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

Gastric freezing was once a recommended treatment for ulcers in the upper intestine. One experiment compared 82 subjects randomly assigned to gastric freezing and 78 subjects randomly assigned to receive a placebo. The two-way table shows the results of the experiment. Is there convincing evidence of an association between treatment and outcome for subjects like these?

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Given:

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.
${E}_{11}=\frac{{r}_{1}×{c}_{1}}{n}=\frac{58×82}{160}\approx 29.73$
${E}_{12}=\frac{{r}_{1}×{c}_{2}}{n}=\frac{58×78}{160}\approx 28.27$
${E}_{21}=\frac{{r}_{2}×{c}_{1}}{n}=\frac{102×82}{160}\approx 52.27$
${E}_{22}=\frac{{r}_{2}×{c}_{2}}{n}=\frac{102×78}{160}\approx 49.73$
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:
${x}^{2}=\sum \frac{\left(O-E{\right)}^{2}}{E}$
$=\frac{\left(28-29.73{\right)}^{2}}{29.73}+\frac{\left(30-28.27{\right)}^{2}}{28.27}+\frac{\left(54-52.27{\right)}^{2}}{52.27}+\frac{\left(48-49.73{\right)}^{2}}{49.73}$
$\approx 0.322$
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 :
$P>0.25$
Command $TI83/84\text{-calculator}:{x}^{2}cdf\left(0.322,1E99,1\right)$ 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:
$P<0.05⇒Reject{H}_{0}$
There is convincing evidence of an association between treatment and outcome for subjects like these.