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

Khadija Wells 2021-01-30 Answered

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?
 Gastric freezing Placebo Total  Improved 283058 Didn't improve5448102 Total 8278160

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Expert Answer

okomgcae
Answered 2021-01-31 Author has 93 answers

Given:
 Gastric freezing Placebo Total  Improved 283058 Didn't improve5448102 Total 8278160
α=Significance level=0.05 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.
H0 : There is no association between treatment and outcome. Ha : 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.
E11=r1×c1n=58×8216029.73
E12=r1×c2n=58×7816028.27
E21=r2×c1n=102×8216052.27
E22=r2×c2n=102×7816049.73
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:
x2=(OE)2E
=(2829.73)229.73+(3028.27)228.27+(5452.27)252.27+(4849.73)249.73
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 x2-value in the row df=(r1)(c1)=(21)(21)=1 :
P>0.25
Command TI83/84-calculator:x2cdf(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<0.05RejectH0
There is convincing evidence of an association between treatment and outcome for subjects like these.

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