Complete Question
Suppose that the amount of time teenagers spend weekly working at part-time jobs is normally distributed with a standard deviation of 40 minutes. A random sample of 15 teenagers was drawn and each reported the amount of the time spent at part-time jobs (in minutes). These are listed here 180, 130, 150, 165, 90, 130, 120, 60, 200, 180, 80, 240, 210, 150, 125. Determine the 95% confidence interval estimate of the population mean.
Answer:
The 95% confidence interval is  [tex] 127.09  <  \mu <  167.57 [/tex]
Step-by-step explanation:
From the question we are told that
  The standard deviation is  [tex]\sigma = 40[/tex]
  The sample size is  n  = 15
   The data given is  180, 130, 150, 165, 90, 130, 120, 60, 200, 180, 80, 240, 210, 150, 125
Generally the sample mean is mathematically represented as
       [tex]\= x = \frac{ \sum x_i}{n }[/tex]
=> Â Â Â Â Â [tex]\= x = \frac{ 130 + 150 + \cdots + 125 }{ 15 }[/tex]
=> Â Â Â Â Â [tex]\= x = 147.33[/tex]
From the question we are told the confidence level is  95% , hence the level of significance is  Â
   [tex]\alpha = (100 - 95 ) \%[/tex]
=> Â [tex]\alpha = 0.05[/tex]
Generally from the normal distribution table the critical value  of  [tex]\frac{\alpha }{2}[/tex] is Â
  [tex]Z_{\frac{\alpha }{2} } =  1.96[/tex]
Generally the margin of error is mathematically represented as Â
   [tex]E = Z_{\frac{\alpha }{2} } *  \frac{\sigma }{\sqrt{n} }[/tex]
=> Â Â [tex]E = 1.96 Â * Â \frac{40}{\sqrt{15} }[/tex]
=> Â Â [tex]E =20.24 [/tex]
Generally 95% confidence interval is mathematically represented as Â
   [tex] 147.33 -20.24<  \mu <  147.33 + 20.24 [/tex]
   [tex] 127.09  <  \mu <  167.57 [/tex]
 Â
