Question

Sleep – College Students (Raw Data, Software Required):
Suppose you perform a study about the hours of sleep that college students get. You know that for all people, the average is about 7.0 hours per night. You randomly select 35 college students and survey them on the number of hours of sleep they get per night. The data is found in the table below. You want to construct a 99% confidence interval for the mean hours of sleep for all college students. You will need software to answer these questions. You should be able to copy the data directly from the table into your software program.

(a) What is the point estimate for the mean nightly hours of sleep for all college students? Round your answer to 2 decimal places. hours (b) Construct the 99% confidence interval for the mean nightly hours of sleep for all college students. Round your answers to 2 decimal places. < μ < (c) Are you 99% confident that the mean nightly hours of sleep for all college students is below the average for all people of 7.0 hours per night? Why or why not? Yes, because 7.0 is above the upper limit of the confidence interval for college students. No, because 7.0 is below the upper limit of the confidence interval for college students.   Yes, because 7.0 is below the upper limit of the confidence interval for college students. No, because 7.0 is above the upper limit of the confidence interval for college students. (d) We are never told whether or not the parent population is normally distributed. Why could we use the above method to find the confidence interval? Because the sample size is greater than 30. Because the sample size is less than 100.     Because the margin of error is positive. Because the margin of error is less than 30.

Student Sleep    1 5.0 2 6.6 3 7.3 4 6.4 5 7.3 6 4.6 7 7.8 8 5.7 9 8.7 10 4.8 11 3.6 12 4.8 13 5.3 14 9.3 15 9.9 16 6.5 17 8.8 18 5.1 19 4.7 20 7.0 21 5.0 22 7.4 23 4.4 24 7.5 25 5.7 26 5.8 27 7.3 28 6.4 29 6.1 30 7.6 31 7.1 32 9.7 33 7.3 34 7.1 35 5.2

Answer 1

a) Sample Mean,    x̅ = ΣX/n =    6.54

b)

Level of Significance ,    α =    0.01          
degree of freedom=   DF=n-1=   34          
't value='   tα/2=   2.728   [Excel formula =t.inv(α/2,df) ]      
                  
Standard Error , SE = s/√n =   1.5808/√35=   0.2672          
margin of error , E=t*SE =   2.7284   *   0.2672   =   0.729
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    6.54   -   0.7290   =   5.8081
Interval Upper Limit = x̅ + E =    6.54   -   0.7290   =   7.2662
99%   confidence interval is (   5.81   < µ <   7.27   )

c)

No, because 7.0 is below the upper limit of the confidence interval for college students.  

d)

Because the sample size is greater than 30.