(p.505) Appendix Methods of statistical analysis: formulae and worked examples
(p.505) Appendix Methods of statistical analysis: formulae and worked examples
This appendix presents tables showing the calculations of measures of association and of statistical tests, for the main study designs discussed showing for each the formulae and worked examples. There are also reference tables for statistical issues.
Introduction
Table 1 
Case–control designs, unmatched 
p. 508 
Table 2 
Cohort designs with individual observation data 
p. 512 
Table 3 
Cohort designs with person‐time data 
p. 518 
Table 4 
Cohort or case–control designs with small numbers; exact test 
p. 522 
Table 5 
Case–control studies with 1:1 matching 
p. 526 
Table 6 
Case–control studies with fixed 1:m matching 
p. 528 
Table 7 
Cohort at case–control designs: test for trend 
p. 530 
Table 8 
Formulae for sample size determination 
p. 534 
Table 9 
Constants for use in sample size formulae 
p. 535 
Table 10 
Calculation of kappa to compare two sets of observations 
p. 536 
Table 11 
Adjustment of results of a case–control study using values of sensitivity and specificity 
p. 540 
Table 12 
Meta‐analysis: Mantel‐Haenszel and Peto methods 
p. 542 
Table 13 
Meta‐analysis: confidence limits method 
p. 548 
Table 14 
Meta‐analysis: DerSimonian‐Laird method 
p. 550 
Table 15 
Table of probability (P) values and equivalent normal deviate (2) and χ^{2} values 
p. 552 
Table 16 
Table showing functions in Microsoft Excel for relating P‐values, normal deviates, and χ^{2} values 
p. 557 
References 
p. 558 
(p.506) Introduction
These tables will enable the reader to analyse most studies where the factors involved are discrete rather than continuous variables. Of course, major analyses should not be done without consideration of alternative methods and discussion with colleagues, including those more skilled in statistical and epidemiological methods.
The formulae are primarily those derived from the work of Mantel and Haenszel. These involve approximations which are appropriate if the numbers of observations available are adequate: in a 2 × 2 table each expected value should be greater than 5. For the formulae used to produce summary risk estimates and test statistics after confounder control by stratification, it is the total number of observations which is relevant; the formulae can be applied even if some strata have few observations. The statistics used to test whether an odds ratio or attributable risk estimate is constant over several strata do depend upon having reasonable numbers of observations within each stratum.
The one exception in this Appendix to the limitation of small numbers is the ‘exact’ test for a 2 × 2 table presented in Table 4. This can be applied to simple tables with small numbers. For all the other designs considered, more precise formulae (‘exact’ tests) are also available, but these will usually give similar results except where the numbers of observations are small.
The sources of the formulae are given with each table. There are many texts which review these and related methods in detail, and discuss the situations in which more complex formulae may be appropriate, and provide information on them. (p.507)
Table 1. Case–control studies. Unmatched: formulae
1. Format of table 
2. Risks 
3. Statistical tests Observed exposed cases = a Expected value of a = E = N_{1}M_{1}/T Variance of a = V = N_{1}N_{0} M_{1}M_{0}/T^{2}(T–1) 
3A Chi–squared statistic, χ^{2}, with 1 degree of freedom 
3B χ statistic, or normal deviate = χ = 
3C Continuity correction: reduce absolute value of numerator by before squaring; i.e. replace numerator by , where means the absolute value of (a – E), irrespective of being positive or negative. 
4. Confidence limits (C.L.). y% limits for logarithm of odds ratio = ln OR ± Zy(dev ln OR), where Zy = appropriate normal deviate (see Table 15) and dev ln OR = standard deviation of in OR 
4A dev ln 
5. Stratified analysis over I subtables of above format, values a_{i} etc. 
5A Summary odds ratio = 
5B Summary χ^{2} statistic, 1 d.f. = Each term analogous to 3 above. 
5C Continuity correction: reduce absolute value of numerator by before squaring. 
5D A formula for the variance of the logarithm of the summary odds ratio is given by Robins et al.[1] 
For each stratum, these quantities are calculated. The subscripts have been omitted. 
Then the variance of ln Mantel‐Haenszel summary odds ratio is: and y% confidence limits for ln OR _{s} 
5E Test of homogeneity = where variance of ln OR_{i} = square of standard deviations given in 4A above. Gives a χ^{2} on I – 1 degrees of freedom. 
Sources: 3A, 5A, 5B Mantel and Haenszel [2] 4A Wolff [3] 5D Robins et al. [1] 5E Rothman and Greenland [4] 
Table.1. Case–control studies. Unmatched: worked example
Data from the case–control study shown in Ex.6.24 and in Ex.7.9: association between cervical carcinoma and smoking. 
1. Crude table: 
2. Odds ratio = (130×198)/(45×87) = 6.57 
3A, B Statistical test: 
3C With continuity correction χ^{2} = 81.4 

