<p>Yes, it is (1 - binomcdf(w/e you put in)).
I’m glad I got that right!</p>
<p>NYEM, Independence is not the same as Homogeneity. Association is the same as Independence; Homogeneity is when you have more than one sample compared on several variables whereas Independence is one large sample (can be split up) compared on two variables.</p>
<p>^ really? idk… just buy an 89 and your problems will be solved
the test was ok i thought. i spent the last 40 minutes making a checkerboard out of my graph on the green insert =/
our teacher definitely overprepared us
and you don’t need to justify the binomial thing. but you can’t use calculator notation right.</p>
<p>Shouldn’t it be (5, 0.3, 1)? Since binom is less than and equal to, P(at least a 2) = 1 - P(x = 1)?</p>
<p>I don’t think you can use calculator notation, thats why i wrote out the binomcdf thing in formula form:
1 - [(5 Cr 2)(2/5)^(.7)(3/5)^.3]</p>
<p>Wait no…not the percentiles, its the probability that its raised to.</p>
<p>Nyem, there’s a distinct difference between the chi-square homogeneity and independence tests. Although the procedures are similar, the two are answering different questions. For the test of homogeneity, you’re testing the equality of three different populations of categorial data. For the test of independence, you just want to see if two categories are independent.</p>
<p>The probability that it is greater than the 70th percentile is .3. And it would be P(at least 2), so you wouldn’t need to find the complement if you did binomial cdf from 2 to 5.]</p>
<p>(5 C x)(.3)^x(.7)^(5-x) for x=2,3,4,5</p>
<p>@alicimoo</p>
<p>That is what I did, 1-binomcdf(5, .3, 1).</p>
<p>This is because we are finding the probability of ATLEAST 2, so it would be 1 - the probability of ATMOST one. which is P(0 or 1)</p>
<p>^_^</p>
<p>Also, I wrote my binomial probability in sigma notation:
<a href=“http://www.codecogs.com/eq.latex?\sum_{n=2}^{5}%20\binom{5}%20{n}%20(.3)^n%20(.7)^{5-n}[/img]”>http://www.codecogs.com/eq.latex?\sum_{n=2}^{5}%20\binom{5}%20{n}%20(.3)^n%20(.7)^{5-n}
</a></p>
<p>Crap.
Yep, @user3725 I think thats correct.
Wow I messed that up. Hope for partial credit on that… x_x</p>
<p>my teacher says that the graders hate calc notation so i wrote it out and then checked it via g-calc</p>
<p>peachsnapple, are you joking?
It’s still an equivalent form; they can’t take points off for it.</p>
<p>I did the same summation as you, portishead.</p>
<p>Ok so for number 2 what do you guys remember as the answers?</p>
<p>2a. I got like 128ish? I just used the z=(y-u)/s
2b. I got like .47ish? I did the 1-(p=1)-(p=0)
2c. I was really confused… I remember getting something like 0.08 but I was totally unsure.</p>
<p>I did like (130-125)/(6.5/sqrt(5)) and used a t-test…6.5 was the given SD and 125 was the given mean…but it was only 5 cars so I was totally confused. What was the answer? Thanks.</p>
<p>btw I did (max-q3)/(min-q1) in absolute value for 6d.</p>
<p>crap…i definetly forgot the 1- thing for the binomcdf…but i did do the 5, .3, 1, instead of 2…i hope i get nice graders…:(</p>
<ol>
<li>a) I think I got 130.</li>
<li>b) Its the [1 - binomcdf(5,.3,1)] that we were just discussing.</li>
<li>c) Its a normalcdf problem, but make sure you use the new st.dev (divide by sqrt.5)</li>
</ol>
<p>huh so what were the exact answers? I think i got 2b right…i plug that in.</p>
<p>What was the formula for 2c (numbers plugged in)? I didnt used normal cdf. Where would you get the new st.dev? But then again, I was totally confused.</p>
<p>2a. 128 something.
2b. didn’t do.
2c. yep, did (x - u) / (sd/sqrt(5)), dont remember the #.</p>
<p>The rationale behind dividing sigma by sqrt(5) because you are examining the distribution of all possible sample means with a sample size of 5 under the CLT.</p>
<p>Sample mean = 125
Sample st.dev = (6.5)/(sqrt.5) = 14.53</p>
<p>normalcdf (-1E99, 130, 125, 14.53)</p>
<p>i don’t remember using normal cdf either…
for the type i/ii…i remember writing type “blank” then justifying it, thinking i would fill it in after i justified it…well, i can’t remember if i ended up putting it down or not. b/c i had kinda run out of time at that point!!!..i’m prety sure i justified it as a type I error…but just say my justification was <em>somehow</em> right…would i beable to get any points?!</p>