Junior Statistician Qualification Examination study materials
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Junior Statistician Qualification Examination: Descriptive Statistics and Data Presentation

Junior Statistician Qualification Examination · Work through your exam, one topic at a time.

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Descriptive Statistics and Data Presentation

This domain checks whether you can summarise a data set correctly and display it without distortion. Items cover measurement scales, measures of centre and spread, distribution shape, and the construction and reading of frequency tables, histograms, boxplots and ogives.

Scales and averages

  • Nominal data support only the mode; ordinal data add the median; only interval and ratio data justify a mean.
  • The mean is dragged toward skewness while the median resists it, so right-skewed data typically give mean greater than median greater than mode.
  • A combined mean must be weighted by group size; a plain average of group means is wrong unless the groups are equal in size.

Spread

  • The range uses only two observations. The interquartile range, Q3 minus Q1, covers the middle 50% and resists outliers.
  • Population variance divides by N, the sample variance s-squared by n minus 1; mixing the two is the commonest arithmetic error in this domain.
  • The coefficient of variation, s divided by x-bar times 100%, compares dispersion across different units or very different means.

Display

  • Classes must be mutually exclusive and exhaustive; state boundaries such as 9.5 to 19.5 so no value can fall into two classes.
  • Histogram bars touch because the horizontal axis is continuous; bar charts for categories leave gaps.
  • An ogive plots cumulative frequency against upper class boundaries, and the median is read where cumulative frequency reaches n/2.
  • On a boxplot, whiskers run to the most extreme values within 1.5 times the IQR; anything beyond is plotted as a flagged outlier.

Worked example

Daily sales: 4, 6, 6, 8, 10, 12, 20. The sum is 66, so the mean is 9.43; the ordered median is 8. Deviations from the mean are minus 5.43, minus 3.43, minus 3.43, minus 1.43, 0.57, 2.57 and 10.57; their squares total 173.65, giving s-squared equal to 173.65/6 = 28.94 and s = 5.38. Then CV = 5.38/9.43 = 57.1%. Since Q1 = 6 and Q3 = 12, IQR = 6 and the upper fence is 12 + 9 = 21, so 20 is not an outlier. A mean above the median confirms right skew.

Traps

  • Dividing by n when a sample standard deviation is requested.
  • Averaging group means without weighting by group size.
  • Reading a median from histogram bar heights instead of from the ogive.
  • Treating an ordinal code such as a 1 to 5 satisfaction scale as ratio data.