Gastric freezing was once a recommended treatment for ulcers in the upper intestine.

Given:
$\begin{array}{cccc}& \text{ Gastric freezing }& \text{Placebo}& \text{ Total }\\ \text{ Improved }& 28& 30& 58\\ \text{ Didn't improve}& 54& 48& 102\\ \text{ Total }& 82& 78& 160\end{array}$
$\alpha =\text{Significance level}=0.05\text{assumption}$ The null hypothesis states that there is no association between the variables. The alternative hypothesis states that there is an association between the variables.
$H_0$ : There is no association between treatment and outcome. $H_a$ : There is an association between treatment and outcome.
The expected frequencies E are the product of the column and row total,divided by the table total.
$E_{11}=\frac{r_1\times c_1}{n}=\frac{58\times 82}{160}\approx 29.73$
$E_{12}=\frac{r_1\times c_2}{n}=\frac{58\times 78}{160}\approx 28.27$
$E_{21}=\frac{r_2\times c_1}{n}=\frac{102\times 82}{160}\approx 52.27$
$E_{22}=\frac{r_2\times c_2}{n}=\frac{102\times 78}{160}\approx 49.73$
Conditions
The conditions for performing a chi-square test of homogeneity /independence are: Random, Independent (10%), Large counts. Random: Satisfled, because the sub jects were randomly assigned to a treatment... Independent: Satisfied, because the 160 subjects are less than 10% of all people (since there are more than 1600 people). Large counts: Satisfied, because all expected counts are at least 5. Since all conditions are satisfied, it is appropriate to carry out test of homogeneity /independence.
Hypothesis test
The chi-square subtotals are the squared differences between the observed and expected frequencies, divided by the expected frequency.
The value of the test-statistic is then the sum of the chi-square subtotals:
$x^2=\sum \frac{(O-E{)}^2}{E}$
$=\frac{(28-29.73{)}^2}{29.73}+\frac{(30-28.27{)}^2}{28.27}+\frac{(54-52.27{)}^2}{52.27}+\frac{(48-49.73{)}^2}{49.73}$
$\approx 0.322$
The P-value is the probability of obtaining the value of the test statistic, or a value more extreme. The P-value is the number (or interval) in the column title of the chi-square distribution table in the appendix containing the $x^2\text{-value in the row }df=(r-1)(c-1)=(2-1)(2-1)=1$ :
$P>0.25$
Command $TI83/84\text{-calculator}:x^{2}cdf(0.322,1E99,1)$ which results in a P-value of 0.570. Note: You can replace 1E99 by any other very large number.
If the P-value is less than or equal to the significance level, then the null hypothesis is rejected:
$P<0.05\Rightarrow Reject{H}_0$
There is convincing evidence of an association between treatment and outcome for subjects like these.