Can anyone help with this?
did you see confidence intervals? There's a formula saying \[C.I.=(\mu_1-\mu_2)\pm c\sqrt{\frac{\sigma_1^2+\sigma_2^2}{n}}\]
UGH i suck at those! :( can't they even make this easy?!
=\ that's the reality of confidence intervals lol
ok how do i do this?
\[C.I.=(5.96-4.12)\pm (1.645)\sqrt{\frac{2.49^2+1.85^2}{40}}\]
1.645 is a critical value you find from a standard normal table. You actually look at 0.9500 in the table and find out it has a critical value of 1.645 . But a confidence interval has 2 tails, so that's why you look at the 0.9500 level, since the other 0.05 tail on the left is taken off, resulting in a 90% C.I. value
I got 5.96 &-2.28
Is that right?
i got 2.65 and 1.036
U positive?
um lemme retry i might have mistyped it
2.65 and 1.033
so you write the range is 1.033 hours to 2.65 hours
Yah i did it again using Mathway and got the same as u. i was solving it by hand before so i was getting the wrong answer.
So i put 1.033 in the first box and 2.65 in the 2nd?
yes, although they say round to 2 decimals, so the first box is 1.03
ok how about the last part?
D .. you conclude there is a significant difference because zero 0 is not in the interval (1.03, 2.65)
ok Thanks again for the help! :D
np =]
now i just got to check over a few then i can send my homework in!
ok maybe ask your teacher what kind of "technology" they expect you to use for these problems since I think using R is a bit overboard
well this is my finals homework so i don't need to worry about that as long as i pass this homework. But if u have time can u check and see if i got some questions right? I think i did but i have to be sure or i have to take this class again and i REALLY don't want to do that.
ok
i can try
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