Question

S. Chi-Squared Test of Independence Tests for the adverse reactions to a new drug vielded the results given in the table. At the 0.05 significance level, test the claim that the treatment is independent of the reaction (headache or no headache experienced). DrugPlacebo 7 Headache No headache 73 91 Hypothesis Ho: На: points Expected Values points Decision points Conclusion points

Answer 1

ANSWER:

Given that,

5)

null hypothesis : Assume that there is no association between adverse reaction and new drug

alternative hypothesis : Assume there is an association between these two variables.

How to calculate the chi-square statistic by hand. First we have to calculate the expected value of the two nominal variables. We can calculate the expected value of the two nominal variables by using this formula:

Where

= expected value

= Sum of the i_th column

= Sum of the k_th row

N = total number

After calculating the expected value, we will apply the following formula to calculate the value of the Chi-Square test of Independence:

= Chi-Square test of Independence
= Observed value of two nominal variables
= Expected value of two nominal variables

Degree of freedom is calculated by using the following formula:
DF = (r-1)(c-1)
Where
DF = Degree of freedom
r = number of rows
c = number of columns

here r = c = 2

Expected values are :

E₁₁ = (84*18) / 182 = 8.308

E₁₂ = (18*98) / 182 = 9.692

E₂₁ = (164*84) / 182 = 75.692

E₂₂ = (164*98) / 182 = 88.308

Decision:

then $\chi$² = 1.798(observed)

tabulated  $\chi$²_{0.05 ; 1} = 3.841

observed  $\chi$² < tabulated  $\chi$​​​​​​​²

​​​​​​​We failed to reject the null hypothesis . According to the given data the two variables are independent.

Conclusion:

it is concluded that null hypothesis Ho is not rejected.Therefore there is NOT enough evidence to claim that two variables are depedent at the 0.05 significance level.