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. Please show written work for question a to c below. a) Find critical value(s). b) "Find the Margin of Error" c) Find confidence interval for the problem.

Answer 1

Answer:

a) z = 2.327

b) The margin of error is of 0.065.

c) The 98% confidence interval for the true population proportion of all New York State union members who favor the Republican candidate is (0.3083, 0.4383).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$, and a confidence level of $1-\alpha$, we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$.

For this problem, we have that:

$n = 300, \pi = \frac{112}{300} = 0.3733$

a) 98% confidence level

So $\alpha = 0.02$, z is the value of Z that has a pvalue of $1 - \frac{0.02}{2} = 0.99$, so $z = 2.327$ is the critical value.

b)

The margin of error is:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

So, applying to this question:

$M = 2.327\sqrt{\frac{0.3733(1-0.3733)}{300}} = 0.065$

The margin of error is of 0.065.

c) Find confidence interval for the problem.

$\pi - M = 0.3733 - 0.065 = 0.3083$

$\pi + M = 0.3733 + 0.065 = 0.4383$

The 98% confidence interval for the true population proportion of all New York State union members who favor the Republican candidate is (0.3083, 0.4383).