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