Question

If the margin of error in an estimate for the mean weight of a shipment is + or -2 pounds at a confidence level of 95 percent, what will be the margin of error at a confidence level of 98 percent? Be sure to show how you arrived at your answer.

Answer 1

Recall that you'll use this table to find the critical z values

https://www.ltcconline.net/greenl/courses/201/estimation/smallConfLevelTable.htm

Answer 2

The margin of error M is equal to

\[\Large M = z*\frac{s}{\sqrt{n}}\]

The part \(\Large \frac{s}{\sqrt{n}}\) will stay the same. Only the value of z changes as the confidence level changes

Answer 3

i see that 1.96 is the confidence interval at 95%, so do you just put what is the interval at 98?

Answer 4

what is n?

Answer 5

at 95% confidence, z = 1.96

s and n are unknown but that doesn't matter. All that matters is that they are fixed and don't change

Answer 6

at 95% confidence, z = 1.96 and M = 2, so
\[\Large M = z*\frac{s}{\sqrt{n}}\]
\[\Large 2 = 1.96*\frac{s}{\sqrt{n}}\]
agreed so far?

Answer 7

yes

Answer 8

so then you replace it with 2.33?

Answer 9

ok let's isolate the \(\Large \frac{s}{\sqrt{n}}\) part

Answer 10

\[\Large 2 = 1.96*\frac{s}{\sqrt{n}}\]
\[\Large \frac{2}{1.96} = \frac{s}{\sqrt{n}}\]
\[\Large 1.020408 = \frac{s}{\sqrt{n}}\]
\[\Large \frac{s}{\sqrt{n}} = 1.020408\]
agreed?

Answer 11

yes!

Answer 12

so we don't know what s or n is, but we know that
\[\Large \frac{s}{\sqrt{n}} = 1.020408\]

Answer 13

okay so then you multiply that with the 2.33?

Answer 14

yes you'll multiply that with 2.33
\[\Large M = z*\frac{s}{\sqrt{n}}\]
\[\Large M = 2.33*1.020408\]
\[\Large M = ???\]

Answer 15

this works because s & n are fixed, so the quantity \(\Large \frac{s}{\sqrt{n}}\) is also fixed

Answer 16

2.37766064?

Answer 17

correct

Answer 18

so if you rounded to 2 decimal points, then it would be M = 2.38

Answer 19

yay!! thank you so much

Answer 20

no problem