Posted by **Shyra** on Thursday, December 2, 2010 at 12:59pm.

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**, Thursday, April 14, 2011 at 7:43pm
YES!

- Statistics -
**Visha**, Monday, October 29, 2012 at 2:25pm
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?

- Data Analysis -
**Jeannine**, Thursday, May 16, 2013 at 2:44am
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**, Wednesday, June 26, 2013 at 10:05pm
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**, Sunday, June 30, 2013 at 4:47pm
Because Shawn, you guys are doing the problems wrong

- Statistics -
**rizwan**, Monday, November 25, 2013 at 1:27am
(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**, Friday, August 14, 2015 at 11:20pm
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.

- Statistics -
**Agatamudi**, Tuesday, April 26, 2016 at 9:56am
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**, Thursday, October 6, 2016 at 8:23pm
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.

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