IAI Actuarial Core Principles · Actuarial Statistics · Exploratory data analysis
Six pairs of values have no ties. The rank differences d between the x-ranks and y-ranks are 1, -1, 2, -2, 0, 0. Using the no-ties formula, what is Spearman's rank correlation coefficient?
Spearman's coefficient is 0.714. The squared rank differences sum to 10, and with n = 6 the formula gives one minus 6 times 10 divided by 6 times 35, which is one minus 0.2857, or 0.714.
- A0.714
- B0.829Correct
- C0.857
- D0.571
- 0.900
Explanation
Σd² = 1+1+4+4+0+0 = 10. rs = 1 - 6×10/(6×35) = 1 - 60/210 = 0.714. Check: 6/210=0.2857, so 1-0.2857=0.714. Hence the key is 0.714, which is the first option.
Did you get it right without looking?
One question tells you little. A timed set on Exploratory data analysis shows your real accuracy, how long you take and where you lose marks.
More Exploratory data analysis questions
- An insurer's motor claims dataset contains the following fields: (i) vehicle registration state, (ii) number of claims in the last year, (ii…
- For paired data, Σx = 30, Σy = 60, Σx² = 200, Σy² = 460, Σxy = 380, n = 5. What is the sample correlation coefficient?
- A sample of 8 monthly claim counts for a small Pune motor portfolio is: 4, 7, 7, 9, 10, 12, 15, 20. What is the sample mean and the median r…
- For a sample, the variances are sx² = 16 and sy² = 25, and the sample covariance is 15. Each x is then transformed to u = 3 - 2x, while y is…
- Which feature is a genuine advantage of a stem-and-leaf plot over a histogram for a small data set of 30 observations?
- Daily premium collections (₹ lakh) over 5 days are 2, 4, 6, 8 and 30. Which statement about the outlier rule 'flag values beyond 1.5 × IQR f…