Estimating proportions using resampling

  • Context: Undergrad 
  • Thread starter Thread starter moonman239
  • Start date Start date
Click For Summary
SUMMARY

The discussion focuses on estimating the number of tabbed pins in a bowling competition scenario involving Joe and Bob. They hit a total of 18 pins, with 5 identified as tabbed. The initial estimate of the proportion of tabbed pins is calculated as 5/18, with a confidence interval of approximately +/- 19.794239%. The problem is framed within the context of a hypergeometric distribution, suggesting that a more accurate estimate can be derived using this statistical model, leading to a 99% confidence interval ranging from 5 to 15 tabbed pins.

PREREQUISITES
  • Understanding of hypergeometric distribution
  • Basic knowledge of confidence intervals
  • Familiarity with statistical estimation techniques
  • Experience with probability theory
NEXT STEPS
  • Study hypergeometric distribution applications in real-world scenarios
  • Learn about constructing confidence intervals in statistical analysis
  • Explore advanced statistical estimation techniques
  • Investigate the implications of sampling without replacement
USEFUL FOR

Statisticians, data analysts, and anyone interested in probability theory and statistical estimation methods.

moonman239
Messages
276
Reaction score
0
I just thought of this problem: Suppose Ann holds a bowling competition for Joe and Bob. She has 30 pins. She puts tabs on some pins, but hides them and tells neither Joe nor Bob how many pins have been tabbed. So Joe and Bob decide to make this a fun experiment. Joe steps up, rolls the ball and hits 9 pins, two of which are tabbed. Then Bob rolls the ball and hits 11 pins, three of which are pinned. Two of the non-tabbed pins that Bob hits have also been hit by Joe. Knowing this, how can they go about estimating how many pins Ann tabbed.

My solution: We know that altogether, they hit 18 pins, 5 of which were tabbed. Knowing this, we can be 99% confident that the actual number is 5/18, +/- approx. 19.794239%.

However, I'd still like to know if there's a way to get a better estimate than that.
 
Physics news on Phys.org
The bowling is essentially drawing without replacement, i.e. from a hypergeometric distribution. With this, the 99% CI is 5 to 15.
 

Similar threads

  • · Replies 35 ·
2
Replies
35
Views
10K
  • · Replies 9 ·
Replies
9
Views
7K
  • · Replies 94 ·
4
Replies
94
Views
9K
  • · Replies 9 ·
Replies
9
Views
2K
Replies
2
Views
3K
Replies
3
Views
3K
  • · Replies 29 ·
Replies
29
Views
3K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K