Mathematical tools for natural sciences

We sometimes sample from two independent populations to estimate the proportion of their favourable outcomes.
Let $n_1$ and $n_2$ be the random samples from two independent distributions. Let $Y_1$ and $Y_2$ be their respective favourable outcomes. From this we compute the proportions \(\small{p_{s1} = \dfrac{Y_1}{n_1} }\) and \(\small{p_{s2} = \dfrac{Y_2}{n_2} }\) of the favourable outcomes. If $p_1$ and $p_2$ are the proportions of favourable outcomes in the two populatons respectively, we can test the equlity of the parent population proportions using a test statistic created out of observed proportions. Here, we distinguish between two cases : one in which the sample sizes $n_1$ and $n_2$ are large, and the second case in which the samples sizes are small.

\(\small{ \dfrac{\dfrac{Y_1}{n_1}- \dfrac{Y_2}{n_2}~-~(p_1-p_2)}{\sqrt{\dfrac{p_1(1-p_1)}{n_1} + \dfrac{p_2(1-p_2)}{n_2} } }~~~~~ }\) approximately will be \(\small{N(0,1)}\).

When we test the null hypothesis that $p_1 = p_2$, the appropriate statistic is given from the above expression by,

\(\small{Z = \dfrac{\dfrac{Y_1}{n_1}- \dfrac{Y_2}{n_2}}{\sqrt{\dfrac{p_1(1-p_1)}{n_1} + \dfrac{p_2(1-p_2)}{n_2} } }~~~~~ }\) approximately will be \(\small{N(0,1)}\).

If the sample sizes \(\small{n_1}\) and \(\small{n_2}\) are large enough, we can replace the probabilities \(\small{p_1}\) and \(\small{p_2}\) by their estimates \(\small{p_{s1}}\) and \(\small{p_{s2}}\) respectively in the denominator of the above expression. The test statistic becomes,

\(\small{Z = \dfrac{p_{s1}-p_{s2}}{\sqrt{\dfrac{p_{s1}(1-p_{s1})}{n_1} + \dfrac{p_{s2}(1-p_{s2})}{n_2} } }~~~~~ }\) approximately will be \(\small{N(0,1)}\).

We proceed with the hypothesis testing as follows:

- Compute the sample fractions \(\small{p_{s1}}\) and \(\small{p_{s2}}\) from the data.
- For the given sample sizes $n_1$ and $n_2$, compute the Z statistics using the above expression.
- For a given significance level of \(\small{\alpha}\), we reject the null \(\small{H_0 : p_1 = p_2 }\) and accept the two sided alternative \(\small{H_A : p_1 \ne p_2 }\) if the computed Z value is outside the range \(\small{(-Z_{1-\alpha/2}, Z_{1-\alpha/2}) }\).

For an one sided alternative \(\small{H_A : p_1 \gt p_2 }\), reject the null if \(\small{Z \gt Z_{1-\alpha} }\). Similarly, for an one sided alternative \(\small{H_A : p_1 \lt p_2 }\), reject the null if \(\small{Z \lt -Z_{1-\alpha} }\)

In order to assess whether the male and female students at the Indian Universities have significantly different opinion on the abolition of death penalty, a survey was carried out across the Indian universities. 720 out of 2100 male students and 440 out of 1200 female students interviewed supported the abolition. Test whether a significance difference exists between the proportion of men and women supporting the abolition.

Let $p_{s1}$ and $p_{s2}$ be the proportion of those supporting the abolition of death penalty among the male and female students interviewed. Let the corresponding proportions in the populations be $p_1$ and $p_2$.

We set up the null and alternate hypothesis for the two sided test as follows:

\(~~~~~~~~~~~\small{H_0 : p_1~=~p_2 } \)

\(~~~~~~~~~~~\small{H_A : p_1~\neq~p_2 } \)

Let us compute the proportion of success (fraction favouring the abolition) in the two samples. We get,

\(~~~~~~~~~\small{p_{s1} = \dfrac{720}{2100} = 0.343}\)

\(~~~~~~~~~\small{p_{s1} = \dfrac{440}{1200} = 0.367}\)

The Z statistic for the test is given by,

\(\small{Z~ = ~ \dfrac{p_{s1}-p_{s2}}{\sqrt{\dfrac{p_{s1}(1-p_{s1})}{n_1} + \dfrac{p_{s2}(1-p_{s2})}{n_2}}}~=~\dfrac{0.343-0.367}{\sqrt{\dfrac{0.343(1-0.343)}{2100} + \dfrac{0.367(1-0.367)}{1200} } }~~=~~ -1.383}\)

Under null hypothesis that $p_1 = p_2$, the Z statistic should follow a unit normal distribution.

