1.

Three bus companies, Speedy, Ahead and Green, are competing for a single tender from the local authority. The local authority asks for details of punctuality over all their services at all bus stations served in the area. The following evidence is offered for the population of services in the previous year (you may assume that a * indicates that this bus station is not served by the bus company)

Percentage (%) punctual journeys in the previous year by bus station served

Bus Company / Bus Station
1 / 2 / 3 / 4 / 5 / 6 / 7 / 8 / 9 / 10 / 11 / 12 / 13 / 14 / 15 / 16 / 17 / 18 / 19 / 20
Speedy / 95 / 88 / 91 / 89 / 77 / 83 / * / 93 / 99 / 84 / 78 / * / 94 / 87 / 96 / 77 / * / 76 / 85 / *
Ahead / 89 / 86 / 91 / 99 / 95 / 83 / 89 / 93 / * / 88 / 85 / 84 / 87 / 88 / * / 88 / 89 / 90 / 89 / 89
Green / * / 83 / 82 / * / 94 / 88 / * / * / * / * / 95 / * / 96 / 80 / * / 98 / * / 94 / * / *
  1. What chart would you choose to show the differences in punctuality between the three companies (assuming that you do not reduce the data)?
  2. Calculate the mean, median and mode for the punctuality of each bus company. Specify the degree of accuracy of your answer
  3. The local authority feels that low punctuality should be weighted more heavily. If the following weights are prescribed:

Percentage Punctuality (%)
Below 80 / 80-89 / 90-100
Weight / 3 / 2 / 1

What is the weighted mean?Give your answers to 2 decimal places.

  1. Explain the differences in results between the mean calculated in b and the weighted mean in c.

Answer Guidelines

  1. Using bar charts to show the different performance by Bus Station will probably show most clearly the differences between the bus companies. However, it will still be difficult to take in the information about 20 different bus stations and so summary statistics will be essential for decision making.
  2. Mean =

Speedy / Ahead / Green
/ 1392 / 1602 / 810
N / 16 / 18 / 9
Mean / 87.0 (1 dp) / 89.0 (1 dp) / 90.0(1 dp)

Median and Mode are easiest to see if the data is put in rank order:

Speedy / 76 / 77 / 77 / 78 / 83 / 84 / 85 / 87 / 88 / 89 / 91 / 93 / 94 / 95 / 96 / 99
Ahead / 83 / 84 / 85 / 86 / 87 / 88 / 88 / 88 / 89 / 89 / 89 / 89 / 89 / 90 / 91 / 93 / 95 / 99
Green / 80 / 82 / 83 / 88 / 94 / 94 / 95 / 96 / 98
Speedy / Ahead / Green
Median (middle value) / / / 94.0 (1dp)
Mode (most frequent value) / 77 / 89 / 94
  1. Weighted Mean =

Speedy / Ahead / Green
/ 2524 / 2736 / 1143
/ 30 / 31 / 13
Weighted
Mean / 84.13 (2 dp) / 88.26 (2 dp) / 87.92 (2 dp)
  1. The mean gives the value if all the values in the data set were equally shared out. The weighted mean gives more weight to certain values – in this case to greater unpunctuality. Whilst Speedy is still the most unpunctual (its weighted mean % punctuality lowest), the weighted mean shows Ahead to be better than Green.

Three bus companies, Speedy, Ahead and Green, are competing for a single tender from the local authority. The local authority asks for details of punctuality over all their services at all bus stations served in the area. The following evidence is offered for the population of services in the previous year (you may assume that a * indicates that this bus station is not served by the bus company)

Percentage (%) punctual journeys in the previous year by bus station served

Bus Company / Bus Station
1 / 2 / 3 / 4 / 5 / 6 / 7 / 8 / 9 / 10 / 11 / 12 / 13 / 14 / 15 / 16 / 17 / 18 / 19 / 20
Speedy / 95 / 88 / 91 / 89 / 77 / 83 / * / 93 / 99 / 84 / 78 / * / 94 / 87 / 96 / 77 / * / 76 / 85 / *
Ahead / 89 / 86 / 91 / 99 / 95 / 83 / 89 / 93 / * / 88 / 85 / 84 / 87 / 88 / * / 88 / 89 / 90 / 89 / 89
Green / * / 83 / 82 / * / 94 / 88 / * / * / * / * / 95 / * / 96 / 80 / * / 98 / * / 94 / * / *
  1. Calculate the population mean absolute deviation, the variance and the range for each of the three bus companies. Give your answers to 1 decimal place.
  2. Comparing your answers from question 1 and question 2, which company would you select to win the tender?

