confidence interval

Next we evaluate the margin of error which is mathematically represented as

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

substituting values

         $MOE = 2.58 * \frac{8 }{\sqrt{29} }$

           $MOE = 3.8328$

The 99% confidence level is constructed as follows

      $\= x - MOE < \= x < \= x + MOE$

substituting values

    $41 - 3.8328 < \= x < 41 + 3.8328$

     $37.167< \= x < 44.833$