Hypothesis Test concerning difference in proportion- Where did i go wrong???

#1
Hey all,
The answer I get is not amoung the answers offered. I'd appreciate some help in telling me where I went wrong!!

In order to determine whether or not a particular medicine was effective in curing the common cold, one group of patients was given the medication, while another group received sugar pills. The results of the study are shown below.

/ Patients cured / Patients not cured
Received medication / 70 / 10
Received sugar pills / 20 / 50

We are interesting in determining whether or not the medication was effective in curing the common cold.

The test statistic is
a. 10.08
b. 54.02
c. 1.96
d. 1.645


This was my solution:

Hypothesis:

H0: p1 - p2 <= 0
Ha: p1 - p2 > 0

/ Group 1- med / Group 2- sugar pill
Sample size / 80 / 70
Response of interest / cured / cured
Count of response / 70 / 20
Sample proportion / 0.875 / 0.286


Hypothesized value 0
Point estimate difference 0.589

Pooled estimate of p 0.6
Standard error 0.080

Test statistic (z) 7.483 (my answer)


As you can see, my answer is not amoung those offered. Can anyone here please explain why?
Thanks for your help!!!
 

mp83

TS Contributor
#2
The statistic is

e0)/{pe(1-pe)(1/n1+1/n2)}^1/2​

where Δe is the sample difference Δ0 the difference under null hypothesis, pe the pooled sample estimate
 
#3
Thanks for your response!

I used exactly that formula:

(Δe-Δ0) = (0.589 - 0) = 0.589

pe(1-pe) = (0.6*(1-0.6)) = 0.24

(1/n1 + 1/n2) = 0.0268

= 0.589 / SQRT(0.24*0.0268)

= 0.589 / 0.0802

= 7.346

Do you see where I went wrong?
 

mp83

TS Contributor
#4
My,my,my...what are we doing here?!

We need to perform a x2-test

Code:
> prop.test(a1,correct=F)

        2-sample test for equality of proportions without continuity
        correction

data:  a1 
[B][COLOR="Red"]X-squared = 54.0179[/COLOR][/B], df = 1, p-value = 1.987e-13
alternative hypothesis: two.sided 
95 percent confidence interval:
 0.4610220 0.7175494 
sample estimates:
   prop 1    prop 2 
0.8750000 0.2857143
 
#5
Again, mp83, thanks for the help.
In addition to knowing the answer, I was hoping to learn how solve the problem (where's the fun in just having the answer if I cannot do it under different circumstances?).

Earlier you said the formula for this was:

(Δe-Δ0)/{pe(1-pe)(1/n1+1/n2)}^1/2

which I agreed with, but somehow I got a different answer. Since I've gotta do it by calculator (not the superhuman stat program you have) and if it's not too much trouble, could I get a step-by-step look at the solution? And how is what you did different from that in the above calculation?

Again, thanks!
 
#7
ok, now what's on the link looks much more familiar. My book calls it the "Goodness of Fit," and the "X2" you were referring to was "chi-squared".
Thanks again for your help on this!
 

mp83

TS Contributor
#8
Sorry I didn't write more myself and throw the link but I'm bit stressed in time.A Chi-Square test can be used to test independence,fit,homogeneity.It's usual of statisticians to write X2,as "chi" is the greek letter X (or χ)