Question

See attachment for question

Answer 1

@amistre64 can u help with this?

Answer 2

prolly .... i find the way they present the long tiny text to be, disorientating ...

Answer 3

lol yah me to

Answer 4

\[\Large Z=\frac{u_1-u_2}{\sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}}\]
or it may be a T score thing, sounds familiar

Answer 5

yah that looks familiar

Answer 6

or i spose: a more confidence interval setup would be:
\[[u_1-u_2]\pm Z_{.95}\sqrt{that~stuff}\]

Answer 7

no i think it's the other way cause that looks familiar to me

Answer 8

the other was is just a zscore presentation. confidence intervals take on a slightly modified version of it

Answer 9

\[[5.37-4.23]\pm(2.575)\sqrt{\frac{(2.33)^2+(1.96)^2}{40}}\]

Answer 10

:( that looks familiar to!

Answer 11

OMG

Answer 12

now u r just confusing me :(

Answer 13

do you recall confidence intervals?\[u\pm Z\sqrt{\frac{\sigma^2}{n}}\]

Answer 14

yes

Answer 15

its just a version of that is all, using the mean/variance properties of 2 samples

Answer 16

1.96 for Z on a 95% CI ... i read it as ((% to start with

Answer 17

if i did it right, lets try: .20 and 2.08 as the range

Answer 18

ok one sec. i'm looking through my notes

Answer 19

FOUND IT!!!

Answer 20

i can never find anything in my notes ... prolly since i dont take notes

Answer 21

lmao well i had what week it was from and the question # so i could find it again.

Answer 22

that looks to use the t distribution values instead of the z distributions ... which of course will produce similar but different values

Answer 23

yup

Answer 24

if we know the population variance ... which is seldom.... we could use the Z stuff, it used to be taught that n>30 would allow you to substitute Z for T

Answer 25

from what i got the problem will look like this \[(5.37-4.23)\pm2.023\times \sqrt{\frac{ 2.33^{2} }{ 40 }+\frac{ 1.96^{2} }{40}}\]

Answer 26

The lower bound is 0.17
The upper bound is 2.11