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.

Answers

Answer 1

Answer:

The 93% confidence interval is   [tex] 26.52   <  \mu < 33.48 [/tex]

Step-by-step explanation:

From the question we are told that

    The sample size is  n =  41

     The mean is  [tex]\= x = 30[/tex]

     The standard deviation is  [tex]\sigma = 12.3[/tex]

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

      [tex]\alpha = (100 - 93 ) \%[/tex]

=>   [tex]\alpha = 0.07[/tex]

Generally from the normal distribution table the critical value  of  [tex]\frac{\alpha }{2}[/tex] is  

   [tex]Z_{\frac{\alpha }{2} } =  1.81[/tex]

Generally the margin of error is mathematically represented as  

      [tex]E = Z_{\frac{\alpha }{2} } *  \frac{\sigma }{\sqrt{n} }[/tex]

=>   [tex]E =  1.812  *  \frac{12.3}{\sqrt{41} }[/tex]  

=>     [tex]E =  3.48 [/tex]  

Generally 93% confidence interval is mathematically represented as  

      [tex]\= x -E <  \mu <  \=x  +E[/tex]

=>    [tex] 30  -3.48  <  \mu < 30  + 3.48 [/tex]

=>     [tex] 26.52   <  \mu < 33.48 [/tex]

     


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Answers

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Answers

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              Im new so If this Did not help please get another person                  

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