A professor at a university wants to estimate the average number of hours of sleep the students get during exam week. The professor wants to find a sample mean that is within 4.266 hours of the true average for all college students at the university with 99% conconfidence. If the professor knows the standard deviation of the sleep times for all college students is 2.539, what sample size will need to be taken

Respuesta :

Answer:

The sample size is [tex]n = 2[/tex]

Step-by-step explanation:

From the question we are told that

   The margin of error is  E =4.266

    The standard deviation is  [tex]\sigma = 2.539[/tex]

From the question we are told the confidence level is  99% , hence the level of significance is    

      [tex]\alpha = (100 - 99 ) \%[/tex]

=>   [tex]\alpha = 0.01[/tex]

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

   [tex]Z_{\frac{\alpha }{2} } =  2.58[/tex]

Generally the sample size is mathematically represented as  

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

=>  [tex]n = [\frac{2.58  *  2.539  }{4.266} ] ^2[/tex]

=>  [tex]n = 2[/tex]