In: Statistics and Probability
Some colleges have pushed back the starting time of morning classes because students were not getting enough sleep. A random sample of 46 college students had an average of 6.54 hours of sleep with a standard deviation is 1.90 hours. Compute and interpret a 98% confidence interval for the population mean amount of sleep that college students get per night. Also determine the critical values and calculate the margin of error.
Solution :
Given that,
Point estimate = sample mean =
=6.54
Population standard deviation =
= 1.90
Sample size = n =46
At 98% confidence level the z is ,
= 1 - 98% = 1 - 0.98 = 0.02
/ 2 = 0.02/ 2 = 0.01
Z/2 = Z0.01 = 2.326 ( Using z table )
Margin of error = E = Z/2
* (
/n)
= 2.326 * ( 1.90 / 46
)
E= 0.6516
At 98% confidence interval estimate of the population mean
is,
- E <
<
+ E
6.54 - 0.6516 <
<6.54 + 0.6516
5.8884 <
< 7.1916
( 5.8884 , 7.1916)
Margin of error = E= 0.6516