1 answer

S. Chi-Squared Test of Independence Tests for the adverse reactions to a new drug vielded the...

Question:

S. Chi-Squared Test of Independence Tests for the adverse reactions to a new drug vielded the results given in the table. At
S. Chi-Squared Test of Independence Tests for the adverse reactions to a new drug vielded the results given in the table. At
Practice Statistic Final for Spring 2019 B Name: Do 1, 4,5,6,7 Or 2, 3, 5,6,7 6. Perform the indicated Goodness of Fit Test U
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
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
Practice Statistic Final for Spring 2019 B Name: Do 1, 4,5,6,7 Or 2, 3, 5,6,7 6. Perform the indicated Goodness of Fit Test Using the data below and a 0.05 significance level, test the claim that the re with percentages of 15%, 20% 25% 25% and S% respectfully responses occur Fr 12 lis 16 18 I 19 Hypothesis Ho На: Expected Values 5 points Decision Conclusion 5 points ANOVA e four brands have the same mean Ar the 00S significance level test the claim that th 20 20 23 19 28 20 28 Hypothesis Ho На: 5 points. Decision 5 points Conclusion 5 points.

Answers

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 ith column

= Sum of the kth 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 \chi2 = 1.798(observed)

tabulated  \chi20.05 ; 1 = 3.841

observed  \chi2 < 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.

.

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