Question

The state agriculture department asked random samples of Indiana farmers in each county whether they favored a mandatory corn check off program to pay for corn product marketing and research. In Tippecanoe County, 263 farmers were in favor of the program and 252 were not. In neighboring Benton County, 260 were in favor and 377 were not.

Answer 1

b. Compute a 95% confidence interval for the difference between the proportions of farmers favoring the program in Tippecanoe County and in Benton County. Do you think opinions differed in the two counties?

Answer 2

@reemii ?!?!?!

Answer 3

Do you know a statistic that is helpful for this? What kind of testing are we doing?

Answer 4

95% confidence intervl?

Answer 5

What statistic are you going to use to compute the p-value?

Answer 6

idk?

Answer 7

Have you already done testing for "comparison between proportions" ?

Answer 8

Yes

Answer 9

is ths is formula? : (p1hat-p2hat) plus/minus z* sqrt (p1hat (1-p1hat)/n1 plus (p2hat (1-p2hat/n2)

Answer 10

idk, i'm reading on some document that a statistic to use is: \(
\frac{f_1-f_2}{S} \sim \mathcal N(0,1)
\)
where
\(f_1,f_2\) are the appriximations of the proportions:
T:(263+, 252-) -> estimation of the proportion: 263/(263+252)=\(f_1\)
B:(260+, 377-) -> estimation of the proportion: 260/(260+377)=\(f_2\)
and

Answer 11

\(S=\sqrt{f(1-f)}\sqrt{\frac 1{n_1}+\frac 1{n_2}} \) where
\(f=\frac{n_1f_1+n_2f_2}{n_1+n_2}\).

Answer 12

ok can u help me solve for the 95 percent confidence int

Answer 13

I can do something about rejecting or not the hypothesis that the means are the same.
\(\frac{f_1-f_2}{S}\in [-z_\alpha,z_\alpha]\) with probability 95% , with \(z_\alpha=...\) oops i don't have the table..

Answer 14

aw.. Is this your statistic?
\((\hat p_1-\hat p_2) \pm z \sqrt{ ( \hat p_1(1-\hat p_1)/n_1 + (\hat p_2 (1-\hat p_2/n_2)}\)

Answer 15

i think yea

Answer 16

oops \((\hat p_2(1-\hat p_2))/n_2 \) <- there's an error in the big formula.

Answer 17

ok

Answer 18

what do i do

Answer 19

it is possible that my stat is the same as your stat (im too tired to compute now).
Fine, when you know which stat you must use you're almost at the end.

Answer 20

you must choose the value of "\(\alpha\)". If you want a 99% confidence interval, choose \(\alpha=0.005\) (half of 1-0.99) and take a look in the table for the value of the quantile \(z_{\alpha}\).

Answer 21

then compute everything + replace \(z_\alpha\) by what you read in the table of quantiles of the normal(0,1) distribution.
is it ok?

Answer 22

umm i dont really understand.

Answer 23

can u work it out plzzz? lol

Answer 24

Look at my old msg: \((f_1-f_2)/S \sim \mathcal N\).
This means, the quantity has a probability of 99% to be contained in the interval \([z_{0.005},z_{1-0.005}]\). Ok for that?

Answer 25

um

Answer 26

ok then wat

Answer 27

?

Answer 28

from \((f_1-f_2)/S \in [z_1,z_2]\) we get to the interval \((f_1-f_2) \in [Sz_1,Sz_2]\)
and then to the interval \([(f_1-f_2)+Sz_1, (f_1-f_2)+Sz_2]\).
In your formula, \(z=z_{0.995}\), and this interval becomes \([(f_1-f_2)\pm zS]\) because \(z_{0.005}=-z_{0.995}\).

Answer 29

it's not a perfect explanation, i hope it helps.

Answer 30

but do u know what the final answer will be?

Answer 31

no, you have to compute.
take your stat and use the statement:
\(n_1=263+252\), \(n_2=260+377\), \(\hat p_1=263/n_1\) etc..
and \(z_{0.995}\) must be looked up in a table.

Answer 32

n1= 515, n2=637

Answer 33

soo

Answer 34

?

Answer 35

have you done something?

Answer 36

no im not sure what to do

Answer 37

I just need the conf interval!

Answer 38

I made a little mistake. you want a 95% conf.int. so \(\alpha=0.975\).
Then \(z_{\alpha}=1.96\). Look, you need to replace everything in your stat:
\((\hat p_1−\hat p_2)±z\sqrt{(\hat p_1(1−\hat p_1)/n_1+(\hat p_2(1−\hat p_2/n_2)}\).
\(n_1=515, n_2=637, \hat p_1 = 262/n_1 = 0.51.., \hat p_2=0.408..\).
Therefore, the conf.int. is
\[
[0.00637.., 0.199]
\]
Try it and tell me if you understand.

Answer 39

oops

Answer 40

so the official conf interval is (0.00637,0.199]

Answer 41

(0.006, 0.199)<---

Answer 42

wat

Answer 43

wait huh!

Answer 44

is it (0.006, 0.199) or ?

Answer 45

it is rather [0.045, 0.16]

Answer 46

[0.045, 0.16] is the answer?

Answer 47

did yo understand how we got there?

Answer 48

so (0.05,0.16) is the answer?!??!

Answer 49

so (0.05,0.16) is the answer? ru positive

Answer 50

using your formula, I obtain that indeed.

Answer 51

ok thanks for ur help. and idk if thats the fomrula i just thought..

Answer 52

yw, gl