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A large on-demand, video streaming company is designing a large-scale survey to determine the mean amount of time corporate eA large on-demand, video streaming company is designing a large-scale survey to determine the mean amount of time corporate executives watch on-demand television. A small pilot survey of 10 executives indicated that the mean time per week is 13 hours, with a standard deviation of 2.5 hours. The estimate of the mean viewing time should be within 30 minutes. The 98% level of confidence is to be used. (Use z Distribution Table.) How many executives should be surveyed? (Round the z-score to 2 decimal places and final answer to the next whole number.)

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Answer #1

Here n = 10

xbar = 13 hours

s = 2.5 hours

E = margin of error = 0.5 hours

c = confidence level = 0.98

we have to find n sample size

n = ( Zc * \sigma ^ )2/ E2

for c = 0.98 , Zc = Z0.98 = 2.33

n = ( 2.33 * 2.5 )2 / 0.52

n = 136

Number of executives = 136

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