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A new vaccination is being used in a laboratory experiment to investigate whether it is effective....

A new vaccination is being used in a laboratory experiment to investigate whether it is effective. There are 203 subjects in the study. Is there sufficient evidence to determine if vaccination and disease status are related?

Vaccination Status   Diseased   Not Diseased   Total
Vaccinated 42 37 79
Not Vaccinated 58   66 124
Total 100   103 203

Step 1 of 8: State the null and alternative hypothesis.

Step 2 of 8: Find the expected value for the number of subjects who are vaccinated and are diseased. Round your answer to one decimal place.

Step 3 of 8: Find the expected value for the number of subjects who are not vaccinated and are not diseased. Round your answer to one decimal place.

Step 4 of 8: Find the value of the test statistic. Round your answer to three decimal places.

Step 5 of 8: Find the degrees of freedom associated with the test statistic for this problem.

Step 6 of 8: Find the critical value of the test at the 0.05 level of significance. Round your answer to three decimal places.

Step 7 of 8: Make the decision to reject or fail to reject the null hypothesis at the 0.05 level of significance.
Step 8 of 8: State the conclusion of the hypothesis test at the 0.05 level of significance.

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Answer #1

to test independence between two variables we have to use chi-square independence test as follows

Actual Values:

42 37

58 66

step1: null hypothesis is two variables are independent

alternative hypothesis: two variables are not independent

step 2:- expected value for the number of subjects who are vaccinated and are diseased

100 x 79/203= 38.9163

step3 :- expected value for the number of subjects who are not vaccinated and are not diseased

124*103/203 = 62.9163

Expected Values:

38.9163 40.0837

61.0837 62.9163

Chi-Squared Values:

0.244357 0.23724

0.155679 0.151145

step 4:- test statistic is

Chi-Square = 0.788422

step 5:- degrees of freedom is

Degrees of Freedom = 1

step 6:- Right-tail p-value is 0.05, df = 1

critical value is 3.841

R command: qchisq(0.05, 1, lower.tail=FALSE) or qchisq(1-0.05, 1)

p value is p = 0.374578

step 7) fail to reject the null hypothesis at the 0.05 level of significance.

step8) insufficient evidence to conclude that variables are dependent.

please like ??

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