Question

A student wants to estimate the mean score of all college students for a particular exam. First use the range rule of thumb to make a rough estimate of the standard deviation of those scores. Possible scores range from 500 to 2200. Use technology and the estimated standard deviation to determine the sample size corresponding to a 95​% confidence level and a margin of error of 100 points. What​ isn't quite right with this​ exercise?

Answer 1

70

Explanation:

it is given that score varies from 500 to 2200

so range =2200-500=1700

standard deviation
`s=(range)/(4)=(1700)/(4)=425`

Error E=100

Confidence level =95%=0.95

significance level α=1-0.95=0.05

`z_{(\alpha )/(2)}=z_{(0.05)/(2)}=z_(0.025=1.96)` from the z table

sample size
`n\geq\left ( z_(\alpha )/(2)* (s)/(E) \right )^2`

`n\geq\left ( 1.96* (425)/(100) \right )^2`

`n\geq69.3889`

so n will be 70