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Lesechka [4]
3 years ago
14

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.
Mathematics
1 answer:
Luden [163]3 years ago
3 0

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).

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