_{1}

^{*}

**Objective:** Cervical cancer is the second most prevalent cancer in females worldwide. Infection with human papillomavirus (HPV) is regarded as the main risk factor of cervical cancer. One objective of this study was to conduct a qualitative systematic review of some case-control studies and to examine the role of human papillomavirus (HPV) in the development of human cervical cancer (CC) beyond any reasonable doubt.
**Methods:** We conducted a systematic review and re-analysis of some impressive key studies aimed to answer the following question. Is there a cause-effect relationship between human papillomavirus and cervical cancer? The method of the conditio sine qua non relationship was used to proof the hypothesis whether the presence of human papillomavirus guarantees the presence of cervical carcinoma. In other words, if human cervical cancer is present, then human papillomavirus is present too. The mathematical formula of the causal relationship
k was used to proof the hypothesis, whether there is a cause-effect relationship between human papillomavirus and cervical carcinoma. Significance was indicated by a
p-value of less than 0.05.
**Result:** The studies analyzed (sample size
N
= 7657) were able to provide strict evidence that human papillomavirus is a necessary condition (a conditio sine qua non) of cervical carcinoma. Furthermore, the studies analyzed provide impressive evidence of a cause-effect relationship (
k = +0.723669245,
p value < 0.00001) between human papillomavirus and cervical carcinoma.
**Conclusion:** Human papillomavirus is the cause of human cervical carcinoma.

Malignant (cancer) cells can be formed in the tissues of the cervix, the lower, narrow end of the uterus to result in cervical cancer. Cervical cancer, predominantly attributable to infection, usually develops slowly over time and is the second [

Human cervical cancer can be a deadly disease too. Identifying the cause of cervical cancer of is of strategic importance in public health.

For the questions addressed in this paper, was searched Pubmed for case-control studies conducted in any country and published in English which investigated the relationship between human papilloma virus and cervical cancer at least by polymerase chain reaction (PCR). The search in Pubmed was performed while using medical key words like “case control study” and “human papilloma virus” and “cervical carcinoma” and “PCR DNA” et cetera. The articles found where saved as a *.txt file while using Pubmed support (Menu: Send to, Choose Radio Button: File, Choose Format: Abstract (text). Click button “create file”). The created *.txt file was converted into a *.pdf file. The abstracts where studied within the *.pdf file. Those articles were considered for a review which provided access to data without any data access barrier.

Novel laboratory techniques [_{t}, b_{t}, c_{t}, d_{t}, N_{t} of

Author | Year | Country | a_{t} | b_{t} | c_{t} | d_{t} | a_{t} + b_{t} + d_{t} | a_{t} + b_{t} + c_{t} + d_{t} = N_{t} | (a_{t} + b_{t} + d_{t})/N_{t} |
---|---|---|---|---|---|---|---|---|---|

Sample size | (SINE) | ||||||||

Eluf-Neto et al., 1994 [ | 1994 | Brazil | 167 | 38 | 32 | 187 | 392 | 424 | 0.924528302 |

Ngelangel et al., 1998 [ | 1998 | Philippines | 333 | 35 | 23 | 346 | 714 | 737 | 0.968792402 |

Chichareon et al., 1998 [ | 1998 | Thailand | 356 | 42 | 21 | 219 | 617 | 638 | 0.967084639 |

Chaouki et al., 1998 [ | 1998 | Morocco | 176 | 38 | 10 | 147 | 361 | 371 | 0.973045822 |

Rolón et al., 2000 [ | 2000 | Paraguay | 109 | 18 | 4 | 73 | 200 | 204 | 0.980392157 |

Franceschi et al., 2003 [ | 2003 | India | 204 | 59 | 1 | 154 | 417 | 418 | 0.997607656 |

Asato et al., 2004 [ | 2004 | Japan | 311 | 333 | 45 | 2916 | 3560 | 3605 | 0.987517337 |

Bernal et al., 2008 [ | 2008 | Spain | 56 | 210 | 4 | 990 | 1256 | 1260 | 0.996825397 |

Total | 1712 | 773 | 140 | 5032 | 7517 | 7657 | 0.981716077 |

Author | Year | Country | a_{t} + b_{t} + d_{t} | N_{t} | (a_{t} + b_{t} + d_{t})/N_{t} | X^{2} (Sine) | k | p value (k) |
---|---|---|---|---|---|---|---|---|

