STATISTICAL INFERENCE

The procedure used for this example was:

1. Take as the null hypothesis that taking vitamin C and having colds are statistically independent events
   H₀ : vitamin C and colds are independent

   Estimate the probability p_v of taking vitamin C (and hence q_v = 1 - p_v) and the probability p_c of having no colds (and hence q_c = 1 - p_c).

   Use H₀ to calculate probabilities for each cell in the table, e.g. for vitamin C and no cold, prob = p_v × p_c because independent

   = × = 0.28

   Calculate expected frequencies by multiplying these probabilities by total frequency n = 100,

e.g. for vitamin C and no cold, e = 100 × 0.28 = 28

For this example p_v and p_c were estimated from the data (but q_v and q_c were calculated from these) so

degrees of freedom = # categories - # parameters estimated - 1

= 4 - 2 - 1 = 1.

Hence the p-value is P( c² ₁ > 9.72).

From the first row of Table 3 P( c²₁ > 9.72) is between 0.0025 and 0.001 so the p-value is less than 0.0025 (p<.0025).

Observed frequencies are significantly different from expected frequencies, if the expected frequencies are based on the hypothesis that vitamin C and colds are independent. So we reject this hypothesis and conclude there is some association.

General case:

Estimate probabilities for A₁, .... , A_r by , ... ,

(this involves estimating (r-1) independent parameters since their total must be 1 because they are probabilities).

Similarly estimate probabilities for B₁ ,..., B_c by

(this involves (c-1) independent estimates).

Null hypothesis H0: A and B are independent Alternative hypothesis H1 : A and B are not independent

If H₀ is true, probability for cell A_i B_j is

p_ij = for i = 1, .... , r ; j = 1, .... , c

So the expected frequency for cell A_i B_j is n × p_ij , i.e.

e_ij = n × =

Calculate all the expected frequencies using

Then calculate

Degrees of freedom = # cells - # estimates - 1

= (r×c) - [ (r-1) + (c-1) ] - 1

= rc - r - c + 1 = (r - 1)(c - 1).

Example - An homogeneous group of people

Suppose the results were

Vitamin C (amount in gms)

Colds	0 g < 1	1 g < 2	g 2	Total
none	64	62	24	150
at least 1	36	6	8	50
	100	68	32	200

Do these data suggest that vitamin C protects against colds?

Null hypothesis H₀: colds and vitamin C are independent.

Assuming H₀ is correct, the expected frequencies are calculated as follows: e.g. for no colds and 0 g < 1.

e₁₁ = = 75

and similarly for all other cells.

Expected frequencies are often shown in the table (e.g. in brackets)

Vitamin C (amount in gms)

Colds	0 g < 1	1 g < 2	g 2	Total
none	64 (75)	62 (51)	24 (24)	150
1	36 (25)	6 (17)	8 (8)	50
	100	68	32	200

Notice the row and column totals are the same for observed and expected frequencies (this is a good check on your arithmetic)

degrees of freedom = (3 - 1) × (2 - 1) = 2
p-value = P( c² > 15.9)

From Table 3, second row, this is less than 0.0005. Since the p-value is small, you would reject the null hypothesis of independence and conclude there is some association between amount of vitamin C and incidence of colds.

Notice, however, that the pattern of differences between observed and expected frequencies is not consistent with a dose-response relationship.

MINITAB commands for chi-squared analysis

where frequencies are stored in columns C1, C2 and C3.

MTB > chis c1 c2 c3

Expected counts are printed below observed counts

  0<=g<1 1<=g<2  g>=2 Total
 1  64  62  24  150
   75.00 51.00 24.00

 2  36  6  8  50
   25.00 17.00  8.00

Total  100  68  32  200

ChiSq = 1.613 + 2.373 + 0.000 +
   4.840 + 7.118 + 0.000 = 15.944
df = 2

If the data are stored as category codes you can create frequencies in a cross-classified table using, e.g.

TABLE C1 BY C2

To do a chi-squared analysis for this table use the subcommand

This says print observed and expected frequencies for each cell as well as doing the chi-squared calculation.