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

The sample size is
`n = 2`

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
`(\alpha )/(2)` is

`Z_{(\alpha )/(2) } =  2.58`

Generally the sample size is mathematically represented as

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

=>
`n = [(2.58  *  2.539  )/(4.266) ] ^2`

=>
`n = 2`