Posted by Shyra on .
A survey is being planned to determine the mean amount of time corporation executives watch television. A pilot survey indicated that the mean time per week is 12 hours, with a standard deviation of 3 hours. It is desired to estimate the mean viewing time within one quarter hour. The 0.95 degree of confidence is to be used. How many executives should be surveyed?
Is the following soluntion correct?
[(1.96*3)/0.25]^2 = 553.1 => 554

Statistics 
matt,
YES!

Statistics 
Visha,
A survey is being planned to determine the mean amount of time corporation executives watch television. A pilot survey indicated that the mean time per week is 14 hours, with a standard deviation of 2.5 hours. It is desired to estimate the mean viewing time within onequarter hour. The 95 percent level of confidence is to be used.
How many executives should be surveyed? 
Data Analysis 
Jeannine,
Population mean = 14
Population standard deviation = 2.5
Margin error = 1/4 = 0.25
level of confidence = 0.95
z = 1.96
n =[(1.96*3)/0.25]^2
=553.19
=554 
Statistics 
Shawn,
Statistical Techniques in Business and Economics 15 edition three years after this post STILL has all the answers wrong for this section.
19. Correct answer is 97
21 correct answer is 196
23 correct answer is 554
25. E=.04,Z=1.96,p=.6, find N.
N=576.24=577(round up) 
Statistics 
Ryan,
Because Shawn, you guys are doing the problems wrong

Statistics 
rizwan,
(1.96*2.5)/0.25)^2=384.16
that's right or wrong??
i'm not understand why the population standard deviation round up to 3??
anyone help? this for question post by shyra.. 
Statistics 
jie,
A survey is being planned to determine the mean amount of time corporation executives watch television. A pilot survey indicated that the mean time per week is 13 hours, with a standard deviation of 2.0 hours. It is desired to estimate the mean viewing time within onequarter hour. The 90% level of confidence is to be used.

Statistics 
Agatamudi,
Here Population mean = 13, S.D = 2
Maximum error = 1/4 = 0.25
Level of significance = 90%
therefore zalfa/2 value = 1.645
n = (Zalfa/2*sigma/E)2 = (1.645*2/0.25)2=1082.41 
Statistics 
ahmeyer,
The U.S. Dairy Industry wants to estimate the mean yearly milk consumption. A sample of 16 people reveals the mean yearly consumption to be 60 gallons with a standard deviation of 20 gallons.