Eluf-Neto et al., 1994 [ | 1994 | Brazil | 392 | 424 | 0.924528302 | 4.53082192 | 0.66941066 | 3.18104E−43 |

Ngelangel et al., 1998 [ | 1998 | Philippines | 714 | 737 | 0.968792402 | 1.37195122 | 0.84304507 | 6.2931E−116 |

Chichareon et al., 1998 [ | 1998 | Thailand | 617 | 638 | 0.967084639 | 1.75104167 | 0.79508743 | 1.04251E−89 |

Chaouki et al., 1998 [ | 1998 | Morocco | 361 | 371 | 0.973045822 | 0.57484076 | 0.74972995 | 2.8642E−47 |

Rolón et al., 2000 [ | 2000 | Paraguay | 200 | 204 | 0.980392157 | 0.15909091 | 0.78631137 | 2.87949E−29 |

Franceschi et al., 2003 [ | 2003 | India | 417 | 418 | 0.997607656 | 0.0016129 | 0.7432314 | 3.7933E−52 |

Asato et al., 2004 [ | 2004 | Japan | 3560 | 3605 | 0.987517337 | 0.66877744 | 0.60055085 | 1.0309E−284 |

Bernal et al., 2008 [ | 2008 | Spain | 1256 | 1260 | 0.996825397 | 0.01232394 | 0.39572399 | 8.05848E−45 |

Total | 7517 | 7657 | 0.981716077 | 9.07046076 | 0.723669245 | |||

Alpha = | 0.05 | |||||||

Degrees of freedom = | 8 | Degr. of fr. = | 1 | |||||

X^{2} (Critical) SINE = | 15.5073131 | Chi crit. k = | 3.841458821 | |||||

X^{2} (Calculated) SINE = | 9.07046076 | X² calc. (k) = | 4009.949276 | |||||

K = | 0.723669245 |

Conditioned B_{t} (Human cervical carcinoma) | ||||
---|---|---|---|---|

Yes = +1 | Not = +0 | Total | ||

Condition A_{t} (HPV PCR DNA pos.) | Yes = +1 | a_{t} | b_{t} | A_{t} |

Not = +0 | c_{t} | d_{t} | A_{t} | |

Total | B_{t} | B_{t} | N_{t} |

All statistical analyses were performed with Microsoft Excel version 14.0.7166.5000 (32-Bit) software (Microsoft GmbH, Munich, Germany). In order to simplify the understanding of this article, to increase the transparency for the reader and to correct some of the misprints of former publications, several of the following lines are repeated word by word and taken from former publications.

The 2 × 2 table in this article is defined [

In general it is ( a t + b t ) = A t ,_{ ( c t + d t ) = A _ t }, ( a t + c t ) = B t , ( b t + d t ) = B _ t and a t + b t + c t + d t = N t . Equally, it is B t + B _ t = A t + A _ t = N t . In this context, it is p ( a t ) = p ( A t ∩ B t ) , p ( A t ) = p ( a t ) + p ( b t ) or in other words p ( A t ) = p ( A t ∩ B t ) + p ( b t ) = p ( A t ∩ B t ) + p ( A t ∩ B _ t ) while p(A_{t}) is not defined as p(a_{t}). In the same context, it is p ( B t ) = p ( a t ) + p ( c t ) = p ( A t ∩ B t ) + p ( c t ) and equally that p ( B _ t ) = 1 − p ( B t ) = p ( b t ) + p ( d t ) . Furthermore, the joint probability of A_{t} and B_{t} is denoted by p ( A t ∩ B t ) . Thus far, it is p ( A t ∩ B t ) = p ( A t ) − p ( b t ) = p ( B t ) − p ( c t ) or in other words it follows that p ( B t ) + p ( b t ) − p ( c t ) = p ( A t ) . Define Λ = p ( b t ) − p ( c t ) , the famous Einstein’s term under conditions of probability theory and we obtain p ( B t ) + Λ = p ( A t ) . In general, it is p ( a t ) + p ( c t ) + p ( b t ) + p ( d t ) = 1 . These relationships are viewed by the table (

In the case of independence of A_{t} and B_{t} it is

p ( A t ∩ B t ) ≡ p ( A t ) × p ( B t ) (1)

