Calculate the p- value

Question

Medical tests were conducted to learn about drug-resistant tuberculosis. Of 284 cases tested in New Jersey, 18 were found to be drug- resistant. Of 536 cases tested in Texas, 10 were found to be drugresistant. Do these data indicate that New Jersey has a statistically significant higher outbreak of drugresistant tuberculosis cases? Use a .03 level of significance. What is the p- value, and what is your conclusion? Is the conclusion any different under critical-value approach?

Answer 1

Data

Let P1' denote observed proportion of drug resistant TB in New Jersey population and P2' is observed proportion of drug resistant TB in Texas, then

P1'= 18/284   = 0.0633803

P2' = 10/536  = 0.0186567

Hypothesis Formation

Null Hypothesis H₀:    P1 - P2 = 0

Alternative Hypothesis H₁:    P1 - P2 > 0

Z Statistic

Z = (P1' - P2')/SQRT(P(1-P)/(1/n1+1/n2))

Where P = (18+10)/(284+536)

                 = 0.0341463

Critical Region

Reject null hypothesis in favor of alternative if Z is greater than Z critical value of 1.88

Computation

Z = (0.0633803 - 0.0186567)/SQRT(0.0341463*(1-0. 0.0341463)(1/284+1/536))

   = 0.0447236/SQRT(0.0329803*0.0053868)

   = 0.0447236/SQRT(0.0001777)

   = 0.0447236/0.01333

   = 3.36

Decision

Null hypothesis is rejected in favor of alternative as Z value is greater than Z critical value. So we can say that New Jersey has statistically greater outbreak of drug resistant TB.