# Statistics

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

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3. 👁 1,203
1. YES!

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posted by matt
2. 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?

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posted by Visha
3. 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

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4. Statistical Techniques in Business and Economics 15 edition three years after this post STILL has all the answers wrong for this section.
25. E=.04,Z=1.96,p=.6, find N.
N=576.24=577(round up)

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posted by Shawn
5. Because Shawn, you guys are doing the problems wrong

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posted by Ryan
6. (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..

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posted by rizwan
7. 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.

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posted by jie
8. 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

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9. 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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posted by ahmeyer
10. xbar ± t*s/vn

= 60 ± 1.75*20/4
= ( 51.25, 68.75)

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posted by Srt

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