A journal article reports that a sample of size was used as a basis for calculating a CI for the true average natural frequency (Hz) of delaminated beams of a certain type. The resulting interval was . You decide that a confidence level of is more appropriate than the level used. What are the limits of the interval

The limit of the (228.533, 234.735)

Step-by-step explanation:

The values are missing in the given question. Therefore, in order to attempt this question, we will make assumptions.

So, let's assume that:

sample size of the journal article that was reported = 5

which is applied for determining a 95% CI

Also, assuming that the resulting interval = (229.764, 233.504)

However, if we are to use a 99% CI which we deemed to be more appropriate;

Then, our objective is to find the limits of this particular interval in question.

To do that:

We need to first find the sample mean at 95% CI by using the formula:

Since; df = n - 1

df = 5 - 1

df = 4

Then;

Sample mean = 231.634

Using the same formula to determine the standard deviation, we have:

At t = 0.025 and df = 4; = 2.776

s = 1.5063

The 99% CI is:

At t =0.005 and df =4; = 4.604

answer: e 9, 5.75

step-by-step explanation:

step 1

find the variation ab in the x coordinate

delta x=(x2-x1)=10-> 8 units

find the variation ab  in the y coordinate 

delta y=(y2-y1)=6-> 2 units

step 2

find the coordinates point c

  x coordinate 

cx=ax+delta x*2/> cx=2+8*(2/> cx=3

  y coordinate 

cy=ay+delta y*2/> cy=4+2*(2/> cy=4.25

the coordinates of point c are (3,4.25)

step 3

find the coordinates point d

  x  coordinate 

dx=ax+delta x*4/> dx=2+8*(4/> dx=4

  y  coordinate 

dy=ay+delta y*4/> dy=4+2*(4/> dy=4.5

the coordinates of point d are (4, 4.5)

step 4

find the coordinates point e

  x  coordinate 

ex=ax+delta x*4/> ex=2+8*(6/> ex=5

  y  coordinate 

ey=ay+delta y*4/> ey=4+2*(6/> ey=4.75

the coordinates of point e are (5, 4.75)

step 5

find the coordinates point f

  x  coordinate 

fx=ax+delta x*4/> fx=2+8*(8/> fx=6

  y  coordinate 

fy=ay+delta y*4/> fy=4+2*(8/> fy=5

the coordinates of point f are (6,5)

step 6

find the coordinates point g

  x  coordinate 

gx=ax+delta x*4/> gx=2+8*(10/> gx=7

  y  coordinate 

gy=ay+delta y*4/> gy=4+2*(10/> gy=5.25

the coordinates of point g are (7,5.25)

step 7

find the coordinates point h

  x  coordinate 

hx=ax+delta x*4/> hx=2+8*(12/> hx=8

  y  coordinate 

hy=ay+delta y*4/> hy=4+2*(12/> hy=5.5

the coordinates of point h are (8,5.5)

step 8

find the coordinates point i

  x  coordinate 

ix=ax+delta x*4/> ix=2+8*(14/> ix=9

  y  coordinate 

iy=ay+delta y*4/> iy=4+2*(14/> iy=5.75

the coordinates of point i are (9,5.75)

- r3ktforgood ☕