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Table 1 Participants’ characteristics for the total sample and by age subgroups

From: Cost-effectiveness analysis of a multifactorial fall prevention intervention in older home care clients at risk for falling

 

Total sample, all ages

Young-old group, 75–84 years

Old-old group, 85–95 years

Participant characteristic

Intervention (n = 49)

Usual care

(n = 43)

Intervention (n = 27)

Usual care

(n = 22)

Intervention (n = 22)

Usual care

(n = 21)

Baseline mean number of falls ± SD

−1.8 ± 2.7

−1.7 ± 3.7

−1.1 ± 1.0

−1.0 ± 1.1

−2.6 ± 3.8

−2.3 ± 5.2

Follow-up mean number of falls ± SD

−1.4 ± 2.7

−1.3 ± 2.2

−0.9 ± 1.4

−1.8 ± 2.5

−2.1 ± 3.7

−0.8 ± 1.9

Change in mean number of falls ± SDa

0.3 ± 2.6

0.3 ± 3.3

0.1 ± 1.7

−0.8 ± 2.4

0.5 ± 3.4

1.5 ± 3.8

Baseline mean cost in CAD ± SD

20,154 ± 21,068

26,150 ± 28,132

17,533 ± 18335*

37,615 ± 32833*

23,372 ± 24,056

14,139 ± 15,298

Follow-up mean cost in CAD ± SD

5126 ± 3914

4800 ± 4305

4789 ± 3988

5559 ± 5359

5540 ± 3873

4004 ± 2733

Change in mean cost in CAD ± SDa

−15,028 ± 20,518

−21,350 ± 27,359

−12,743 ± 17498*

−32,056 ± 32204*

−17,831 ± 23,837

−10,135 ± 14,992

Mean age in years ± SD

84.1 ± 5.0

83.2 ± 5.1

80.5 ± 2.8

78.9 ± 2.5

88.6 ± 3.0

87.7 ± 2.7

Female (%)

67%

77%

70%

73%

64%

81%

Fear of falling (%)

41%

49%

41%

45%

41%

52%

Fall in the last 6 months (%)

73%

67%

67%

77%

82%

57%

  1. aAs fall is a bad outcome, we have added a negative sign to indicate a bad effect. Change refers to the difference between six-month follow-up and baseline. For mean change in number of falls, a high estimate implies more falls prevented. For change in mean number of falls, a positive estimate means that there were fewer falls in the “Follow-up” period than in the “Baseline” period. For change in mean cost, a lower estimate indicates fewer resources used
  2. Intervention = multifactorial fall prevention intervention; Usual care = usual home care services; n = number of participants; SD = standard deviation. * denotes p < 0.05 using Welch’s t-test