Answer Guidelines

  1. MAD = Variance = Range =

Speedy / Ahead / Green
/ 98 / 46 / 54
/ 826 / 250 / 373
N / 16 / 18 / 9
MAD / 6.1 (1 dp) / 2.6 (1 dp) / 6.0 (1 dp)
Variance / 51.6 (1 dp) / 13.9 (1 dp) / 41.6 (1 dp)
Range / 23 / 16 / 18
  1. Probably Ahead although a case could be made for Green. The problem here is to try and resolve the information given by the measures of location (mean, median and mode) and the measures of dispersion (MAD, variance and range). Whilst Green might look to marginally ‘beat’ Ahead, Green only visits 9 of the 20 bus stations and so there is more information in the Ahead statistics.

3.

The following table gives a sample of food items bought by a household during the years indicated:

Item / 2000 / 2001 / 2002
Price (£) / Quantity / Price (£) / Quantity / Price (£) / Quantity
Bread (loaves) / 0.80 / 400 / 1.00 / 385 / 1.35 / 350
Milk (litres) / 0.22 / 500 / 0.25 / 550 / 0.40 / 500
Eggs (doz) / 1.10 / 60 / 1.30 / 55 / 1.65 / 50
Meat (Kg) / 3.00 / 100 / 5.50 / 90 / 10.00 / 70
  1. Calculate a simple price index for bread in 2001 and 2002 relative to the base year 2000.
  2. If the weights for the various elements were as follows:

Item / Weight
Bread (loaves) / 125
Milk (litres) / 165
Eggs (doz) / 100
Meat (Kg) / 130
520

Calculate the weighted food price index for 2002 relative to the base year 2000.

  1. Calculate the Laspeyre price index for 2002 relative to the base year 2000.
  2. Calculate the Paasche price index for 2002 relative to the base year 2000.
  3. Explain the difference in results between c and d above.

Answer Guidelines

  1. Simple price index =

Item / Weight / Base year price / 2002 price / Ratio 2002 price/base year price / weighted price index
Bread (loaves) / 125 / 0.8 / 1.35 / 1.70 / 210.9375
Milk (litres) / 165 / 0.22 / 0.4 / 1.82 / 300
Eggs (doz) / 100 / 1.1 / 1.65 / 1.50 / 150
Meat (Kg) / 130 / 3 / 10 / 3.33 / 433.33
520 / Sum = / 1094.27
Q0Pn / Q0P0 / QnPn / QnP0
Bread (loaves) / 540 / 320 / 472.5 / 280
Milk (litres) / 200 / 110 / 200 / 110
Eggs (doz) / 99 / 66 / 82.5 / 55
Meat (Kg) / 1000 / 300 / 700 / 210
Sum / 1839 / 796 / 1455 / 655

Weighted price index =

c and d.

Laspeyre =

Paasche =

  1. The Laspeyre index works out the index based on a past base period whilst the Paasche index works out the index based on current prices

4.

The following table gives the total weekly hours worked by gender for full time (31 hours per week, including paid overtime) employees working in the UK (male employees aged 16-64 and female employees aged 16-59) in 20XX.

Total Weekly hours / Males / Females
31-35 / 319 / 599
36-40 / 2487 / 1768
41-45 / 1403 / 285
46-50 / 1148 / 85
51 or more / 1020 / 60
  1. What can you infer from the open class interval used for 51 or more hours?
  2. Calculate the mean and median for this data.
  3. Calculate the relative cumulative frequencies showing ‘more than’ number of hours worked for both male and female data.
  4. Graphically represent the data in c above.
  5. What percentage of male employees work more than 48 hours?

Answer Guidelines

  1. That if the interval was closed at say, 55, then some observations would not fit into a class. So the class width must be larger than the previous classes. It is also likely that the data is widely spread beyond 51 and so for calculation purposes, whilst an upper limit is not required, the class midpoint must be chosen to be ‘representative’ of the data in the class. For the calculations which follow, a midpoint of 60.5 hours has been selected. Other values above this would also be appropriate but given that there are likely to be fewer very high observations, the midpoint should not creep up too much.

