下表显示了 6 月份 9 天内，从中午到下午 5 点的平均气温 $x$ （摄氏度）以及在某海滩咖啡馆售出的冰淇淋甜筒数量 $y$ 。

((i))
绘制这些数值的散点图，并清晰地标注坐标轴。
((ii))
计算该数据的积差相关系数，并对其数值进行评价。
((iii))
咖啡馆老板在平均气温为 $33.3^\circ\text{C}$ 的一天雇用了一名临时工来协助他。
1. ((a))
  使用合适的回归线估算当天售出的冰淇淋甜筒数量，并对该估算结果的可靠性进行评价。
2. ((b))
  账目检查显示，该临时工报告当天售出了 108 个冰淇淋甜筒。结合你上面的解答，说明老板是否有理由怀疑该临时工在处理当天利润时存在不诚实行为。
((iv))
通过计算 $\ln y$ 与 $x$ 之间的积差相关系数的值，说明 $y = a + bx$ 和 $\ln y = a + bx$ 中哪一个是更好的模型。由此，假设你所选的模型保持有效，求当温度升高 $2^\circ\text{C}$ 时，售出的冰淇淋甜筒数量增加的百分比的最佳估计值。

它不会直接告诉你答案，而是问你卡在哪一步看看 AI 是怎么陪你一步步弄懂这道题的

完整解法

9((iii))((b))

1. Recall the estimate and its reliability status from (a)

From part (iii)(a) the number of cones was estimated using the regression line of $y$ on $x$ , but the temperature $33.3^\circ\text{C}$ lies outside the range of the data values of $x$ . Hence the estimate is obtained by extrapolation and is therefore not reliable.

2. Compare with the reported figure and draw the conclusion

Comparing the reported sale of $108$ cones with the regression estimate would only be meaningful if the estimate were reliable. Because the estimate is an unreliable extrapolation, any discrepancy with $108$ cannot be used to justify a charge of dishonesty. Therefore the owner does not have a sound reason, based on this estimate alone, to suspect that the temporary staff was dishonest.

答案

The owner does not have strong reason to suspect dishonesty. The estimate in (a) was obtained by extrapolation (33.3°C is outside the range of the recorded temperatures), so it is unreliable. Since the predicted figure itself cannot be trusted, the difference between it and the reported sale of 108 cones is not a sound basis for concluding that the temporary staff was dishonest.

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下表显示了 6 月份 9 天内，从中午到下午 5 点的平均气温 xxx（摄氏度）以及在某海滩咖啡馆售出的冰淇淋甜筒数量 yyy。

完整解法

9((iii))((b))

同一知识点的其他题

下表显示了 6 月份 9 天内，从中午到下午 5 点的平均气温 $x$ （摄氏度）以及在某海滩咖啡馆售出的冰淇淋甜筒数量 $y$ 。