STATISTICAL INFERENCE

This is statistically significant (because the p-value is much less than 0.05) so we reject H₀ and conclude the sex ratio differed from 1:1.

Notice that the observed proportion of boys, 1/7, was the same for both the above examples. The p-value was affected by the total number, n, of babies or equivalently, by the expected numbers of boys and girls, np and n(1-p).

Example - 50 coin tosses

Let X = number of heads. Assume X ~ Binomial (50,½).

In this case, the expected number of heads, assuming the coin is balanced, is 50 × ½ = 25. The p-value is the probability of getting the observed value of 20 or a more extreme value ( 20 heads) in either direction (ie. 30 heads)

p-value = P(X 20 or X 30) which is tedious to calculate directly.

Instead we can use the Normal approximation to a binomial distribution, since n is large

i.e. if X ~ Binomial (n,p) then approximately X ~ Normal (np,npq)

For n = 50 and p = ½, np = 25, npq = 12.5 and Z =

p-value = P(X 20 or X 30) and use the continuity correction

= P(Z or Z )

P(Z - 1.273 or Z 1.273)

= 2 × (1 - 0.898)

0.20

As this p-value is not statistically significant we conclude that the data are consistent with the hypothesis that P(head) = ½ i.e. that the coin may be balanced.

Testing more than 2 categories

c is the Greek letter chi (pronounced ki).

Remarks

1. This takes into consideration differences between the observed frequencies for the categories and the expected frequencies if the null hypothesis were true.
2. However it ignores the direction of the differences (because they are squared) so it corresponds to a two-tailed test.
3. It adjusts for the sizes of the expected values - as suggested by the first two examples in this section.
4. In the two category case this method is identical to the normal approximation to the binomial distribution:
   

   Hence, when there are just two categories and the total frequency is constrained to sum to a fixed number n, (so that there is only one r.v., X), then c² = Z².

P-values

To calculate p-values we need the probability distribution for c²

This is the chi-squared distribution with 1 degree of freedom - written c²₁ - see Table 3.

e.g. P(Z < - 1.96 or Z > 1.96) = 0.05

This is equivalent to

P(Z² > 1.96²)

= P(Z² > 3.84)

= P ( c²₁ > 3.84)

= 0.05 from Table 3 (first row)

More generally consider a frequency table for k categories

categories	:	A1	A2	...	Ak
observed frequencies	:	o1	o2	...	ok
expected frequencies	:	e1	e2	...	ek

Let e₁, e₂,...,e_k be expected frequencies predicted by some (null) hypothesis or model.

To see how well the data (observed frequencies) correspond to the hypothesis (and hence the expected frequencies) calculate

If the data correspond closely to the hypothesis then (o_i- e_i) will be small for every category, so c² will be small.

If the data are not consistent with the hypothesis some or all of the (o_i-e_i) values will be large, so c² will be large.

To determine how large c² could be (just by chance) if the hypothesis were true, we need to know its sampling distribution. This is the chi-squared distribution with (k-1) degrees of freedom. So for k = 3 categories, use 3 - 1 = 2 degrees of freedom.

Table 3 gives probabilities of observing a value of c² greater than various tabulated values. Its use is illustrated in the following example.

Example - 60 throw fair die tester

Null Hypothesis: die is fair so the probability for each value is .

If this is true, for n = 60 tosses the expected frequency for each x_i is 60 × = 10.

spots xi :	1	2	3	4	5	6	Total
observed freq.oi :	12	7	10	17	5	9	60
expected freq.ei :	10	10	10	10	10	10	60

If the die is fair, c² will be small. If the die is not fair, c² will be large. How do you decide whether c² = 8.8 is large or small?

There are k = 6 categories so use the chi-squared distribution with (k-1) = 5 degrees of freedom.

Large values of c² provide evidence against the hypothesis so "extreme" values are those greater than or equal to the calculated c² value. Hence

From the 5th row of Table 3, the required probability is between 0.1 and 0.15, so p-value > 0.1. Since this p-value is not small, we do not reject H₀.

We conclude the data are consistent with the hypothesis that the die is fair.

Comments

1. Small c² values show that the data are consistent with H₀.

   Large c² values show that the data are inconsistent with H₀ and if the p-value is very small you would reject H₀.

2. The sampling distribution of c² is approximately the chi-squared distribution provided none of the expected values is too small (it is usual to require e_i 5). If the e_i's are too small you can combine adjacent categories until expected frequencies are 5.
3. If there are k categories and the expected frequencies are completely specified by the hypothesis H₀ (i.e. no parameters need to be estimated from the data) then c² has (k - 1) degrees of freedom.
4. If the expected frequencies are not completely specified by H₀ and m parameters have to be estimated from the data then c² has (k - m - 1) degrees of freedom.