The mathematical formula of the necessary condition relationship (conditio sine quam non) [

p ( A t ← B t ) ≡ a t + b t + d t N t ≡ + 1 (2)

and used to proof the hypothesis: without A_{t} no B_{t} . In particular it is

Conditioned B_{t} | Total | |||
---|---|---|---|---|

Yes = +1 | No = +0 | |||

Condition A_{t} | Yes = +1 | p(a_{t}) = p(A_{t} Ç B_{t}) | p(b_{t}) | p(A_{t}) |

No = +0 | p(c_{t}) | p(d_{t}) | p(A_{t}) | |

Total | p(B_{t}) | p(B_{t}) | 1 |

p ( A t ← B t ) ≡ p ( a t ) + p ( b t ) + p ( d t ) p ( A t ← B t ) ≡ p ( A t ∩ B t ) + p ( B _ t ) p ( A t ← B t ) ≡ p ( A t ∩ B t ) + ( 1 − p ( B t ) ) p ( A t ← B t ) ≡ + 1 (3)

Scholium.

The study design and other factors can have an impact on bias with respect to the necessary condition. A different question worth asking concerns the relationship between the independence of an event A_{t} (a condition) and another event B_{t} (conditioned) and the necessary condition relationship. A fundamental question worth considering at this stage is whether is it possible that an event A_{t} is a necessary condition of event B_{t} an even if the event A_{t} (a necessary condition) is independent of an event B_{t} (the conditioned). In this context, the conditio sine qua non was defined as

p ( A t ← B t ) ≡ p ( A t ∩ B t ) + p ( B _ t ) ≡ + 1 (4)

or as

p ( A t ← B t ) ≡ p ( A t ∩ B t ) + ( 1 − p ( B t ) ) ≡ + 1 (5)

Under conditions where an event A_{t} is independent of an even B_{t} it is equally true that

p ( A t ∩ B t ) ≡ p ( A t ) × p ( B t ) (6)

Rearranging equation before it is

p ( A t ) × p ( B t ) + ( 1 − p ( B t ) ) ≡ + 1 (7)

or

p ( A t ) × p ( B t ) ≡ p ( B t ) (8)

or

p ( A t ) ≡ + 1. (9)

Only under conditions where p(A_{t}) = 1, theoretically it is possible to treat A_{t} as a necessary condition of B_{t} even if A_{t} is independent of B_{t} and vice versa, otherwise not. In other words, it is very difficult to treat a statistically significant conditio sine qua non relationship as very convincing if at the same time an event A_{t} is independent of and event B_{t} and vice versa. While discussing the statistical significance of results with respect to a necessary condition (or a sufficient condition or a necessary and sufficient condition), such or similar arguments should be considered. Due to an inappropriate study design or other sources of possible bias, the statistical significance of a conditio sine qua non relationship should be treated with very great cautious if evidence is provided that at the same time the same investigated parameters are independent of each other.

Under some circumstances, the rule three and other methods can be used to test the significance of a necessary condition. In this publication, the chi-square [

H_{0}: The sample distribution does agree with the hypothetical (theoretical) distribution of a necessary condition.

H_{A}: The sample distribution does not agree with the hypothetical (theoretical) distribution of a necessary condition.

The X^{2} Goodness-of-Fit Test can be shown schematically as

χ 2 ≡ ∑ t = + 1 t = + N ( ( Observed t − Expected t ) 2 Expected t ) (10)

The degrees of freedom are calculated as N − 1. Interestingly, if there is no discrepancy between an observed and a theoretical distribution at all, then the value of the calculated X^{2} = 0. As the discrepancy between an observed and the theoretical distribution of a necessary condition becomes larger, the X^{2} becomes larger. This X^{2} values are evaluated by the known X^{2} distribution. An adjustment (Yate’s correction for continuity) can be used when there is one degree of freedom. When there is more than one degree of freedom, the same adjustment is not used. Applying this to the formula above, we find the X^{2} Goodness-of-Fit Test with continuity correction shown schematically as

χ 2 ≡ ∑ t = + 1 t = + N ( ( | Observed t − Expected t | − ( 1 2 ) ) 2 Expected t ) (11)

Under circumstances, where the term (|Observed_{t} − Expected_{t}|) is less than 1/2, the continuity correction should be omitted. The theoretical (hypothetical) distribution of a necessary condition is shown schematically by the 2 × 2 table (

