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Mr. Norton owns two appliance stores. In store 1 the number of TV sets sold by a salesperson is, on average, 13 per week with a standard deviation of 5. In store 2, the number of TV sets sold by a salesperson is, on average, 7 with a standard deviation of 4. Mr. Norton has a position for a person to sell TV sets. There are two applicants. Mr. Norton asked one of them to work in store 1 and the other in store 2, each for 1 week. The salesperson in store 1 sold 10 sets, and the salesperson in store 2 sold 6 sets. Based on this information, which person should Mr. Norton hire? At a glance, I would choose the person who sold more TV sets. But again, the answer would make no sense without proving. Can I assume that I have, for example, a normal distribution? I will calculate the area of both. The bigger the area, the better the selling, right?
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CaptainBlack said: See post #4, it wasn't a joke. Click to expand... I didn't take it as a joke. ...the sample sizes used to compute the means and standard deviations are too small for any meaningful conclusion to be drawn Click to expand... Under this scenario, the negative binomial would fit fine when \(\displaystyle \mu
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SupremeCookie said: What if store 1 just has more traffic? In a better location? Click to expand... See post #4, it wasn't a joke.
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george357 said: Store 1: \(\displaystyle Z_1 = \frac{10 - 13}{5} = -\frac{3}{5} = -0.6\) Store 2: \(\displaystyle Z_2 = \frac{6 - 7}{4} = -\frac{1}{4} = -0.25\) It seems the closest to the mean, the better the selling, so store 2 wins. Store 1: \(\displaystyle A_1 = \frac{1}{5\sqrt{2\pi}}\int_{9.5}^{10.5} e^{-\frac{1}{2}\left(\frac{x - 13}{5}\right)^2} = 0.06657\) Store 2: \(\displaystyle A_2 = \frac{1}{4\sqrt{2\pi}}\int_{5.5}^{6.5} e^{-\frac{1}{2}\left(\frac{x - 7}{4}\right)^2} = 0.09643\) Therefore, store 2 wins again. Yeah, I agree with you. Remarks: Although store 2 won 2 times, I am not convinced with the answer. Logically, the more she sells, the better the salesperson. I would recommend for Mr. Norton to hire the female in store 1. Click to expand... What if store 1 just has more traffic? In a better location?
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SupremeCookie said: Compute and compare the z-scores. No need to assume normal distribution. Click to expand... Store 1: \(\displaystyle Z_1 = \frac{10 - 13}{5} = -\frac{3}{5} = -0.6\) Store 2: \(\displaystyle Z_2 = \frac{6 - 7}{4} = -\frac{1}{4} = -0.25\) It seems the closest to the mean, the better the selling, so store 2 wins. Jomo said: Assume normal distribution. Click to expand... Store 1: \(\displaystyle A_1 = \frac{1}{5\sqrt{2\pi}}\int_{9.5}^{10.5} e^{-\frac{1}{2}\left(\frac{x - 13}{5}\right)^2} = 0.06657\) Store 2: \(\displaystyle A_2 = \frac{1}{4\sqrt{2\pi}}\int_{5.5}^{6.5} e^{-\frac{1}{2}\left(\frac{x - 7}{4}\right)^2} = 0.09643\) Therefore, store 2 wins again. CaptainBlack said: Neither, he should call a forensic accountant, the fraud squad or a statistician. One would anticipate the number of TVs sold in a week, in a store, would have a Poison distribution. In which case the SD should be equal to the square root of the mean number sold, that is not the case here. So one would suspect either the sample sizes used to compute the means and standard deviations are too small for any meaningful conclusion to be drawn, or some sort fraud is taking place. Click to expand... Yeah, I agree with you. Remarks: Although store 2 won 2 times, I am not convinced with the answer. Logically, the more she sells, the better the salesperson. I would recommend for Mr. Norton to hire the female in store 1.
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CaptainBlack said: Neither, he should call a forensic accountant, the fraud squad or a statistician. One would anticipate the number of TVs sold in a week, in a store, would have a Poison distribution. In which case the SD should be equal to the square root of the mean number sold, that is not the case here. So one would suspect either the sample sizes used to compute the means and standard deviations are too small for any meaningful conclusion to be drawn, or some sort fraud is taking place. Click to expand... I was thinking negative binomial would be a good fit for $\mu
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george357 said: Mr. Norton owns two appliance stores. In store 1 the number of TV sets sold by a salesperson is, on average, 13 per week with a standard deviation of 5. In store 2, the number of TV sets sold by a salesperson is, on average, 7 with a standard deviation of 4. Mr. Norton has a position for a person to sell TV sets. There are two applicants. Mr. Norton asked one of them to work in store 1 and the other in store 2, each for 1 week. The salesperson in store 1 sold 10 sets, and the salesperson in store 2 sold 6 sets. Based on this information, which person should Mr. Norton hire? At a glance, I would choose the person who sold more TV sets. But again, the answer would make no sense without proving. Can I assume that I have, for example, a normal distribution? I will calculate the area of both. The bigger the area, the better the selling, right? Click to expand... Neither, he should call a forensic accountant, the fraud squad or a statistician. One would anticipate the number of TVs sold in a week, in a store, would have a Poison distribution. In which case the SD should be equal to the square root of the mean number sold, that is not the case here. So one would suspect either the sample sizes used to compute the means and standard deviations are too small for any meaningful conclusion to be drawn, or some sort fraud is taking place.
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Assume normal distribution.
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Compute and compare the z-scores. No need to assume normal distribution.
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