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Table 2 Health related behavioral characteristics of NHP user groups; n = 1707

From: Usage of Natural Health Products (NHPs) for respiratory diseases: user characteristics and NHP-Consumption behavior during the Covid-19 pandemic in Germany

Variables

Total sample

RD-NHP user

nRD-NHP user

Non-NHP user

X²(df);

p-value

Cramer’s V

Total (%)

100% (n = 1707)

46.5% (n = 794)

39.3% (n = 670)

14.2% (n = 243)

  

Covid-19 infection

      

Yes

16.4% (n = 280)

23.0% (n = 183)

9.6% (n = 64)

13.6% (n = 33)

X²(2) = 0.49.91, p = < 0.001

 

No

83.6% (n = 1427)

77.0% (n = 611)

90.4% (n = 606)

86.4% (n = 210)

0.171

Influenza vaccination

      

Yes

34.8% (n = 594)

32.7% (n = 260)

35.5% (n = 238)

39.5% (n = 96)

X²(2) = 1.25, p = 0.264

 

No

65.2% (n = 1113)

67.3% (n = 534)

64.5% (n = 432)

60.5% (n = 147)

0.048

Covid-19 vaccination

      

Yes

85.4% (n = 1458)

85.8% (n = 681)

85.4% (n = 572)

84.4% (n = 205)

X²(2) = 0.30, p = 0.862

 

No

14.6% (n = 249)

14.2% (n = 113)

14.6% (n = 98)

15.6% (n = 38)

0.013

Tetanus vaccination

      

Yes

66.7% (n = 1138)

69.8% (n = 554)

64.6% (n = 433)

62.1% (n = 151)

X²(2) = 6.94, p = 0.031

 

No

33.3% (n = 569)

30.2% (n = 240)

35.4% (n = 237)

37.9% (n = 92)

0.064

Preventive check-ups

      

Yes

65.7% (n = 1121)

69.0% (n = 548)

64.6% (n = 433)

57.6% (n = 140)

X²(2) = 11.27, p = 0.004

 

No

34.3% (n = 586)

31.0% (n = 246)

35.4% (n = 237)

42.4% (n = 103)

0.081

Health consultants

      

General practitioner

80.5% (n = 1374)

82.2% (n = 653)

75.7% (n = 507)

88.1% (n = 214)

X²(2) = 20.34, p = < 0.001

0.109

Pharmacist

23.1% (n = 395)

29.7% (n = 236)

20.7% (n = 139)

8.2% (n = 20)

X²(2) = 51.88, p = < 0.001

0.174

Alternative practitioner

6.6% (n = 113)

9.3% (n = 74)

5.7% (n = 38)

0.4% (n = 1)

X²(2) = 25.49, p = < 0.001

0.122

Family/friends

26.1% (n = 445)

31.6% (n = 251)

24.8% (n = 166)

11.5% (n = 28)

X²(2) = 39.92, p = < 0.001

0.153

Nobody

5.6% (n = 96)

4.3% (n = 34)

7.5% (n = 50)

4.9% (n = 12)

X²(2) = 7.18, p = 0.028

0.065