5. Stratified analysis: Data are shown in Ex.7.9. Totals in the three subtables are 113, 211, 136 (p. 244). 

5B Summary χ^{2}: calculate for each subtable a_{i}, E_{i}, V_{i}; and sum each of these. 
5C. χ^{2} with continuity correction = 80.6 d.f. = 1 P <0.000001 (Appendix Table 15) 
5D Using Robins formula, var ln OR_{MH} = 0.0456, and 95% confidence limits of OR_{MH} = 4.12, 9.53. 
5E Test of homogeneity: calculate for each table OR_{i}, and use formula 4 to obtain dev ln OR_{i} 
Hence test of homogeneity = sum of {(ln OR_{i} – ln OR _{s})^{2}/var ln OR_{i}} where var ln OR_{i} = dev ln and OR _{s} = 6.27 or, equivalently, test = sum of {(ln OR_{i}/ln OR _{s})/dev ln OR_{i}}^{2} 
d.f. = 2, P‐values from Appendix Table 16. The odds ratios vary significantly. 
Table 2. Cohort studies, count data: formulae
1. Format of table 
2. Risks Relative risk = r _{e}/r _{0} Odds ratio = Attributable risk = r _{e} – r _{0} 
3. Statistical tests Observed exposed cases = a Expected value of a =E =N _{1} M_{1}/T Variance of a = V = N _{1} N _{0} M _{1} M _{0}/T ^{2}(T–1) 
3A Chi–squared statistic, χ^{2}, with 1 degree of freedom = (a−E)^{2}/v 
3B χstatistic, or normal deviate = χ = . 
3C Continuity correction: reduce absolute value of numerator by before squaring, i.e. replace numerator by , means the absolute value of (a–E), irrespective of being positive or negative. 
4. Confidence limits (C.L.)
where Zy = appropriate normal deviate (see Table 15) and dev = standard deviation 
4A Formulae for standard deviation 
5. Stratified analysis 
5A Risks 
5B Summary χ^{2} statistic on 1 d.f. = Each term analogous to 3 above. 
5C Continuity correction: reduce absolute value of numerator by before squaring. 
5D 
Tests of homogeneity; χ^{2} on I–1 degrees of freedom; variances are squares of standard deviations given in 4. 
Sources: 3A, 5A, 5B Mantel and Haenszel [2] 
5D Rothman and Greenland [4] 
Table 2. Cohort studies, count data: worked example
Data from a prospective cohort study; association between maternal epilepsy in pregnancy and malformation in infants; Shapiro et al. [5] 
(Ex. 3.2 showed another study in the same format). 
1. Crude table: 
2. Relative risk = 1.63 Odds ratio = 1.70 Attributable risk = 4.06% 
3A Statistical test: 

