Respuesta :

Answer:

The  90%  confidence interval for population mean is   [tex]14.7 < \mu < 19.1[/tex]

Step-by-step explanation:

From the question we are told that

   The sample mean is  [tex]\= x = 16.9[/tex]

    The confidence level is  [tex]C = 0.90[/tex]

     The sample size is  [tex]n = 45[/tex]

     The standard deviation

Now given that the confidence level is  0.90 the  level of significance is mathematically evaluated as

       [tex]\alpha = 1-0.90[/tex]

       [tex]\alpha = 0.10[/tex]

Next we obtain the critical value of  [tex]\frac{\alpha }{2}[/tex]  from the standardized normal distribution table. The values is  [tex]Z_{\frac{\alpha }{2} } = 1.645[/tex]

The  reason we are obtaining critical values for [tex]\frac{\alpha }{2}[/tex]  instead of  that of  [tex]\alpha[/tex]  is because [tex]\alpha[/tex]  represents the area under the normal curve where the confidence level 1 - [tex]\alpha[/tex] (90%)  did not cover which include both the left and right tail while [tex]\frac{\alpha }{2}[/tex]  is just considering the area of one tail which is what we required calculate the margin of error

  Generally the margin of error is mathematically evaluated as

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

substituting values

         [tex]MOE = 1.645* \frac{ 9 }{\sqrt{45} }[/tex]

         [tex]MOE = 2.207[/tex]

The  90%  confidence level interval is mathematically represented as

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

substituting values

     [tex]16.9 - 2.207 < \mu < 16.9 + 2.207[/tex]

    [tex]16.9 - 2.207 < \mu < 16.9 + 2.207[/tex]

     [tex]14.7 < \mu < 19.1[/tex]

         

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