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.

  5.   y=6; de=7; fg=8

  6.   x=4.2; jl=14.2; mn=13

step-by-step explanation:

when chords cross inside a circle, the product of the segment lengths is a constant.

5. dh×eh = fh×gh

  y = gh = dh×eh/fh = 3×4/2 . . . divide by the coefficient of gh

  y = 6

of course, the total chord length is the sum of the lengths of its segments.

  de = dh +eh = 3+4 = 7

  fg = fh +gh = 2+6 = 8

6. this problem works the same way as the previous one.

  x = lk = mk×nk/jk = 6×7/10 = 4.2

  jl = jk +lk = 10 +4.2 = 14.2

  mn = mk +nk = 6 +7 = 13

comment on chords and secants

you can think of the points h (problem 5) and k (problem 6) as points where the chords meet. a chord is part of a secant line, a line that intersects the circle in two points. one can also consider these points (h or k) to be the points where the respective secant lines meet.

then, the product that is a constant is the product of the distance from the meeting point (h or k) to one circle intersection with the distance from that meeting point to the other circle intersection. for problem 5, the constant is the product hd×he = hf×hg.

it turns out that this rule regarding the products of lengths to points of intersection is also true when the secants intersect outside the circle.

that is, you only need to remember one rule to work both kinds of problems.

answer: on kudzu, being female and being green are independent events.

step-by-step explanation: