## Relative Risk

- In a study to judge the effectiveness of inoculation in preventing a virus infection,the following results were obtained :

*A. Estimate the relative risk of infection.*

B. By how much does inoculation reduce the risk of infection?

C. Estimate the odds ratio.**Interpret your estimate.**

D. State the appropriate null and alternative hypotheses for this problem.

E. Is inoculation effective in preventing the virus infection? Justify your answer.

**Ans.**R-software

> epi.2by2(tab1,method = "cohort.count",conf.level = 0.95)

Point estimates and 95 % CIs:

X2 test statistic: 7.669 p-value: 0.006

Wald confidence limits

* Outcomes per 100 population units

A) Relative risk of infection = 0.22

B) Inoculation Reduces Risk of infection by( 1 - 0.22 )= 0.78 = .78 times

C) Odds Ratio = 0.20

The odds of Inoculation getting infected are 0.20 times the odds of a not inoculated getting infected

D) **Null Hypothesis:** Their is no difference between Inoculated and Not Inoculated across Infected and Not infected

**Aternative Hypothesis:** Their is a significant difference between Inoculated and Not Inoculated across Infected and Not infected

E) Yes, Inoculation is effective in preventing the virus infection as the p-value of the chi-square test is 0.006 which is less then 0.05. Hence,We reject null hypothesis an conclude that inoculation is effective.

- After a customer service course, 30customers were asked whether or not they were completely stisfied with their service. A year later same experiment is conducted to see whether or not the effects of the customer service course have waned. the results are :

*A. Have the effects of the course waned ?*

B. State clearly the null and alternative hypotheses tested in part (a).

C. Analyze the residuals and interpret your results.

**Ans.**> tab2

Point estimates and 95 % CIs:

X2 test statistic: 4.32 p-value: 0.038

Wald confidence limits

* Outcomes per 100 population units

A) Yes, The effects of the course waned by 0.21 times. from (1 - ( 1/1.27)) = 0.21

B) **Null Hypothesis:** Their is no difference between 1st sample and 2nd_sample accross Response

**Alternative Hypothesis:** Their is significant difference between 1st sample and 2nd_sample accross Response

C) Since the Computed P-value is 0.038, which is less than 0.05, we reject null hypotheisis and conclude that their is a significant difference between 1st and 2nd sample accross Response.