Some colleges have pushed back the starting time of morning classes because students were not getting...

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.

Solutions

Expert Solution

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 * ( / $\sqrt$ n)

= 2.326 * ( 1.90 / $\sqrt$ 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

orchestra answered 1 year ago

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