Skip to main content
. 2024 Aug 27;82:138. doi: 10.1186/s13690-024-01379-1

Table 4.

Participants’ knowledge about sickle cell disease prevention stratified by age and sex for an exploratory study at University of Cape Coast from 2nd February to 22nd December 2022

Total Sex P-value Age (years) P-value
Female Male 15–19 20–29 30–39 ≥40
Sickle cell condition testing must be mandatory for all couples before marriage 0.020 0.747
 Disagree 27 (7.0) 12 (6.4) 15 (7.6) 1 (3.2) 21 (7.1) 4 (7.7) 1 (14.3)
 Uncertain 7 (1.8) 0 (0.0) 7 (3.5) 0 (0.0) 7 (2.4) 0 (0.0) 0 (0.0)
 Agree 352 (91.2) 176 (93.6) 176 (88.9) 30 (96.8) 268 (90.5) 48 (92.3) 6 (85.7)
It should be mandatory for all newborns to be tested for sickle cell condition 0.268 0.384
 Disagree 21 (5.4) 9 (4.8) 12 (6.1) 0 (0.0) 18 (6.1) 2 (3.8) 1 (14.3)
 Uncertain 26 (6.7) 9 (4.8) 17 (8.6) 3 (9.7) 22 (7.4) 1 (1.9) 0 (0.0)
 Agree 339 (87.8) 170 (90.4) 169 (85.4) 28 (90.3) 256 (86.5) 49 (94.2) 6 (85.7)
Two people who have sickle cell condition should be stopped from marrying 0.071 0.890
 Disagree 51 (13.2) 19 (10.1) 32 (16.2) 3 (9.7) 40 (13.5) 8 (15.4) 0 (0.0)
 Uncertain 47 (12.2) 19 (10.1) 28 (14.1) 3 (9.7) 38 (12.8) 5 (9.6) 1 (14.3)
 Agree 288 (74.6) 150 (79.8) 138 (69.7) 25 (80.6) 218 (73.6) 39 (75.0) 6 (85.7)
If I know that I am at risk of giving birth to a sickle cell child, it will change my pregnancy plans 0.224 0.533
 Disagree 71 (18.4) 28 (14.8) 43 (21.7) 6 (19.4) 52 (17.6) 12 (23.1) 1 (14.3)
 Uncertain 112 (29.0) 57 (30.3) 55 (27.8) 8 (25.8) 91 (30.7) 13 (25.0) 0 (0.0)
 Agree 203 (52.6) 103 (54.8) 100 (50.5) 17 (54.8) 153 (51.7) 27 (51.9) 6 (85.7)
Pastors/Imams should advise all prospective couples who have sickle cell disease from marrying 0.165 0.110
 Disagree 97 (25.1) 41 (21.8) 56 (28.3) 8 (25.8) 71 (24.0) 18 (34.6) 0 (0.0)
 Uncertain 53 (13.7) 23 (12.2) 30 (15.2) 3 (9.7) 47 (15.9) 3 (5.8) 0 (0.0)
 Agree 236 (61.1) 124 (66.0) 112 (56.6) 20 (64.5) 178 (60.1) 31 (59.6) 7 (100.0)

The data is presented as frequencies n (%) with statistical significance at p < 0.05 determined using chi-square test