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

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).

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{(\pi(1-\pi))/(n)}`

In which

z is the zscore that has a pvalue of
`1 - (\alpha)/(2)`.

For this problem, we have that:

`n = 300, \pi = (112)/(300) = 0.3733`

a) 98% confidence level

So
`\alpha = 0.02`, z is the value of Z that has a pvalue of
`1 - (0.02)/(2) = 0.99`, so
`z = 2.327` is the critical value.

b)

The margin of error is:

`M = z\sqrt{(\pi(1-\pi))/(n)}`

So, applying to this question:

`M = 2.327\sqrt{(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).