In a sample of 4500 first year college students, a national survey reported that 39% participated in community service or volunteering work. a)Find the margin of error for 99% confidence level. Explain the findings of the study to someone who does not know statistics. [2 pts] b)Suppose a similar study collects a new sample of 4,500 first year students. If a 99% confidence interval is computed from the new sample, will the confidence interval will be exactly the same as the results in a)

Step-by-step explanation:

a)Confidence interval is written 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

p = x/n

Where

n represents the number of samples

x represents the number of success

From the information given,

n = 4500

p = 39/100 = 0.39

q = 1 - 0.39 = 0.61

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

α = 1 - 0.99 = 0.01

α/2 = 0.01/2 = 0.005

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

1 - 0.005 = 0.995

The z score corresponding to the area on the z table is 2.58. Thus, the z score for a confidence level of 99% is 2.58

Margin of error = 2.58√(0.39)(0.61)/4500

Margin of error = 0.0188

The lower limit of the confidence interval is

0.39 - 0.0188 = 0.3712

The upper limit of the confidence interval is

0.39 + 0.054 = 0.4088

Therefore, we are 99% certain or confident that the proportion of the population of first year college students that participated in community service or volunteering work lies between 0.3712 and 0.4088

b) The margin of error would remain the same but the confidence interval would be affected by the value of the sample proportion.

answer: was that a question or answer?

step-by-step explanation:

its c or d   take a lucky guess my dude