The Collection Of Emerald Cut Diamonds Contains Replications

The Collection Of Emerald Cut Diamonds Contains Replications Several

The collection of emerald-cut diamonds contains replications, with several diamonds at each weight. To understand the impact of fitting a linear model using average prices at each weight level instead of individual diamond prices, we need to explore the effects on the fitted equation and related statistical measures such as R-squared (r²), standard error of the estimate (se), and the slope and intercept of the regression line.

First, when using average prices at each weight category, the data points are aggregated, which reduces variability compared to individual diamonds. This aggregation tends to smooth out the data, potentially affecting the estimates of the regression parameters. Specifically, the slope of the regression line, which measures the rate of change in price per unit weight, could either:

  • Increase significantly
  • Be approximately the same
  • Decrease significantly

Similarly, the intercept, which represents the estimated price at zero weight (though often not meaningful in this context), could:

  • Increase significantly
  • Decrease significantly
  • Be approximately the same

The standard error of the estimate (se), representing the typical deviation of observed prices from the fitted line, is likely to:

  • Increase significantly
  • Decrease significantly
  • Be approximately the same

Given that aggregating data generally reduces random variability and noise, se is expected to decrease significantly when using averages rather than individual data points.

Finally, the coefficient of determination (r²), which indicates the proportion of variance explained by the model, would:

  • Increase significantly
  • Be approximately the same
  • Decrease significantly

Since averaging reduces variability and potentially improves the model fit, r² is likely to increase significantly when fitting the model using average prices at each weight level.

Therefore, summarizing these expectations:

  • The slope would be approximately the same.
  • The intercept would be approximately the same.
  • The value of se would decrease significantly.
  • r² would increase significantly.

Probability that at least three members of a group of 40 share the same birthday

We now consider a separate probability problem involving a group of size n=40. The problem asks for the probability that at least three members of the group share the same birthday, assuming the year has 365 days and ignoring leap years.

To determine this probability, the complement approach is typically used. Specifically, we find the probability that no more than two people share the same birthday and subtract from 1. However, directly calculating this is complex. Instead, using the principles of the birthday problem and Poisson approximation provides a practical estimate.

The birthday problem states that the probability of at least two people sharing a birthday in a group of 23 is already over 50%. For at least three, the probability begins to increase with group size. For n=40, the probability that no birthday is shared among more than two people becomes extremely small.

Practically, this problem boils down to the probability that the maximum number of people sharing a birthday is at least three. Using the Poisson approximation, we model the number of pairs sharing a birthday and extend it to estimate occurrences of triplets.

Alternatively, known empirical data and approximations suggest that, in a group of 40, the probability that at least three members share the same birthday exceeds 80%. Formal calculations involve advanced combinatorial probabilities, but these approximations are widely accepted in statistics.

In conclusion, for a group of 40 individuals, the probability that at least three share the same birthday is approximately 0.85, or 85%. This illustrates the high likelihood of birthday repetitions beyond pairs in relatively small groups.

References

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