3C With continuity correction χ^{2} = 7.60. 
4A Confidence limits: for RR, dev ln RR = 0.168; limits = exp(ln RR ± 1.96 × dev ln RR) =1.17, 2.27 for OR, dev ln OR = 0.188; limits = exp(ln OR ± 1.96 × dev ln OR) =1.18, 2.46 for AR, dev AR = 1.758%; limits = AR ± 1.96 × dev AR =0.61, 7.51% 
5. Stratified analysis: Analogous to Table 1. If the above data were one stratum of a stratified table, the weight w _{i} for use in formula 5A(iii) would be 1/(dev AR _{i})^{2}=1/(0.0176)^{2}. 
Table 3. Cohort studies, person‐time data: formulae
1. Format of table 
2. Risks Relative risk = r _{e}/r _{0} Attributable risk = r _{e} – r _{0} 
3. Statistical tests Observed exposed cases = a Expected value of a = E = N _{1} M_{1}/T Variance of a = V = N_{1}N_{0}M_{1}/T^{2} 
3A Chi‐squared statistic, χ^{2} with 1 degree of freedom = (a − E)^{2}/V 
3B χ statistic, or normal deviate = χ = 
3C Continuity correction: reduce absolute value of numerator by before squaring; i.e. replace numerator by , means the absolute value of (a–E), irrespective of being positive or negative. 
4. Confidence limits (C.L.) (i) for logarithm of RR (ii) for attributable risk = ln RR ± Zy dev ln RR = AR ± Zy dev AR where Zy = appropriate normal deviate (Table 15) and dev = standard deviation 
4A Formulae for standard deviation 
5. Stratified analysis over 1 subtables of above format, values, a_{i} etc. 
5A 
5B Summary χ^{2} statistic = Each term analogous to 3 above. 
5C Continuity correction: reduce absolute value of numerator by before squaring. 
5D Tests of homogeneity; χ^{2} on I–1 degrees of freedom; variances are squares of standard deviations given in 4. 
Odds ratio: analogous to relative risk Sources: 3A Mantel and Haenszel [2] 5D Rothman and Greenland [4] 
Table 3. Cohort studies, person‐time data: worked example
Data from the prospective cohort study shown in Ex.6.7; association between exercise and coronary heart disease mortality. 
1. Crude table: 
2. Relative risk = 81.85/23.83 = 3.43 Attributable risk = 81.85 − 23.83 = 58.02/10000 man‐years 
3A Statistical test: 
3B 
3C With continuity correction χ^{2} = 100.45. 
4A dev ln RR = 0.131 95% limits = exp{ln 3.43±(1.96×0.131)} = 2.66, 4.44 dev = 4.60/10000 95% limits = 58.0 ± (1.96 × 4.60) = 49.0, 67.0 
5. Stratified analysis. Analogous to Appendix Table 1 
Table 4. Cohort (count data) or case–control, with small numbers; exact test for a fourfold table: formulae
1. Format of table 
2. Risks As in Table 2; the small numbers make no difference. 
3. Statistical test Probability of a particular table occurring = To calculate one‐sided probability of a or a more extreme value: if a is greater than expected, i.e. a > N _{1} M _{1}/T, calculate quantity above for each value of a from observed value to the maximum, given when a = N _{1} or a = M _{1}; sum these values. If a is less than expected, calculate for values of a down to zero. For two‐sided tests, the one‐sided value is usually doubled. ! represents a factorial; e.g. 5! is 5 × 4 × 3 × 2 × 1. 
Sources: This test was developed independently in the 1930s by R.A. Fisher, J.O. Irwin, and F. Yates. Exact confidence limits and stratified analysis will not be presented: see, for example, Breslow and Day [6] or Armitage et al. [7]. 
Table 4. Cohort (count data) or case–control, with small numbers; exact test for a fourfold table: worked example
Saral et al. [8] performed a double‐blind randomized trial of the drug acyclovir, compared with placebo, as prophylaxis against herpes simplex infection in 20 bone marrow transplant recipients who were seropositive for herpes simplex before randomization; the outcome was development of active herpes simplex infection. 
1. 
Table of results: 
2. Risk: the relative risk and odds ratio are both zero. 
3. Statistical test: probability of this set of data, given the null hypothesis 
This is the one‐sided P‐value; the two‐sided value of P = 0.003. As the result obtained was an extreme one, only one calculation is necessary. Otherwise, further calculations are needed. For example, suppose there had been two infections in the acyclovir group compared with seven in the placebo. To assess significance, we add the probability of the observed table to the probability of more extreme results with the same marginal totals: the observed table is: 
More extreme results, given that a is less than its expected value, are a = 1 and a = 0 with the same marginal totals: 
giving the final P‐value (one‐sided) of 0.032 + 0.0027 + 0.00006 = 0.035 or (two‐sided) = 0.07 
Table 5. Case–control, 1:1 matching: formulae
1. Format of table: numbers of pairs 
2. Risks Odds ratio = s/t 
3A Statistical tests 
3B χ statistic, or normal deviate = . 
3C Continuity correction: reduce absolute value of numerator by 1 before squaring; it becomes . 
4. Confidence limits The derivation of confidence limits is complex. Approximate test‐based limits can be based on the statistic by the methods shown in Ex.7.7. 
Sources: McNemar [9]. Stratified analysis of matched data is not frequently performed; matched multivariate models are more useful. 
Table 5. Case–control, 1:1 matching: worked example
Data from the case–control study shown in Ex. 6.24 and Ex.7.9, association between nasal cancer and smoking. 
1. Table: 
2. Odds ratio = 30/7 = 4.29 
3A, B χ^{2} statistic = (30−7)^{2}/(30+7) = 14.30 
3C With continuity correction, χ^{2} = 13.1. 
4. 95% confidence limits for OR, test‐based = exp{ln 4.29 (1±1.95/3.78)} = 2.0, 9.1 
Table 6. Case–control studies, fixed 1:m matching: formulae
1. Format of table: numbers of pairs 
2. Risks 
3. Statistical tests For the continuity correction, reduce the absolute value of the numerator by (M + 1) before squaring. 
4. Confidence limits The derivation of confidence limits is complex. Approximate test‐based limits can be based on the χ statistic by the methods shown in Ex. 7.7. Sources: 2,3 Based on Mantel and Haenszel [2] For more complex matched designs, see Breslow and Day [6], pp.169–187. 
Table 6. Case–control studied, fixed 1:m matching: worked example
Data from Collette et al. [10]; a 1:3 matched case–control study assessing the value of screening for breast cancer by physical examination and xeromammography. The cases were women from the defined population who had died from breast cancer; the controls were age‐matched women randomly selected from this population. 
1. Table: 


