Question

Each year, more than 2 million people in the United States become infected with bacteria that are resistant to antibiotics. In particular, the Centers of Disease Control and Prevention have launched studies of drug-resistant gonorrhea.† Suppose that, of 189 cases tested in a certain state, 12 were found to be drug-resistant. Suppose also that, of 429 cases tested in another state, 8 were found to be drug-resistant. Do these data suggest a statistically significant difference between the proportions of drug-resistant cases in the two states? Use a 0.02 level of significance. (Let p1 = the population proportion of drug-resistant cases in the first state, and let p2 = the population proportion of drug resistant cases in the second state.) State the null and alternative hypotheses. (Enter != for ≠ as needed.)
H0:
Ha:
Find the value of the test statistic. (Round your answer to two decimal places.)
What is the p-value? (Round your answer to four decimal places.)
p-value =
What is your conclusion?
Reject H0. There is a significant difference in drug resistance between the two states.Do not reject H0. There is a significant difference in drug resistance between the two states. Reject H0. There is not a significant difference in drug resistance between the two states.Do not reject H0. There is not a significant difference in drug resistance between the two states.

Answer 1

Answer:

Reject H₀. There is a significant difference in drug resistance between the two states.

Step-by-step explanation:

In this case we need to determine whether the data suggest a statistically significant difference between the proportions of drug-resistant cases in the two states.

The significance level of the test is, α = 0.02.

(1)

The hypothesis can be defined as follows:  

H₀: There is no difference between the proportions of drug-resistant cases in the two states, i.e. $p_{1} - p_{2}= 0$.  

Hₐ: There is a statistically significant difference between the proportions of drug-resistant cases in the two states, i.e. $p_{1} - p_{2}\neq 0$.  

(2)

Compute the sample proportions and total proportion as follows:

$\hat p_{1}=\frac{12}{189}=0.063\\\\\hat p_{2}=\frac{8}{429}=0.019\\\\\hat p=\frac{12+8}{189+429}=0.032$

Compute the test statistic value as follows:

$Z=\frac{\hat p_{1}-\hat p_{2}}{\sqrt{\hat p(1-\hat p)\times [\frac{1}{n_{1}}+\frac{1}{n_{2}}]}}$

   $=\frac{0.063-0.019}{\sqrt{0.032(1-0.032)\times [\frac{1}{189}+\frac{1}{429}]}}\\\\=2.86$

The test statistic value is 2.86.

(3)

The decision rule is:

The null hypothesis will be rejected if the p-value of the test is less than the significance level.

Compute the p-value as follows:

$p-value=2\cdot P(Z>2.86)=2\times 0.00212=0.00424$

p-value = 0.00424 < α = 0.02.

The null hypothesis will be rejected at 0.02 significance level.

Reject H₀. There is a significant difference in drug resistance between the two states.