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 Â 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 Â
      [tex] \^ p = \frac{k}{n}[/tex]
=> Â Â Â Â [tex] \^ p = \frac{112}{300}[/tex]
=> Â Â Â Â [tex] \^ p = 0.3733 [/tex] Â
From the question we are told the confidence level is  98% , hence the level of significance is  Â
   [tex]\alpha = (100 - 98 ) \%[/tex]
=> Â [tex]\alpha = 0.02[/tex]
Generally from the normal distribution table the critical value  of  [tex]\frac{\alpha }{2}[/tex] is Â
  [tex]Z_{\frac{\alpha }{2} } =  2.33[/tex]
Generally the margin of error is mathematically represented as Â
   [tex]E =  Z_{\frac{\alpha }{2} } * \sqrt{\frac{\^ p (1- \^ p)}{n} } [/tex]
=> Â [tex]E = 2.33 * \sqrt{\frac{0.3733 Â (1- 0.3733)}{300} } [/tex]
=> Â [tex]E = Â 0.06508 Â [/tex]
Generally 95% confidence interval is mathematically represented as Â
   [tex]\^ p -E <  p <  \^ p +E[/tex]
=> Â [tex]0.3733 -0.06508 < Â p < 0.3733 + 0.06508 [/tex]
=> Â [tex]0.308 < Â p < 0.4038 [/tex]