The theoretical distribution of a necessary condition (conditio sine qua non) is determined by the fact that c_{t} = 0. The X^{2} Goodness-of-Fit Test with continuity correction of a necessary condition (conditio sine qua non) is calculated as

Conditioned | Total | |||
---|---|---|---|---|

Yes = +1 | No = +0 | |||

Condition | Yes = +1 | a_{t} | b_{t} | (a_{t} + b_{t}) |

No = +0 | c_{t} = 0 | d_{t} | (c_{t} + d_{t}) | |

Total | (a_{t} + c_{t}) | (b_{t} + d_{t}) | (a_{t} + b_{t} + c_{t} + d_{t}) |

χ 2 ( SINE ) ≡ ( ( | ( a + b ) − ( a + b ) | − ( 1 2 ) ) 2 ( a + b ) ) + ( ( | ( d ) − ( c + d ) | − ( 1 2 ) ) 2 ( c + d ) ) = 0 + ( ( | d − ( c + d ) | − ( 1 2 ) ) 2 ( c + d ) ) (12)

or more simplified as

χ 2 ( SINE ) ≡ ( ( | − c | − ( 1 2 ) ) 2 ( c + d ) ) + 0 (13)

Under these circumstances, the degree of freedom is d.f. = N − 1 = 2 − 1 = 1 . The conditio sine qua non model can be used widely and is one of the new and appropriate methods of analysis of binary outcome variables. In this context, meta-analysis and systematic reviews aims to combine effects estimated from several studies to achieve greater precision of the conclusions drawn and can provide us with more convincing and reliable evidence of some special aspects of medicine. In meta-analysis the heterogeneity between the studies can be modelled via the additive properties of the chi square distribution too. In general, let X_{t} denote n independent random variables which follow a chi-square distribution. The sum of these independent chi-square variates is itself a chi-square variate which is known as the additive property of independent chi-squares. There may be disadvantages in the use of the chi-square-goodness-of-fit test. Still, the chi square distribution, a continuous probability distribution, is related to the standard normal distribution and is a simple and good measure of model adequacy. However, a particular concern with the use of the chi-square-goodness-of-fit test is a priori justified if expected cell frequencies of a 2 × 2 table are too small (all are less than one).

Huxley [

k ( U R t , W 0 t ) ≡ ( p ( U R t × W 0 t ) − ( p ( U R t ) × p ( W 0 t ) ) ) ( p ( U R t ) × p ( U _ R t ) ) × ( p ( W 0 ) × p ( W _ 0 t ) ) 2 (14)

where _{R}U_{t} denotes the cause and _{0}W_{t} denotes the effect while the chi-square distribution [

k ( U R t , W 0 t ) ≡ N × N × ( p ( U R t × W 0 t ) − ( p ( U R t ) × p ( W 0 t ) ) ) N × N × ( p ( U R t ) × p ( U _ R t ) ) × ( p ( W 0 ) × p ( W _ 0 t ) ) 2 (15)

or that

k ( U R t , W 0 t ) ≡ ( N × N × p ( U R t × W 0 t ) − ( N × p ( U R t ) × N × p ( W 0 t ) ) ) ( N × p ( U R t ) × N × p ( U _ R t ) ) × ( N × p ( W 0 ) × N × p ( W _ 0 t ) ) 2 (16)

or at the end

k ( U R t , W 0 t ) ≡ ( ( N × a t ) − ( U R t × W 0 t ) ) ( U R t × U _ R t ) × ( W 0 × W _ 0 t ) 2 (17)

where N is the sample size, a t = N × p ( U R t ∩ W 0 t ) , R U t = N × p ( R U t ) , R U _ t = N × p ( R U _ t ) , 0 W t = N × p ( 0 W t ) , 0 W _ t = N × p ( 0 W _ t ) . Several factors can have an impact on the calculated causal relationship k with the potential of bias.

Scholium.

Firstly, the relationship between condition and cause has an impact on the causal relationship k. A proper and deeper analysis of the relationship between cause and condition is beyond the scope of this article and can be found in literature [_{t} can be a (necessary, sufficient, necessary and sufficient) condition of an event B_{t} even if both are independent of each other, is already answered few lines before. Still, under which circumstances can we treat an event as a cause or as the cause of another event? Can an event be a cause of another event without being a (necessary, sufficient, necessary and sufficient et cetera) condition of the same event? The concept of this article is restricted on its capacity to bring high degrees of conceptual exactness and rigour to questions like these but not incapable. Most authors who have written on the question of the relationship between condition and cause came to different conclusions. Currently still worthy of consideration is the remark of von Bar.

“Die erste Voraussetzung, welche erforderlich ist, damit eine Erscheinung als die Ursache einer anderen bezeichnet werden könne, ist, daß jene eine der Bedingungen dieser sein. Würde die zweite Erscheinung auch dann eingetreten sein, wenn die erste nicht vorhanden war, so ist sie in keinem Falle Bedingung und noch weniger Ursache. Wo immer eine Kausalzusammenhang behauptet wird, da muß er wenigstens diese Probe aushalten… Jede Ursache ist nothwendig auch eine Bedingung eines Ereignisses; aber nicht jede Bedingung ist Ursache zu nennen.” [

Translated into English:

“The first requirement, which is required, thus that something could be called as the cause of another, is that the one has to be one of the conditions of the other. If the second something had occurred even if the first one did not exist, so it is by no means a condition and still less a cause. Wherever a causal relationship is claimed, the same must at least withstand this test… Every cause is necessarily also a condition of an event too; but not every condition is cause too.”

From this statement, it could appear that there is a gap between what is a cause and what is a condition. Is it possible to generalize this finding? In probabilistic approaches to causation, it is obvious that a cause of an event is equally a condition of the same event too. Clearly, the same relationship must not be given the other way too. A condition of an event must not equally be a cause of the same event. In summary, the objections build on the contradiction between condition and cause are no longer justified. A cause is a condition of an event too but not necessarily vice versa. A condition of an event must not be equally the cause of the same event. Thus far, like other fundamental concepts, the concepts of necessary conditions, the concepts of sufficient conditions and the concepts of necessary and sufficient conditions can be one of the handy tools to determine precisely whether a causal relationship is significant or not. A study which provides evidence of a significant causal relationship k without at the same time providing evidence of a significant necessary condition, or of a significant sufficient condition or of a significant necessary and sufficient condition should be treated with some cautious.

Secondly, a proper study design is necessary to use the mathematical formula of the causal relationship k with confidence otherwise bias is possible. For example, the probabilities of two events within a population are know precisely and shown schematically by the 2 × 2 table (

The causal relationship k (

Now we perform a study A with a sample size of n = 10,000. The verum group is n Verum = n / 2 = 5000 and the placebo group is n Placebo = n / 2 = 5000 . We do expect that the probabilities within the sample are the same like in the population. Under these conditions we obtain the following picture (

Thus far, this study has provided the following data (

Calculating the causal relationship k under these conditions, we obtain

k ( U R t , W 0 t ) ≡ ( ( ( ( 10000 ) × ( 4955 ) ) ) − ( ( ( 5000 ) × ( 4955 ) ) ) ) ( ( 4955 ) × ( 5045 ) ) × ( ( 5000 ) × ( 5000 ) ) 2 = + 0 .991031209 (18)

Effect | Total | |||
---|---|---|---|---|

Yes = +1 | No = +0 | |||

Cause | Yes = +1 | 0.99 | 0.009 | 0.999 |

No = +0 | 0 | 0.001 | 0.001 | |

Total | 0.99 | 0.1 | 1 |

Effect | Total | |||
---|---|---|---|---|

Yes = +1 | No = +0 | |||

Cause | Yes = +1 | (0.99/0.999) | (0.009/0.999) | 1 |

No = +0 | 0 | (0.001/0.001) | 1 | |

Total | (0.99/0.999) | (2-(0.99/0.999)) | 2 |

Effect | Total | |||
---|---|---|---|---|

Yes = +1 | No = +0 | |||

Cause | Yes = +1 | (0.99/0.999) × 5000 | (0.009/0.999) × 5000 | 1 × 5000 |

No = +0 | 0 × 5000 | (0.001/0.001) × 5000 | 1 × 5000 | |

Total | (0.99/0.999) × 5000 | (2-(0.99/0.999)) × 5000 | 2 × 5000 |

Effect | Total | |||
---|---|---|---|---|

Yes = +1 | No = +0 | |||

Cause | Yes = +1 | 4955 | 45 | 5000 |

No = +0 | 0 | 5000 | 5000 | |

Total | 4955 | 5045 | 10,000 |

a causal relationship k of k = +0.991031209. The study with the sample size n = 10,000 should have obtained a causal relationship k = +0.314800094 while the same obtained a causal relationship k = +0.991031209. This example demonstrates that the study design as such can be a source of bias if inappropriate measures are taken. To reduce the bias, it makes sense to consider the prevalence of a factor/an event within the population as much as possible as an essential part of study design. According to the data above, the prevalence of a cause within the population is p(_{R}U_{t}) = 0.999. The sample size of the study is still n = 10,000. Under these assumptions, the sample size of the verum group should be considered as n Verum = p ( R U t ) × n = 0.999 × 10000 = 9900 while the sample size of the placebo group is n Placebo = p ( R U _ t ) × n = 0.001 × 10000 = 1 (

The causal relationship k (

Let us assume that the sample size of this case control study B is n = 1000, with 500 cases and 500 controls. We obtain the following data (

The data of this case control study (

The chi-squared distribution [

Under some circumstances the chi-square [

Effect | Total | |||
---|---|---|---|---|

Yes = +1 | No = +0 | |||

Cause | Yes = +1 | 0.99 × 10000 | 0.009 × 10000 | 9990 |

No = +0 | 0 × 10000 | 0.001 × 10000 | 10 | |

Total | 9900 | 100 | 10,000 |

Effect | Total | |||
---|---|---|---|---|

Yes = +1 | No = +0 | |||

Cause | Yes = +1 | (0.99/0.99) | (0.009/0.01) | (2-(0.001/0.01)) |

No = +0 | 0 | (0.001/0.01) | (0.001/0.01) | |

Total | 1 | 1 | 2 |

Effect | Total | |||
---|---|---|---|---|

Yes = +1 | No = +0 | |||

Cause | Yes = +1 | (0.99/0.99)) × 500 | (0.009/0.01) × 500 | 950 |

No = +0 | 0 × 500 | (0.001/0.01) × 500 | 50 | |

Total | 500 | 500 | 1000 |

p-value | One sided X^{2} | Two sided X^{2} | |
---|---|---|---|

The chi square distribution | 0.1000000000 0.0500000000 0.0400000000 0.0300000000 0.0200000000 0.0100000000 0.0010000000 0.0001000000 0.0000100000 0.0000010000 0.0000001000 0.0000000100 0.0000000010 0.0000000001 | 1.642374415 2.705543454 3.06490172 3.537384596 4.217884588 5.411894431 9.549535706 13.83108362 18.18929348 22.59504266 27.03311129 31.49455797 35.97368894 40.46665791 | 2.705543454 3.841458821 4.217884588 4.709292247 5.411894431 6.634896601 10.82756617 15.13670523 19.51142096 23.92812698 28.37398736 32.84125335 37.32489311 41.82145620 |

test the significance of a causal relationship. Under conditions where the probability of events is constant from trial to trial, we expect a constant causal relationship k_{t}. In other words, at each Bernoulli trial t it is

| k ( U R t , W 0 t ) | ≡ | 1 | (19)

Performing N Bernoulli trials (Sample size N), the basic relationship will not change. It follows that

N × | k ( U R t , W 0 t ) | ≡ N × | 1 | (20)

or that

N × | k ( U R t , W 0 t ) | − N × | 1 | = 0 (21)

Simplifying equation we obtain

N × ( | k ( U R t , W 0 t ) | − | 1 | ) = 0 (22)

Multiplying equation by itself it is

N × ( | k ( U R t , W 0 t ) | − | 1 | ) × N × ( | k ( U R t , W 0 t ) | − | 1 | ) = 0 × 0 (23)

or

N 2 × ( | k ( U R t , W 0 t ) | − | 1 | ) 2 = 0 (24)

Dividing equation by N * |1| = N, we obtain

N 2 × ( | k ( U R t , W 0 t ) | − | 1 | ) 2 N = 0 N = 0 (25)

or

N × ( | k ( U R t , W 0 t ) | − | 1 | ) 2 = 0 (26)

or the X^{2} value as

χ 2 = N × ( | k ( U R t , W 0 t ) | − | 1 | ) 2 = 0 (27)

The chi square (X^{2}) statistic can be used to investigate whether the observed distribution of the causal relationship differ from the theoretical expected distribution of the causal relationship. The table (

Claims.

Null hypothesis:

The presence of human papillomavirus DNA is a necessary condition (a conditio sine qua non) of human cervical cancer. In other words, the sample distribution agrees with the hypothetical (theoretical) distribution of a necessary condition.

Alternative hypothesis:

The presence of human papillomavirus DNA is not a necessary condition (a conditio sine qua non) of human cervical cancer. In other words, the sample distribution does not agree with the hypothetical (theoretical) distribution of a necessary condition.

The significance level (Alpha) below which the null hypothesis will be rejected is alpha = 0.05.

Proof.

The data reviewed by this article which investigated the relationship between the presence of human papillomavirus DNA and human cervical cancer are viewed by the table (^{2} Calculated (SINE) = 9.070460764. The critical Chi square can be obtained (degrees of freedom = 8, alpha = 0.05) as X^{2} Critical = 15.50731306. In particular, due to the data of the studies meta-analyzed, human papillomavirus DNA and human cervical cancer are not independent (degrees of freedom = 1, Chi-square value calculated = 4009.949276) of each other. The detailed calculations are shown by the table (^{2} Calculated (SINE) = 9.070460764) is less than the critical Chi square value (X^{2} Critical =15.50731306). Due to this evidence, we do not reject the null hypothesis in favor of the alternative hypotheses. The data as published by studies presented by the table (

Q. e. d.

Claims.

Null hypothesis: (no causal relationship, k = 0)

There is no causal relationship between human papillomavirus and human cervical cancer.

Alternative hypothesis: (causal relationship, k ¹ 0)

There is a causal relationship between human papillomavirus and human cervical cancer.

Conditions.

Alpha level = 5%. The two tailed critical Chi square value (degrees of freedom = 1) for alpha level 5% is 3.841458821.

Proof.

The data for this hypothesis test are illustrated in the 2 × 2 table (

The causal relationship k (HPV DNA, Human cervical cancer) is calculated [

k ( HPV DNA , CC ) = ( ( 7657 × 1712 ) − ( 2458 × 1852 ) ) ( 1852 × 5805 ) × ( 2485 × 5172 ) 2 = + 0.723669245

The value of the test statistic k = +0.723669245 is equivalent to a calculated [

χ Calculated 2 = 7657 × ( ( ( 7657 × 1712 ) − ( 2458 × 1852 ) ) ( 1852 × 5805 ) × ( 2485 × 5172 ) 2 ) × ( ( ( 7657 × 1712 ) − ( 2458 × 1852 ) ) ( 1852 × 5805 ) × ( 2485 × 5172 ) 2 ) χ Calculated 2 = 7657 × 0.723669245 × 0.723669245 χ Calculated 2 = 4009.949276

The chi-square statistic, uncorrected for continuity, is calculated as X² = 4009.949276 and thus far equivalent to a p value of p < 0.00001. The calculated chi-square statistic exceeds the critical chi-square value of 3.841458821 (

Q. e. d.

The nature of the relationship between HPV and cervical cancer has been exhaustively investigated over more than 20 years. Several studies which have unequivocally shown that HPV-DNA can be detected in about 95% to 100% of adequate

Human cervical cancer | Total | |||
---|---|---|---|---|

Yes | No | |||

Human papillomavirus PCR DNA | Yes | 1712 | 773 | 2485 |

No | 140 | 5032 | 5172 | |

Total | 1852 | 5805 | 7657 | |

p (A_{t} ¬ B_{t})= | 0.981716077 | |||

X² (A_{t} ¬ B_{t})= | 9.070460764 | |||

k (A_{t},B_{t})= | 0.723669245 | |||

p value (k)< | 0.00001 |

specimens of cervical cancer, while there was no significant variation in HPV positivity among countries and support the claim that HPV is a necessary condition cervical cancer. The most reviews available have consistently concluded that there is a strong evidence of an association between HPV and cervical cancer [

Human papillomavirus is the cause of human cervical cancer.

Barukčić, I. (2018) Human Papillomavirus―The Cause of Human Cervical Cancer. Journal of Biosciences and Medicines, 6, 106-125. https://doi.org/10.4236/jbm.2018.64009