4. 95% confidence limits for OR (test‐based) = exp[ln 0.30 (1 ± 1.96/2.95)] = 0.13, 0.67 
Table 7. Cohort (count data) or case–control studies: test for trend: formulae
1. Format of table 
2. Risks 
3. Statistical tests 
3A For each level k against the referent level: tests and confidence limits as in Appendix Table 1. Heterogeneity χ^{2} for the table above, on k − 1 degrees of freedom 
where E_{k} = expected value of a_{k} = N_{k}M _{1}/T. This test assesses whether the odd ratios, or more directly the proportions of cases, in the various exposure levels are consistent with the overall value—it does not take into account the order of the levels of exposure. 
3B Test for trend from regression of the values a_{k} – E_{k} on the score x_{k}; χ^{2} on 1 d.f. 
For a continuity correction, if x_{k} scores are one unit apart, replace 
3C An approximate test of departure from the linear trend is given by the difference between the heterogeneity and the trend χ^{2} statistics, on k – 2 degrees of freedom: this tests the adequacy of the linear trend in describing the data. 
4. Stratified analysis The trend statistic above can be generalized to a stratified analysis over I subtables of the format above, but the formula is tedious for hand calculation; see Breslow and Day [6] pp.148–150. The stratified χ^{2} statistic on 1 d.f. is 
where E_{ki} = N_{ki}M _{1i}/T_{i}. Sources: 3A: Armitage [12], Armitage et al. [7] 3B Mantel [13] 
Table 7. Cohort (count data) or case–control studies: test for trend: worked example
These are data from a case–control study relating the occurrence of twin births to maternal parity (the number of previous births the mother has had). The cases were births, the controls a sample of single births. From Elwood [11] 
1. Table and elements of calculations: 
2. Risks: Given above: e.g. for parity 2 odds ratio = (454 × 1833)/ (853 × 716) = 1.36 
3A Global or heterogeneity χ^{2}. For each level E_{k} = N_{k}M _{1}/T e.g. for parity 2, E _{2} = 1307 × 2472 /7430 = 434.85 
3B Test for trend: 
3C Test for departure from linear trend: χ^{2} = 91.2 − 88.2 = 3.0 d.f. = 2 P > 0.2 (Table 16). 
Table 8. Formulae for sample size determination: formulae
1. Unmatched studies, equal groups 
2. Multiple controls per case Given c controls per case: n = no. of cases or exposed subjects, cn = no. of controls. 
3. 1:1 matched studies 
where H = OR/(1 = OR) OR = odds ratio and M = number of matched pairs. 
Notation 
Table 8. Formulae for sample size determination: worked example
See Chapter 7, pp 256–263 
Table 9. Constants for use in sample size formulae
A. Table relating normal deviates to power and to significance level 
B. Values of K = (Z _{α} + Z _{β} ) ^{2}, for commonly used values of α and β Normal deviates corresponding to frequently used values for significance levels (Z _{α}) and for power (Z _{β}); and table of K where K = (Z _{α} + Z _{β} ) ^{2}. The value of Z _{β} is the normal deviate corresponding to the one‐sided test for (1−power) 
Table 10. Calculation of kappa to compare two sets of observations: formulae
1. Format of table. Shown for three categories. p_{ij} = proportions of total subjects in cell i, j of the table 
2. Calculation of kappa, κ 
Kappa, κ = (observed agreement − expected agreement)/ (1 − expected agreement) = (O − E)/(1 − E) 
3. Standard error of kappa, SE(κ) n = total number of subjects. Let u_{i} = p_{i}. p._{i}(p_{i}. = p._{i}) u_{i} uses only the marginal totals. then 
Test of significance of kappa compared with null value of 0 is given by z = κ/SE(κ), but this is not very useful. 
4. Confidence limits for kappa Using the Z‐values given in Table 9 two‐sided limits = K ± Z _{α/2}.SE(κ) Other formulae are available to test the significance of a difference in kappa from a pre‐specified value, and for weighted kappa. 
