Question

A clinic administers two drugs to two groups of randomly assigned patients to cure the same disease: 70 patients received Drug 1 and 80 patients received Drug 11 . The following table gives the information about the numbers of patients cured and those not cured by each of these two drugs.

Drag	Cured	Net Cured
I	60	10
II	59	21

For a chi-square test of homogeneity for this contingency table, what are the expected frequencies for the second column, rounded to two decimal places?

Answer 1

The expected values are computed in terms of row and column totals. In fact, the formula is E = RXC, where R; corresponds to the total sum of elements in row i, C, corresponds to the total sum of elements in column j, and T is the grand total. The table below shows the calculations to obtain the table with expected values: Expected Values Column 1 Column 2 Total Row 1 119x70 = 55.533 31x70 = 14.467 70 150 000 Row 2 119680 = 63.467 31X80 = 16.533 150 80 Total 119 31 150 Based on the observed and expected values, the squared distances can be computed according to the following formula: (E-02/E. The table with squared distances is shown below: Squared Distances Column 1 Column 2 Row 1 (60-55.533) 55.533 = 0.359 (10-14.467) 14.467 = 1.379 Row 2 (59-63.467) = 0.314 63.467 (21-16.533) 16.533 = 1.207

The two variables are Drug and status of cure (Crured or not cured)

Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: H: The two variables are independent H: The two variables are dependent This corresponds to a Chi-Square test of independence. Rejection Region Based on the information provided, the significance level is a = 0.05, the number of degrees of freedom is df = (2-1) x (2-1) = 1, so then the rejection region for this test is R= {x:x 3.841} > Test Statistics The Chi-Squared statistic is computed as follows: 72 = - (Oij - Eij) W Pijl Ej = 0.359 +0.314+ 1.379 + 1.207 = 3.259 Decision about the null hypothesis Since it is observed that x2 = 3.259 < x2 = 3.841, it is then concluded that the null hypothesis is not rejected. Conclusion It is concluded that the null hypothesis Ho is not rejected. Therefore, there is NOT enough evidence to claim that the two variables are dependent, at the 0.05 significance level.

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Answer 2

Test Statistic

برج = (0 - 3)

O : Observed Frequency

E: Expected Frequency

Row Total x Column Total Grand Total

Given,

O : Observed frequency

Drug	Cured	Not Cured
I	60	10
II	59	21

E: Expected Count/Frequency

Drug	Cured	Not Cured
I	(70*119) / 150	(70*31) / 150
II	(80*119) / 150	(80*31) / 150

Drug	Cured	Not Cured
I	55.5333	14.4667
II	63.4667	16.5333

O	E	O-E	(O-E)2	(O-E)2/E
60	55.5333	4.4667	19.9511	0.359264
10	14.4667	-4.4667	19.9511	1.379109
59	63.4667	-4.4667	19.9511	0.314356
21	16.5333	4.4667	19.9511	1.20672
			Total	3.259449

O-E) x2 => Z = 3.2594

Value of the ch-square test statistic = 3.259