Spearman's rank correlation Problem set A Worked solutions

A theatre records the ticket price and the attendance at eight shows. Calculate Spearman's rank correlation coefficient.

• Ranks of price (£): 3, 5, 1, 8, 7, 4, 6, 2
• Ranks of attendance (thousands): 6, 4, 8, 1, 3, 5, 2, 7
• From the calculator: r_s = −0.976

Eight students record the number of books read in a term and their score on a reading test. Calculate Spearman's rank correlation coefficient.

• Ranks of books (ties averaged): 3.5, 6.5, 2, 6.5, 1, 5, 3.5, 8
• Ranks of score (ties averaged): 2, 6, 3, 8, 1, 6, 4, 6
• From the calculator: r_s = 0.864

Two judges place eight dancers, A to H, in order, from 1st place to 8th place.

1. (a) Write down the rank each judge gave to A, B, C, D, E, F, G and H, in that order.

   • Find each letter in each list. Ranks of A to H for Judge 1: 4, 2, 5, 8, 1, 7, 6, 3
   • Ranks of A to H for Judge 2: 3, 1, 4, 6, 2, 8, 7, 5

2. (b) Calculate Spearman's rank correlation coefficient.

   • From the calculator: r_s = 0.833

A shop records its weekly advertising spend (hundreds of pounds) and its sales (units) over seven weeks. Use Spearman's rank correlation to test, at the 5% significance level, whether there is a positive correlation. The critical value is 0.714.

1. (a) State the hypotheses.

   • H0: there is no correlation between advertising spend and sales
   • H1: there is a positive correlation between advertising spend and sales

2. (b) Calculate r_s.

   • Ranks of advertising: 2, 6, 1, 7, 4, 5, 3
   • Ranks of sales: 4, 5, 1, 7, 3, 6, 2
   • From the calculator: r_s = 0.857

3. (c) State the conclusion in context.

   • r_s = 0.857 > 0.714, so reject H0.
   • There is sufficient evidence, at the 5% significance level, of a positive correlation between advertising spend and sales.

Eight students record their weekly hours of television (x) and their test mark (y). Use Spearman's rank correlation to test, at the 10% significance level, whether there is a negative correlation. The critical value is 0.524.

1. (a) State the hypotheses.

   • H0: there is no correlation between hours of television and test mark
   • H1: there is a negative correlation between hours of television and test mark

2. (b) Calculate r_s.

   • Ranks of tv hours: 3, 6, 1, 8, 4, 7, 2, 5
   • Ranks of mark: 5, 3, 8, 2, 6, 1, 7, 4
   • From the calculator: r_s = −0.952

3. (c) State the conclusion in context.

   • r_s = −0.952 < −0.524, so reject H0.
   • There is sufficient evidence, at the 10% significance level, of a negative correlation between hours of television and test mark.

A park keeper records the age of nine trees (years) and their height (metres). The critical value for a two-tailed test at the 5% significance level is 0.700.

1. (a) Calculate r_s.

   • Ranks of age: 2, 7, 4, 1, 9, 5, 3, 6, 8
   • Ranks of height: 2, 6, 4, 1, 8, 7, 3, 5, 9
   • From the calculator: r_s = 0.933

2. (b) Test, at the 5% significance level, whether there is any correlation. State the hypotheses and the conclusion in context.

   • H0: there is no correlation between age and height. H1: there is a correlation between age and height.
   • |r_s| = 0.933 > 0.700, so reject H0.
   • There is sufficient evidence, at the 5% significance level, of a correlation between age and height.

Two judges score six singers out of 10.

1. (a) Calculate r_s.

   • Ranks of Judge 1: 4, 2, 6, 3, 5, 1
   • Ranks of Judge 2: 3, 4, 6, 1.5, 5, 1.5
   • From the calculator: r_s = 0.783

2. (b) The critical value for a one-tailed test at the 5% significance level is 0.829. Determine whether there is significant evidence that the two judges tend to agree.

   • H0: there is no correlation between the two sets of scores. H1: there is a positive correlation.
   • r_s = 0.783 < 0.829, so do not reject H0.
   • There is insufficient evidence, at the 5% significance level, that the two judges tend to agree.

Eight runners record their weekly training distance (km) and their time for a 2 km run (minutes).

1. (a) Calculate r_s and describe the correlation in context.

   • Ranks of training (km): 1, 4, 2, 8, 5, 3, 7, 6
   • Ranks of time (min): 8, 6, 7, 1, 4, 5, 2, 3
   • From the calculator: r_s = −0.976
   • strong negative correlation: runners who train more tend to run 2 km faster.

2. (b) A coach says that runners who train more always run faster. Does the data support this claim?

   • r_s = −0.976 is close to −1 but not equal to −1.
   • No. r_s is not exactly −1, so the ranks are not perfectly reversed: at least one runner trains more than another but is slower. The data show a tendency, not 'always'.

The scatter diagram shows the height y (mm) of a plant at week x for ten plants. For these data Pearson's r = 0.692 and Spearman's r_s = 0.964. The point (10, 230) was misrecorded. The correct value is (10, 62).

1. (a) Without calculating, say how the new value of Pearson's r compares with the old value.

   • The point was far from the pattern. Once it is moved onto the pattern, the points lie closer to a straight line, so r increases.

2. (b) Without calculating, say how the new value of r_s compares with the old value.

   • The point was the largest y before and is still the largest y after, so no rank changes and r_s does not change.

3. (c) The heights are converted from millimetres to centimetres. How does this affect r_s?

   • Dividing every height by 10 keeps the order of the heights.
   • The ranks are unchanged, so r_s is unchanged.

A culture of bacteria is counted each hour. The number of bacteria is shown.

1. (a) Calculate r_s.

   • Ranks of hour: 1, 2, 3, 4, 5, 6, 7, 8
   • Ranks of count: 1, 2, 3, 4, 5, 6, 7, 8
   • From the calculator: r_s = 1

2. (b) Pearson's r for these data is 0.850. Explain why r_s = 1 although r is less than 1.

   • Every increase in x comes with an increase in y, so the ranks are identical and r_s = 1.
   • The points follow a curve, not a straight line, so Pearson's r is less than 1.