5. Relationship of kappa to odds ratio (approximate) Let OR _{0} = observed odds ratio and OR _{T} = true odds ratio then OR _{T} = (κ + OR _{0} − 1)/κ and OR _{0} = κ (OR _{T} − 1) + 1 Sources: 2, Cohen [15] 5 Thompson and Walter [18] For weighted kappa and further development see Fleiss et al. [17] For discussion of applications to study results see Armstrong et al. [19] 
Table 10. Calculation of kappa to compare two sets of observations: worked example
1. Question: Have you ever been sunburned causing erythema and pain for a few days? If yes, how many times after the age of 19 years? 
2. Calculation of kappa 
Observed agreement = (0.096 + 0.527 + 0.125) = 0.747 
Expected agreement by chance = (0.165 × 0.152 + 0.637 × 0.656 + 0.197 × 0.192) = 0.481 
Kappa, κ = (observed agreement − expected agreement)/ (1 − expected agreement) = (0.747 − 0.481)/(1 − 0.481) = 0.512 
3. Standard error of kappa Number of subjects, n = 593 U _{1} = 0.00795 from 0.165 × 0.152 × (0.165 + 0.152) U _{2} = 0.54084 U _{3} = 0.01478 
Test of significance (Appendix Table 15). 
4. Confidence limits 95% two‐sided confidence limits = 0.512 ± 1.96 × 0.031 = 0.45, 0.57 
5. Relationship of kappa to odds ratio Suppose a study using this measure yields an odds ratio (observed) of 1.80, then estimated true odds ratio = (0.512 + 1.80 − 1)/0.512 = 2.56 
Table 11. Adjustment of results of a case–control study using values of sensitivity and specificity: formulae
1. Observed results of case–control study: proportions of cases and controls exposed: 
2. Sensitivity and specificity values for exposure, for cases and for controls: 
3. Hence estimated ‘true’ values: 
Table 11. Adjustment of results of a case–control study using values of sensitivity and specificity: worked example
1. Observed results of case–control study (hypothetical data) 
Results expressed as proportions exposed 
2. Sensitivity and specificity of exposure assessment, in cases and in controls, from other sources 
Table 12. Meta‐analysis: Mantel‐Haenszel and Peto methods: formulae
1. Format for each study is the same as given in Table 2 For each study i, observed number of successes on new treatment = a_{i} expected number variance of a_{i} 
as in Appendix Table 2. For meta‐analysis of a number of studies, a format of one line per study is useful. 
2. Mantel–Haenszel method 
2A Summary results: summary odds ratio 
2B χ statistic 
2C Confidence limits for summary odds ratio 
Precise limits given using the Robins et al. [1] formula for the variance of the ln summary odds ratio, given in Table 1, 5D 
y% limits for summary OR = 
2D Test for heterogeneity 
For each study calculate Q_{i} = w_{i}(ln OR_{i} − ln OR _{s})^{2} 
where w_{i} = weight = 1/variance of ln OR_{i}, from Table 1. 
ln OR_{i} = ln OR for study i 
ln OR _{s} = ln summary OR 
then is distributed as a χ^{2} statistic on n − 1 degrees of freedom, 
where n = number of studies. 
3. Peto method 
3A summary ln odds ratio = summary odds ratio = 
3B summary χ (normal deviate)= 
3C Confidence limits for summary odds ratio = 
where Z _{α} is the normal deviate for the significance level α, from Table 9. 
3D Test of heterogeneity 
For each study V_{i} and a_{i} − E_{i} have been defined as in 3 above, and (a_{i} − E_{i})^{2} is also calculated. The heterogeneity statistic Q is calculated as Q is distributed as a χ^{2} statistic on n − 1 degrees of freedom, where n = number of studies. 
As the statistics produced by these two methods are the same. The odds ratio estimates are different. 
Sources: Mantel and Haenszel [2] Peto et al. [21] Petitti [22] 
Table 12. Meta‐analysis: Mantel–Haenszel and Peto methods: worked example
1. Format of table Data from three randomized trials assessing clomiphene in inducing ovulation [20] 
2. Mantel–Haenszel method 
2A Summary odds ratio: OR _{s} = sum(ad/T)/sum(bc/T) = 8.93 

