Question

A consultant for an association of personnel directors wants to find the proportion of clerical personnel that change jobs because they are bored with their work. The consultant queries a random sample of 400 clerical workers who recently changed jobs, 200 of which state that they changed jobs because they were bored. The consultant wishes to prepare a 95% confidence interval for the true proportion changing jobs because of boredom. What is the margin of error for this interval

Answer 1

Answer:

the margin of error for this interval is 0.049

Step-by-step explanation:

The computation of the margin of error for this interval is shown below;

Given that

n = 400

x = 200

Now

p = x ÷ n = 200 ÷ 400 = 0.5

Now the margin of error is

= $z \times \sqrt{\frac{p(1-p)}{n}} \\\\1.96 \times \sqrt{\frac{0.5 \times 0.5}{400} }$

= 0.049

Hence the margin of error for this interval is 0.049