<p>You’re probably right.
I just want a 3. Hope I did enough on the MC and rest of FRQ’s to do that.</p>
<p>HelloAll, you’re correct. We were given sample std. devs, thus we use “t”</p>
<p>Yeah I just had a brain fart on that question. I just saw that they gave the standard deviation and immediately thought z test. After the test was done I realized it was the sample st. dev. and not the population. It has to be at least partially wrong, but the confidence indicator has a difference of .049 between z and t so the confidence interval should still be about the same.</p>
<p>Ahhh, same here erasmuss22. Will they still be able to give us partial credit?</p>
<p>does anyone have the 2006 multiple choice test?? i desperately need it</p>
<p>hey just to echo everybody else… can someone pm me 9&34… thanks.</p>
<p>Yeah, it was a 2 independent samples T-interval.</p>
<p>the problem involving an interval (purposely very vague) involved proportions…there are no t tests involving proportions just z tests</p>
<p>I don’t see how it was proportions when means and std. dev were given. explain?</p>
<p>but can’t you use a z test to model a t distribution when n>30 according to the central limit theorem?</p>
<p>can we discuss the stats FRQ now lol?</p>
<p>it was def t, not z</p>
<p>I’m not sure if we can discuss yet, maybe by 4?</p>
<p>It was definitely t, they gave you sample sizes. What did you guys put for the equation at the end?</p>
<p>can we discuss free response questions already?</p>
<p>it is z…thinkin about a different question. Its been 48 hours so we can discuss this i believe. For the last formula I wrote 1.5IQR-MAX=y, and if y is negative then there are outliers that can most definitely induce right skewness on the distribution.</p>
<p>Yep, its exactly 48 hrs since the end of the exam…so technically yes.</p>
<p>I put (Max-Median)/(Median-Min) and if it is greater than 1 it’s skewed right.</p>
<p>I did this:
((Maximum+Minimum)/2)/median
Basically taking the average of the max and the min and dividing that by the median, and if it is greater than one its skewed to the right.</p>
<p>yea that could work as well because perfect normality would have that value be equal to 1 (since 50% of the distribution should be from min-med and another from med-max)</p>