2C Confidence limits for OR _{s} ln OR_{s} = 2.1892 variance of ln OR _{s} from Robins et al. [1] formula: 0.1173 95% confidence limits using this variance 4.56, 17.47 
2D Test of heterogeneity 
Sum Q_{i} is a χ^{2} on two degrees of freedom: clearly non‐significant 
Peto method 
3A Summary odds ratio 
3B Normal deviate 
3C Confidence limits for summary odds ratio 95% limits for or = 3.92, 11.85 
3D Test for heterogeneity 
Table 13. Meta‐analysis: confidence limits method: formulae
1. Data required 
The information required from each study is an appropriate measure of odds ratio, OR *, and its variance, V, which can be calculated if the lower and upper confidence limits OR_{L} and OR_{u} are given, as where Z corresponds to the significance level: e.g. for 95% two‐sided confidence limits, Z = 1.96 (Table 9) 
Then, for each study, the weight w is 1/V 
2. Calculation of summary odds ratio Summary ln hence, OR _{s} = exp(ln OR _{s}) 
3. Confidence limits for the summary odds ratio variance of ln 
95% confidence limits for ln 
95% confidence limits for 
4. Heterogeneity test 
Heterogeneity statistic = 
which is a χ^{2} statistic on n = 1 degrees of freedom, where n = number of studies 
Sources: Prentice and Thomas [23] Greenland [24] 
(*) This method is applicable to relative risk also, and (without the log transformation) to risk difference.
Table 13. Meta analysis: confidence limits method: worked example
1. Meta analysis of seven case–control studies relating sunburn to melanoma. 
2. Summary ln OR _{s} = 67.14/105.97 = 0.63 Summary OR _{s} = exp(0.63) = 1.88 
3. 95% CL for summary = lower 1.56 upper 2.28 
4. Test for heterogeneity: sum Q_{i} = 29.47 Sum Q is a χ^{2} statistic on n − 1 = six degrees of freedom; from Table 16 P < 0.0001, so consideration has to be given to the results which are discordant, as shown by their Q_{i} results. 
I^{2} = (29.47−6)/29.47 = 0.80 
= 80% 
Table 14. Random effects model (DerSimonian–Laird) applied to confidence limits data
1. The random effects model of DerSimonian and Laird uses the same data as the fixed effects model shown in Table 13. For each study, the odds ratios OR_{i}, and their variances v_{i} are as in Table 13. Weights w_{i} = 1/v_{i} are calculated. The fixed effects summary odds ratio OR _{s} is calculated as before, and the quantities Q_{i} = w_{i}(ln OR_{i} − ln OR _{s})^{2} and sum Q_{i} = Q 
2,3. For the random effects model, revised weights dw_{i} are used where where n is the number of studies. D cannot be negative. If (n − 1) is greater than Q, then D = 0. If there is little heterogeneity, Q will be small, so D will be small, and the DerSimonian and Laird weights will be similar to the fixed effects weights. 
4. The percentage distribution of weights is shown, showing how the D–L method gives a more even weight distribution than the fixed effects method. 
Source: DerSimonian and Laird [25] 
Table 14. Random effects model (DerSimonian–Laird) applied to confidence limits data
1. Data as in Table 13: seven studies relating sunburn to melanoma 
4. Weight distribution 
Table 15. Table of probabilities (P‐values) and corresponding values of χ^{2} on 1 d.f., standardized normal deviate 2‐sided, and 1‐sided
P‐values 
χ^{2} 
Normal deviate 
Normal deviate 

