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]
=> [tex]\= x = \frac{ 130 + 150 + \cdots + 125 }{ 15 }[/tex]
=> [tex]\= x = 147.33[/tex]
From the question we are told the confidence level is 95% , hence the level of significance is
[tex]\alpha = (100 - 95 ) \%[/tex]
=> [tex]\alpha = 0.05[/tex]
Generally from the normal distribution table the critical value of [tex]\frac{\alpha }{2}[/tex] is
[tex]Z_{\frac{\alpha }{2} } = 1.96[/tex]
Generally the margin of error is mathematically represented as
[tex]E = Z_{\frac{\alpha }{2} } * \frac{\sigma }{\sqrt{n} }[/tex]
=> [tex]E = 1.96 * \frac{40}{\sqrt{15} }[/tex]
=> [tex]E =20.24 [/tex]
Generally 95% confidence interval is mathematically represented as
[tex] 147.33 -20.24< \mu < 147.33 + 20.24 [/tex]
[tex] 127.09 < \mu < 167.57 [/tex]
