Question

You measure 41 textbooks' weights, and find they have a mean weight of 30 ounces. Assume the population standard deviation is 12.3 ounces. Based on this, construct a 93% confidence interval for the true population mean textbook weight. Round z-values to 2 decimal places; round t-values to 3 decimal places. Round final answers to 2 decimal places.

Answer 1

Answer:

The 93% confidence interval is   $26.52 < \mu < 33.48$

Step-by-step explanation:

From the question we are told that

    The sample size is  n =  41

     The mean is  $\= x = 30$

     The standard deviation is  $\sigma = 12.3$

From the question we are told the confidence level is  93% , hence the level of significance is    

      $\alpha = (100 - 93 ) \%$

=>   $\alpha = 0.07$

Generally from the normal distribution table the critical value  of  $\frac{\alpha }{2}$ is  

   $Z_{\frac{\alpha }{2} } = 1.81$

Generally the margin of error is mathematically represented as  

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

=>   $E = 1.812 * \frac{12.3}{\sqrt{41} }$  

=>     $E = 3.48$  

Generally 93% confidence interval is mathematically represented as  

      $\= x -E < \mu < \=x +E$

=>    $30 -3.48 < \mu < 30 + 3.48$

=>     $26.52 < \mu < 33.48$