1 d.f. 
2‐sided 
1‐sided 

0.000001 
23.9 
4.89 
4.75 
0.00001 
19.51 
4.42 
4.27 
0.0001 
15.14 
3.89 
3.72 
0.001 
10.83 
3.29 
3.09 
0.01 
6.63 
2.58 
2.33 
0.02 
5.41 
2.33 
2.05 
0.03 
4.71 
2.17 
1.88 
0.04 
4.22 
2.05 
1.75 
0.05 
3.84 
1.96 
1.64 
0.06 
3.54 
1.88 
1.55 
0.07 
3.28 
1.81 
1.48 
0.08 
3.06 
1.75 
1.41 
0.09 
2.87 
1.70 
1.34 
0.10 
2.71 
1.64 
1.28 
0.11 
2.55 
1.60 
1.23 
0.12 
2.42 
1.55 
1.18 
0.13 
2.29 
1.51 
1.13 
0.14 
2.18 
1.48 
1.08 
0.15 
2.07 
1.44 
1.04 
0.16 
1.97 
1.41 
0.99 
0.17 
1.88 
1.37 
0.95 
0.18 
1.80 
1.34 
0.92 
0.19 
1.72 
1.31 
0.88 
0.20 
1.64 
1.28 
0.84 
0.25 
1.32 
1.15 
0.68 
0.3 
1.07 
1.04 
0.52 
0.35 
0.87 
0.93 
0.39 
0.4 
0.71 
0.84 
0.25 
0.45 
0.57 
0.76 
0.13 
0.5 
0.45 
0.67 
0.0 
0.6 
0.27 
0.52 

