Question

A survey of 300 union members in New York State reveals that 112 favor the Republican candidate for governor. Construct the 90% confidence interval for the true population proportion of all New York State union members who favor the Republican candidate. Group of answer choices 0.308 < p < 0.438 0.301 < p < 0.445 0.316 < p < 0.430 0.327 < p < 0.419

Answer 1

Complete Question

A survey of 300 union members in New York State reveals that 112 favor the Republican candidate for governor. Construct the 98% confidence interval for the true population proportion of all New York State union members who favor the Republican candidate. Group of answer choices

A 0.308 < p < 0.438

B 0.301 < p < 0.445

C  0.316 < p < 0.430

D 0.327 < p < 0.419

Answer:

The correction option is A

Step-by-step explanation:

From the question we are told that

   The sample size is  n  =  300

     Th number that are in favor is k = 112

   Generally the sample proportion is mathematically represented as  

           $\^ p = \frac{k}{n}$

=>        $\^ p = \frac{112}{300}$

=>        $\^ p = 0.3733$  

From the question we are told the confidence level is  98% , hence the level of significance is    

      $\alpha = (100 - 98 ) \%$

=>   $\alpha = 0.02$

Generally from the normal distribution table the critical value  of  $\frac{\alpha }{2}$ is  

   $Z_{\frac{\alpha }{2} } = 2.33$

Generally the margin of error is mathematically represented as  

     $E = Z_{\frac{\alpha }{2} } * \sqrt{\frac{\^ p (1- \^ p)}{n} }$

=>   $E = 2.33 * \sqrt{\frac{0.3733 (1- 0.3733)}{300} }$

=>   $E = 0.06508$

Generally 95% confidence interval is mathematically represented as  

      $\^ p -E < p < \^ p +E$

=>   $0.3733 -0.06508 < p < 0.3733 + 0.06508$

=>   $0.308 < p < 0.4038$