Question
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
Is the following soluntion correct?
[(1.96*3)/0.25]^2 = 553.1 => 554
Answers
YES!
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 one-quarter hour. The 95 percent level of confidence is to be used.
How many executives should be surveyed?
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
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
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)
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)
Because Shawn, you guys are doing the problems wrong
(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..
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..
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 one-quarter hour. The 90% level of confidence is to be used.
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
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
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.
xbar ± t*s/vn
= 60 ± 1.75*20/4
= ( 51.25, 68.75)
= 60 ± 1.75*20/4
= ( 51.25, 68.75)