Question

I'm a survey of 200 employees, 65% agreed that the management of the cafeteria should be changed. If the confidence interval is 95%, the margin of error for this survey is %, and the maximu

Answer 1

Answer:

Step-by-step explanation:

Confidence interval for population proportion is expressed as

Sample proportion ± margin of error

Margin of error = z × √pq/n

Where

z represents the z score corresponding to the confidence level

p = sample proportion. It also means probability of success

q = probability of failure

q = 1 - p

From the information given,

Sample proportion, p = 65/100 = 0.65

q = 1 - 0.65 = 0.35

n = 200

To determine the z score, we subtract the confidence level from 100% to get α

α = 1 - 0.95 = 0.05

α/2 = 0.05/2 = 0.025

This is the area in each tail. Since we want the area in the middle, it becomes

1 - 0.025 = 0.975

The z score corresponding to the area on the z table is 1.96. Thus, the z score for a confidence level of 95% is 1.96

Margin of error = 1.96√(0.65)(0.35)/200

Margin of error = 0.066 = 6.6%

The maximum boundary of the interval is

0.65 + 0.066 = 0.716