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 174 cases tested in a certain state, 11 were found to be drug-resistant. Suppose also that, of 375 cases tested in another state, 7 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).
A. State the null and alternative hypotheses.
B. Find the value of the test statistic.
C. What is the p-value?
D. What is your conclusion?
1. Reject H0. There is a significant difference in drug resistance between the two states.
2. Do not reject H0. There is a significant difference in drug resistance between the two states.
3. Reject H0. There is not a significant difference in drug resistance between the two states.
4. Do not reject H0. There is not a significant difference in drug resistance between the two states.

Answer 1

Answer:

A)

Null hypothesis:H₀:- There is no significant difference between in drug resistance between the two states

Alternative Hypothesis :H₁:

There is  significant difference between in drug resistance between the two states

B)

The calculated value Z =  2.7261 > 2.054 at 0.02 level of significance

Rejected H₀

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

C)

P - value = 0.0066

P - value = 0.0066 < 0.02

D)

1) Reject H₀  

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

Step-by-step explanation:

Step(i):-

Given first sample size n₁ = 174

Suppose that, of 174 cases tested in a certain state, 11 were found to be drug-resistant.

First sample proportion

                    $p_{1} = \frac{x_{1} }{n_{1} } = \frac{11}{174} = 0.0632$

Given second sample size n₂ = 375

Given data  Suppose also that, of 375 cases tested in another state, 7 were found to be drug-resistant

Second sample proportion

                 $p_{2} = \frac{x_{2} }{n_{2} } = \frac{7}{375} = 0.0186$

Step(ii):-

Null hypothesis:H₀:- There is no significant difference between in drug resistance between the two states

Alternative Hypothesis :H₁:

There is  significant difference between in drug resistance between the two states

Test statistic

          $Z = \frac{p_{1}-p_{2} }{\sqrt{PQ(\frac{1}{n_{1} }+\frac{1}{n_{2} } ) } }$

       Where

        $P = \frac{n_{1}p_{1} +n_{2} p_{2} }{n_{1} +n_{2} }$

       $P = \frac{174 (0.0632) + 375 (0.0186) }{174+375 } = \frac{17.9718}{549} = 0.0327$

      Q = 1 - P = 1 - 0.0327 = 0.9673

Step(iii):-

 Test statistic

          $Z = \frac{p_{1}-p_{2} }{\sqrt{PQ(\frac{1}{n_{1} }+\frac{1}{n_{2} } ) } }$

         $Z = \frac{0.0632-0.0186 }{\sqrt{0.0327 X 0.9673(\frac{1}{174 }+\frac{1}{375 } ) } }$

      Z  =   2.7261

 

Level of significance = 0.02 or 0.98

The z-value = 2.054

The calculated value Z =  2.7261 > 2.054 at 0.02 level of significance

   Reject H₀  

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

 P- value

P( Z > 2.7261) = 1 - P( Z < 2.726)

                        = 1 - ( 0.5 + A (2.72))

                        = 0.5 - 0.4967

                         = 0.0033

we will use two tailed test

2 P( Z > 2.7261)  = 2 × 0.0033

                           = 0.0066

P - value = 0.0066 < 0.02

 Reject H₀  

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