Total Weekly hours / midpoint xi / Males / fixi / Females / fixi
31-35 / 33 / 319 / 10527 / 599 / 19767
36-40 / 38 / 2487 / 94506 / 1768 / 67184
41-45 / 43 / 1403 / 60329 / 285 / 12255
46-50 / 48 / 1148 / 55104 / 85 / 4080
51 or more / 60.5 / 1020 / 61710 / 60 / 3630
Sum / 6377 / 282176 / 2797 / 106916

Mean =

For male dataFor female data =

Cumulative frequency: male / Cumulative frequency: female / Relative Cumulative frequency: male / Relative Cumulative frequency: female
more than 30.5 / 6377 / 2979 / 1 / 1
more than 35.5 / 6058 / 2380 / 0.949976 / 0.798926
more than 40.5 / 3571 / 430 / 0.559981 / 0.144344
more than 45.5 / 2168 / 145 / 0.339972 / 0.048674
more than 50.5 / 1020 / 242 / 0.15995 / 0.081235
more than 70.5 / 0 / 0 / 0 / 0

  1. Estimate from the graph: 22%

5.

A class of students were asked to record their heights on a sheet passed around during a lecture. The following observations were recorded for the male students:

Heights of male students taking quantitative methods in 20XX

5’11” / 5’7” / 5’7” / 5’7” / 6’1” / 5’10” / 5’10.5” / 5’11” / 5’11” / 5’11”
5’10.5” / 5’10.5” / 5’11” / 5’10” / 5’7.5” / 5’7.5” / 5’9” / 5’10.5” / 5’7” / 5’6”
5’10” / 5’10” / 6’1” / 5’9” / 6’2” / 6’0” / 6’0” / 5’10” / 1.85m / 6’1”
6’2” / 5’8” / 5’3” / 5’7” / 5’11” / 5’10” / 5’11” / 5’10” / 5’9.5” / 6’1”
1’3” / 8’7.75” / 6’3” / 6’0” / 5’11” / 5’8” / 5’5” / 5’11” / 5’8” / 6’3”
  1. Reduce the data into a frequency table, using a tally sheet and commenting on any difficulties you encounter.
  2. Briefly justify your choice of class interval
  3. State for your frequency table:
  4. The boundary values
  5. The class intervals
  6. The class midpoints
  1. Calculate the mean and the variance for this data.

Answer Guidelines

  1. Difficulties include the need to convert measurements to a common scale (mixture of imperial and metric). Also some erroneous recording (respondents recording over 8 foot and 1’3”) which should be removed (although care needs to be taken when discounting data only to remove data items which are known to carry error).

Inches / Tally / Frequency
>61.5 to ≤ 63.5 / I / 1
>63.5 to ≤ 65.5 / I / 1
>65.5 to ≤ 67.5 / IIII III / 8
>67.5 to ≤ 69.5 / IIII I / 6
>71.5 to ≤73.5 / IIIIIIIIIIIIIIII I / 21
>73.5 to ≤ 75.5 / IIII II / 7
>61.5 to ≤ 63.5 / IIII / 4
Total / 48
  1. As many of the measurements fall on the inch measurement, the class boundaries should be chosen to ensure there is no ambiguity as to which observation falls into which interval.
  1. Boundary values are non-overlapping: first class: lower: 61.5, upper: 63.5 etc
  2. Class interval: 2 inches
  3. Class midpoints: first class – 62.5 inches (as shown for rest of classes in table below)

Inches / Tally / Frequency / Class midpoint
Xi / fixi / - Xi / ( - Xi)2
>61.5 to ≤ 63.5 / I / 1 / 62.5 / 62.50 / 7.42 / 55.01
>63.5 to ≤ 65.5 / I / 1 / 64.5 / 64.50 / 5.42 / 29.34
>65.5 to ≤ 67.5 / IIII III / 8 / 66.5 / 532.00 / 3.42 / 11.67
>67.5 to ≤ 69.5 / IIII I / 6 / 68.5 / 411.00 / 1.42 / 2.01
>71.5 to ≤73.5 / IIIIIIIIIIIIIIII I / 21 / 70.5 / 1480.50 / -0.58 / 0.34
>73.5 to ≤ 75.5 / IIII II / 7 / 72.5 / 507.50 / -2.58 / 6.67
>61.5 to ≤ 63.5 / IIII / 4 / 74.5 / 298.00 / -4.58 / 21.01
48 / Sum / 3356.00 / Sum / 126.05

Mean = Variance =