<p>"what I did:
A high, B high, C high,
A low, B low, C low,</p>
<p>A control, B control, C control,
control nutrient high, control nutrient low, control nutrient control salinity"</p>
<p>I did basically the same thing, but instead of control, I used the words "medium salinity" and "placebo" (instead of ABC). Would that still be acceptable?</p>
<p>if 3b/c was the M=E+D, part b was at least. So u can do 1 - the probability of none. so 1- binomcdf(3,.0913,0).</p>
<p>3c i believe dealth with CLT. ur population is now 3 so. new SD= old/rad3.
then do a z score with that standard deviation and find p value.
and the treatment thing must have been 12 different treatments. wwhy else would there be 12 bowls? i did put 6 things tho :. i just didnt get the problem when i was at it.</p>
<p>I couldn't get 3C, but I'm pretty sure A was 9.18%, and B was about 25%</p>
<p>for question 3 on FR,</p>
<p>Given
M = D + E, E --> N(0,1.5) normally distributed at 0 with sd of 1.5</p>
<p>a) If true depth (D) is 2 ft, what is prob that our measurement (M) will be negative?
so....
M= 2 + E
if M < 0 --> 2 + E < 0 --> E<-2
so now we want P( E < -2) where E is normally distributed about N(0,1.5)
z-score = (-2-0)/1.5 = -1.333
Now we just do normalcdf(-E99, -1.333) and that is the answer, which should be .0912.</p>
<p>b) suppose three independent measurements are taken where true depth is still 2 feet, what is the probability that at least one of the measurements will be negative?</p>
<p>This question logically extends off of A, but now we get to use probability calculations.
So the prob of at least one measurement being negative is the same as the sum of probabilities that only one is negative, that only two are negative, and that all of the three are negative.
P = (3 choose 1)(.0912)(1-.0912)^2 + (3 choose 2)(.0912)^2(1-.0912) + (3 choose 3)(.0912)^3(1) = .24943
I'm sure there's a way to do it much more simply on the ti-83 using binomialpdf/cdf but I forgot it.</p>
<p>c) What is the prob that the mean of the above mentioned depth measurements will be negative?</p>
<p>Now for this section you can go back to using z-scores rather than probability calculations.
It's actually the same as part a except now we have an n of three.
so now z-score = (-2-0)/(1.5/sqrt(3)) = -2.3094
p= normalcdf(-E99, -2.3094) = .01046</p>
<p>aw man is it 6 treatments manipulating both variables at the same time, or is 12 treatments?!</p>
<p>We're not sure, but I'm pretty sure the guy wanted a controlled experiment. How could it be controlled with only 6 treatments with replication?</p>
<p>oh btw, on the shrimp experiment question, part c and d, what were the statistical advantages and disadvantages?</p>
<p>Yesss i got number three all correct! Now the only one im worried about about are the treatments which no one can seem to agree on a correct answer for. Hopefully it's "accept any legitimate well thought out set-up."</p>
<p>lol my c and d for 5 were pretty much all BS</p>
<p>Also, im studying for my stat final next week....if I did 100 95% confidence intervals would i get the true mean exactly 95 times?</p>
<p>if i was to forget to interpret question 4 with the confidence interval alone and went ahead and did a test stat, do you think i will get partial credit?</p>
<p>my response to the advantage/disadvantage was crazy.</p>
<p>I said if you used multiple species of shrimp, it could introduce lurking variables such as competition between different species or other interactions, which could affect the growth of some shrimp, thus using the same species (tiger shrimp) was an advantage</p>
<p>Disadvantage:
Tiger shrimp aren't representative of all shrimp</p>
<p>:-</p>
<p>Pheonix, I said that multiple shrimp would introduce confounding variables, and one shrimp fails to analyze the possibility of a lurking varible. Of course i went more in depth than that but that was the gist.</p>
<p>tamirms:</p>
<p>On part b, this is the simpler solution that I used (I don't use all of those calculator prob. programs...im old school i look up table values):</p>
<p>p(atleast one negative) = 1 - p(all positive)
p(atleast one negative) = 1-(1-.0918)^3</p>
<p>yea, and i used that notation because i was too lazy to make up variables i hope the graders dont mind</p>
<p>guys, for #6, all i did to find "s" was put the data in a list and do stat, calc, 1varstats, and "s" is the "Sx" that shows up on the screen....</p>
<p>You plug in n=10 and s, get your statistic, look up p-value, make conclusion, and then shade in some areas...routine</p>
<p>how about this?</p>
<p>advantage: narrows the population, thus allowing for more explicit conclusion</p>
<p>disadvantage: narrows the population, cannot generazlie to the general shrimp population</p>
<p>its basically the same thing that turns out to be an advantage and a disadvantage? did i screw up?</p>
<p>I was reading in a book today, each sample you take has a c% chance of having the mean or whatever. Each sample is independent. Thus, it is a binomial random variable. How specific can you get on the MC, w/o "discussing them." Can you talk about the ideas - the exacts.</p>
<p>I said its blocking = takes out a lot of lurking variables = more conclusions
Bad is that it only applies to tiger shrimp and it would be hard to make it for the entire shrimp pop. What you have (above my post) is correct.</p>
<p>What...we aren't talking about multiple choice...we're talking about stuff for our upcoming stat class finals...</p>
<p>thanks for clarifying</p>
<p>ExRunner, you would THEORETICALLY get 95/100 that would capture the true mean</p>
<p>i see it more like this:</p>
<p>the ratio (# intervals that capture true mean)/(all intervals) approaches .95 for a 95% confidence interval as n, the number of intervals, approaches infinity</p>
<p>My class we are doing experiments and do we need SRS so we dont have to repeat the experiment were setting up, or do we accept the results w/o replication.</p>