You are the only vendor for food and merchandise at a concert in the area. Suppo

Question

You are the only vendor for food and merchandise at a concert in the area. Suppose for a random sample of 72 concert goers, the mean amount of money spent on food and merch is $23.84 with a standard deviation of $5.07. What sample size would be necessary to estimate the mean amount of money spent on food and merchandise at a concert in Southern California within .1 hours with 98% confidence? You are the only vendor for food and merchandise at a concert in the area. Suppose for a random sample of 72 concert goers, the mean amount of money spent on food and merch is $23.84 with a standard deviation of $5.07. What sample size would be necessary to estimate the mean amount of money spent on food and merchandise at a concert in Southern California within .1 hours with 98% confidence?

Explanation / Answer

S = 5.07 , ME = 0.1

Z value at 98% = 2.33

ME = z * (s/sqrt(n))

0.1 = 2.33 * (5.07/sqrt(n))

n = 13954.93 = 13955

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