We have taken \(\small{\alpha =0.05}\) to be the probability for rejecting the null hypothesis. The critical value is \(\small{-Z_{1-\frac{\alpha}{2}} = -Z_{0.975} = -1.96 }\). Since this is a two sided test, the null hypothesis will be rejected by Z values less than \(\small{-1.96 }\)

This rejection region is indicated as shaded portion in the figure below:

**Since the computed value Z = -1.383 is in the acceptance region, we accept the null hypothesis to conclude that there is no significnce difference between the proportion of male and female students supporting the abolition of death penalty. The statistical significance of this test is 0.05.**

For the computed Z value of -1.383, the p-value is obtained by computing the area under the unit normal curve to the right of this Zvalue. From the Gussian table or using the R command

"pnorm(-1.383)", we get this value as 0.0833.

We rename the observation as,

\(\small{a = x_1,~~b = x_2,~~c = n_1-x_1,~~d = n_2 - x_2 ~~ }\) to write a $2 \times 2$ contingency table:

\( \begin{array} {|r|r|}\hline & X & Y & Row~sum \\ \hline Favourable & a & b & a+b \\ \hline Not~favourable & c & d & c+d \\ \hline column~sum& a+c& b+d & a+b+c+d = n \\ \hline \end{array} \)

For a given set of values for a,b,c and d, the probability of getting this table follows a hypergeometric distribution under the null hypothesis that the two distributions X and Y have equal fraction of favourable events:

\( \small{P_h(a,b,c,d,n) = \dfrac{\displaystyle \binom{a+b}{a} \binom{c+d}{c}}{\displaystyle \binom{n}{a+c}} = \dfrac{\displaystyle \binom{a+b}{b} \binom{c+d}{d}}{\displaystyle \binom{n}{b+d}} = \dfrac{(a+b)!(c+d)!(a+c)!(b+d)!}{a!b!c!d!n!} } \)

This expression gives th exact hypergeometric probability of observing this contingency table with particular valus of a,b,c,d under the null hypothesis that the two distributions contain equal fraction of favourable events. Since this probability is exactly computed without making approximations to larger sizes, it is called an

For computing the significance of observed data under null hypothesis, we compute the probability P(a,b,c,d,n) for various possible vlues of a,b,c and d while keeping the marginal totals (ie, row and column sums) sme as the observed table. We then sum all these probabilities to see whether this total probability is less than the threshold value. This gives a one tailes test. For a two sided test, tables are contructed with extreme values in the opposite direction.

A very detailed description of this methdology can be found in the wikipedia page here.

In R, the function

The function is defined as,

prop.test(x, n, p= NULL, alternative = "two.sided", correct = TRUE) x = data vector of counts of successes.n = data vector containing number of trials in each case for which successes in x are observed.p = a vector of probabilities of successes.alternative = string value specifying alternate hypothesis. Possible values are "two.sided", "less" or "more"correct = A boolean value indicaing whether Yates continuity corretion to be applied. It is TRUE by default. The function returns a data structure with various results.

The Fisher's exact test can be perfored using the R library function

fisher.test(x, alternative = "two.sided", conf.int = TRUE, conf.level = 0.95) where,x is a 2 by 2 matrix contaning the values of the contingency table of events under categories. All other parameters are as defined before. This function can also perform test for contingency tables of dimension more than 2 by 2. Many more parameters are there. See "help(fisher.test") in R for more details.

Use of these two functions are demonstrated in the Rscript below:

################################################## ## Testing two population proportions with prop.test ## 1. Using prop.test() functon ## In a health survey, 520 out of 600 men and 550 out of 600 women questioned said they use antibiotics whenever fever ## continues for more than 2 days. We want to test whether there is a significant difference in the fraction of men and women ## who start taking antibotics after 2 days of fever. x = c(520, 550) ## favourble response in each category of data n = c(600, 600) ## total samples in each category. res = prop.test(x,n,alternative = "two.sided") print(res) #### Perform Fishers exact test for a contingency table. ## We have the following contingency table for patients belonging to two economic strata admitted for throat cancer in ## a hosital. They say whether they indulge in tobacco abuse or not: ## Higher-income Lower-income ## ## Tobacco abuse 11 17 ## ## No abuse 42 39 ## ##We do Fishers exact test to find whether the patients from higher income group indulge in tobacco abuse ## in a significantly different proportion than the patients from lower income group x = matrix(c(11,42,17,39), ncol=2) res1 = fisher.test(x, alternative = "two.sided", conf.int = TRUE, conf.level = 0.95) print(res1) ###############------------------------------------------------

Executing the above script in R prints the following results and figures of probability distribution on the screen:

2-sample test for equality of proportions with continuity correction data: x out of n X-squared = 7.2552, df = 1, p-value = 0.00707 alternative hypothesis: two.sided 95 percent confidence interval: -0.0867225 -0.0132775 sample estimates: prop 1 prop 2 0.8666667 0.9166667 Fisher's Exact Test for Count Data data: x p-value = 0.2797 alternative hypothesis: true odds ratio is not equal to 1 95 percent confidence interval: 0.2250716 1.5641267 sample estimates: odds ratio 0.6036637