Question

Could someone help me with stats, this is the only question I don't get.

You want to estimate the difference in grade point averages between two groups of university
students to be accurate within 0.2 grade point, with probability approximately equal to 0.95. If thestandard deviation of the grade point measurements is approximately equal to 0.6, how many studentsmust be included in each group? (Assume that the groups will be of equal size)

Answer 1

No, not that question. I posted the wrong one

Answer 2

In an attempt to compare the starting salaries for university graduates who majored in education
and the social sciences, random samples of 100 recent university graduates were selected from each

(a) Do the data provide sufficient evidence to indicate a difference in average starting salaries foruniversity graduates who majored in education and the social sciences? Test using α = 0.05

Answer 3

@satellite73 can you please help?

Answer 4

sorry bro im on an exam right now, if you still need help, ill be there when im done

Answer 5

ok...i will try it again and see....i will write on here if i didnt get it

Answer 6

you should find the difference in means
as a start

Answer 7

Here is the procedure. You will want to use a two-tailed test
http://stattrek.com/hypothesis-test/difference-in-means.aspx

Answer 8

Here is a t-calculator
http://onlinestatbook.com/2/calculators/t_dist.html

Answer 9

I got a difference of mean of 2206 and a standard error of 325.44

Answer 10

ok on the difference in means. I haven't done the standard error calculation yet.
Can you post your work?

Answer 11

ok, I just got 325.44

Answer 12

t= diff/ std error= 2206/325.44

Answer 13

you now need the degrees of freedom DF

Answer 14

my DF was 453.09

Answer 15

t= 6.8
and with a large DF, it is unlikely to get this value by chance

Answer 16

one site states
Some texts suggest that the degrees of freedom can be approximated by the smaller of n1 - 1 and n2 - 1; but the above formula gives better results.

both are 100-1= 99. So according to this, DF=99

But the other site states
The degrees of freedom is the number of independent estimates of variance on which MSE is based. This is equal to (n1 - 1) + (n2 - 1), where n1 is the sample size of the first group and n2 is the sample size of the second group.

that gives 99+99= 198

Off-hand, I don't know what DF is! But it is large, and we will get a small probability of t being 6.8 by chance alone. This means the null hypothesis (difference in means = 0)
can be rejected.

Answer 17

i did it again...i got 49.3

Answer 18

as long as DF is much bigger than 5, we can reject the null hypothesis (according to the t-calculator)

Answer 19

oh ok....umm could you explain one part to me that i have a hard time understanding

Answer 20

Since we have a two-tailed test, the P-value is the probability that a t-score having 40 degrees of freedom is more extreme than -1.99; that is, less than -1.99 or greater than 1.99.

We use the t Distribution Calculator to find P(t < -1.99) = 0.027, and P(t > 1.99) = 0.027. Thus, the P-value = 0.027 + 0.027 = 0.054.

Answer 21

I don't get where they get -1.99 and 1.99 in that statement

Answer 22

that is from the example. Let me read it.

Answer 23

ok

Answer 24

First, the null hypothesis is diff in mean =0
if we plotted diff in means from many different samples, we would be a bell shaped curve (roughly speaking)

they found t= -1.99 (which I read as (roughly) , 1.99 std dev below the expected value of 0)

the chance that t (by chance because the samples always give slightly different difference in means) is -1.99 or lower , is 0.027

Answer 25

***I don't get where they get -1.99 and 1.99 in that statement ***
that is the value of t, which they found using their data. Exactly the way you found your t=6.78

Answer 26

oh ok. Umm i have a few other questions like this. I'm going to go try them out, could I post them on here with the answer later? Could you check my work on them?

Answer 27

ok.

Answer 28

thank you

Answer 29

yw

Answer 30

btw, this formula
DF = (s12/n1 + s22/n2)2 / { [ (s12 / n1)2 / (n1 - 1) ] + [ (s22 / n2)2 / (n2 - 1) ] }
gives DF= 197, which is close to the 99+99= 198 estimate of the other site
(but definitely different from the minimum of n1 and n2)

Answer 31

I got 49 with that formula

Answer 32

@phi for this question "An experiment was conducted to test the effect of a new drug on a viral infection. The infection was induced in 100 mice and the mice were randomly split into two groups of 50. The first group, the control group, received no treatment for the infection. The second group received the drug. After a 30-day period, the proportions of survivors, in two groups were found to be 0.36 and 0.60 respectively.
(a) Is there sufficient evidence to indicate that the drug is effective in treating the viral infection?
Use α = 0.05" for the proportion is 0.48 and the SE is 0.009992 and the T value was 4.8.....would i accept the null hypothesis