Question

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

Answer 1

Answer:

The sample size is $n = 2$

Step-by-step explanation:

From the question we are told that

   The margin of error is  E =4.266

    The standard deviation is  $\sigma = 2.539$

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

      $\alpha = (100 - 99 ) \%$

=>   $\alpha = 0.01$

Generally from the normal distribution table the critical value  of  $\frac{\alpha }{2}$ is  

   $Z_{\frac{\alpha }{2} } = 2.58$

Generally the sample size is mathematically represented as  

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

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

=>  $n = 2$