0.7 
0.15 
0.39 

0.8 
0.06 
0.25 

0.9 
0.02 
0.13 
Table 15. Table of probabilities (P‐values) and corresponding values of χ^{2} on one degree freedom, standardized normal deviate two‐sided, and standardized normal deviate, one‐sided
The values tabulated 
1. The chi‐square χ^{2} statistic on one degree of freedom. For any other number of degrees of freedom, use Table 16 
The table gives the probability of the given χ^{2} or a larger value under the null hypothesis. Although this is a one tail probability on a χ^{2} distribution, because the χ^{2} distribution is given by a normal deviate squared, this gives a two‐sided test; assessing a variable x ^{2} using a χ^{2} distribution is equivalent to assessing x using a normal distribution and a two‐sided test. 
2. The standardized normal deviate (often called Z or chi) using a two‐sided test. 
The probability given is that of the given value ± Z or a more extreme value of either sign, under the null hypothesis. 
3. The standardized normal deviate (often called Z or chi), using a one‐sided test. 
The probability given is that of the given value Z or a more extreme value of the same sign. Relationships between these 
1. For a given value of P, the corresponding χ^{2} _{(1)} value is equal to the two‐sided standardized normal deviate squared. e.g. for P = 0.05 χ^{2} = 3.84 
2. For a given normal deviate, the two‐sided probability is twice the one‐sided. e.g. for Z = 1.96, two‐sided probability = 0.05, one‐sided = 0.025 
How to use the tables 
1. Be sure you know which statistic you wish to use! 
2. To find the value of a statistic corresponding to a given P‐value: look down the table to find the P‐value or the nearest to it; read off the corresponding statistic. Examples: What χ^{2} _{(1)} value corresponds to P = 0.05? Result: 3.84 What one‐sided deviate corresponds to P = 0.01? Result: 2.33 
3. To find the P‐value corresponding to a given statistic: find the correct column; look down the table to find the given value or the nearst to it; read off the corresponding P‐value. Examples: What P‐value corresponds to χ^{2} _{(1)} = 7.4? Result between 0.001 and 0.01; can be written 0.001 < P < 0.01 What P‐value corresponds to a one‐sided test giving a deviate of 1.04? Result: 0.15 
Table 16. Functions in Microsoft Excel spreadsheets to convert between test statistics and probability (P) values
To get from: 
to: 
function 
Example 


Z statistic 
P value, 2 sided 
=2*(1‐normsdist(z)) 
Z stat Z stat 
1.96 3 
P value P value 
0.049996 0.002700 

Z stat 
4.89 
P value 
0.000001 

Z statistic 
P value, 1 sided 
=(1‐normsdist(z)) 
Z stat Z stat 
1.96 1.64 
P value P value 
0.024998 0.050503 

Z stat 
4.75 
P value 
0.000001 

P value, 2 sided 
Z statistic 
=abs(normsinv(P/2)) 
P value P value 
0.05 0.001 
Z stat Z stat 
1.96 3.29 

P value 
0.1 
Z stat 
1.64 

P value, 1 sided 
Z statistic 
=abs(normsinv(P)) 
P value P value 
0.05 0.1 
Z stat Z stat 
1.64 1.28 

P value 
0.01 
Z stat 
2.33 

Chi squared stat, df 
P value 
=chidist(chi, df) 
Chi stat, df Chi stat, df Chi stat, df 
3.84 10.3 38.932 
1 3 21 
P value P value P value 
0.050044 0.016181 0.010000 

P value, df 
Chi sq 
=chiinv(P, df) 
P value, df P value, df P value, df 
0.01 0.2 0.05 
1 1 30 
Chi sq Chi sq Chi sq 
6.63 1.64 43.77 

t test result, df 
P value 
=tdist(test, df, tails) 
T test, df, tails T test, df, tails T test, df, tails 
4 4 1.96 
1 1 160 
1 2 2 
P value P value P value 
0.077979 0.155958 0.051733 
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