Does Lab Attendance Improve Student Performance?

[kliker]

a teacher wants to see if the lab helps the student perform better. He took two independent samples, 77 students who attended the lab and 23 who didnt attend the lab. 6 out of 77 that attended the lab said that they had difficulties, 16 out of 23 that didnt attend the lab said that had difficulties

i need to find the 95% confidence interval for the difference of rates of students that had difficulties

i said p1 = 6/77 = 0.0779
p2 = 16/23 = 0.695

then i said that the confidence interval is

(0.0779-0.695-1.96*sqrt(0.0779(1-0.779)/77+0.695(1-0.695)/23),0.0779-0.695+1.96*sqrt(0.0779(1-0.779)/77+0.695(1-0.695)/23) =

= (-0.807532,-0.426668)

hence we can be 95% sure that the lab helps student to understand better what they re doing

am i correct? I am not sure if i got the results right, I mean the first sample is big >=30 but the second one is smaller, so I am not sure if i took the right equation

which in this case is the following

p1-p2+/-Za/2*sqrt(p1(1-p1)/n+p2(1-p2)/m)

thanks in advance

 

[cronxeh]

Ok your confidence interval is negative there, that should be a warning sign that something is not quite right. Perhaps if you can state which distribution you using, how you found the variance, the mean, and how you are going to come up with the confidence interval. You have 2 different samples here, so maybe you should keep them separate?

I.E your total number of students is not 77+23, you have

Group 1 (n=77), 6/77="had difficulties", 71/77="did not have difficulties"

Group 2 (n=23), 16/23="had difficulties", 7/23="did not have difficulties"

So now think about how you would find sample variance, and since you have n>30 for first sample perhaps you can use normal distribution to find confidence interval. For group 2, since sample is small what distribution would you use?

 

[kliker]

since we want to know if lab helps them, doesn't this mean that we want to find p1-p2 confidence interval?

that's what I am doing, i don't find the interval for p1 and then for p2 but for p1-p2

so in this case negative confidence interval should mean that p1-p2<0 => p1<p2

how would I do it with your method? actually yea maybe my method is in correct i don't know

For the second sample only I guess I would have to use the Tn-1 distribution

 

[cronxeh]

Ah, yes, the difference of two proportions then.

SE1=sqrt(0.0779*(1-0.0779))/sqrt(77)=0.0305

SE2=sqrt(0.695*(1-0.695))/sqrt(23)=0.0960

SE of p1-p2 = sqrt(SE1²+SE2²) = 0.1007
p1-p2 +/- z*(SE1²+SE2²)

p1-p2 +/- 2*0.1007

-0.6171+/-2*(0.1007)= -0.4156, -0.8185 or (-41.56%,-81.85%) difference with 95% confidence

 

[kliker]

thanks, i think i did the same thing

now i want to find it using hypothesis testing

actually i want using hypothesis testing to show that students who go to lab have less difficulties than those who don't

hence

H0: p1=p2
H2: p1<p2

also the α must be 0.05

i can find now the z which is (p1-p2)/s (1)

δ=0 hence s=sqrt(pq(1/n+1/m))

p = (6+16)/100 = 26/100 = 0.26

hence

s = sqrt(0.26*0.74(1/77+1/23)) = 0.104230

from (1) = (0.0779-0.695)/ 0.104230 = --5.92056

now H0 will be true if z>za

but za = - 1.65 so H0 is not true => H1 is true

that means that p1<p2 hence students who go to lab have less difficulties than those who dont

what i want to ask is

am I correct? I am not sure about the za though, we have n and m n>30 m<30

im using the N(0,1) distribution, but i know that when its less than <30 i should use the T distribution

but here we have n>30 and m<30, so which one should I use in order to find za?

thanks for your help