<p>guys,</p>
<p>so is it type 1 error or type 2?</p>
<p>type 2m 10 char.</p>
<p>darn. and can we not use calculator notations such as binomcdf?
what if i did it…</p>
<p>hey this is a kinda silly q but for number 1 wut did u put for the discussion? just tht men r higher in this this and this rite?</p>
<p>and if we calculated confidence interval using z (which i kno is wrong) but hav the proper explanation and everything then wut score is that most likely to be? and i didnt show numbers subbed into a formula either so?</p>
<p>um okay…I guess nobody else had Form B…
Maybe Form B is just for international then…? lol</p>
<p>for the FRQ #6, what were we supposed to do?
I calculated the mean by doing (lowest + highest) /2 = 27.5
and it was higher than the median.
so I said if mean - median is positive, then it’s skewed right.
would that be a valid answer</p>
<p>howon92: the mean isn’t necessarily (lowest + highest) / 2. actually, if the distribution is skewed, then that formula will not produce an answer nowhere near the actual mean.</p>
<p>i did (maximum - median) / (median - minimum) and said that, if it is large (or even just greater than 1), this means that there is a wider range to the right of the median than to the left of the median, indicating a skewness to the right.</p>
<p>For #6, I put the formula:
(Max-Q3)/(Q1-Min), is that flawed in any way?</p>
<p>Also for question 1, we were suppose to draw a segmented bar graph right?</p>
<p>bell curves: that formula works well, too. i actually wish i had written: (Q3 - Med) / (Med - Q1)</p>
<p>the maximum/minimum could be an outlier that is not representative of the spread/skewness… the Q1 and Q3 include a good, representative portion of the distribution, so outliers would not change the statistic drastically.</p>
<p>For 2b, I got 80 something percent. I used binocdf. Gah I cant get a 5 anymore.</p>
<p>why can’t you get a 5 anymore?! that one question alone could never make your score drop that much…</p>
<p>and yes, a segmented bar graph would suffice… that’s what i did! i was unsure how to “interpret” or whatever the graph, though. i was just like “10 more males than females have had a job over the summer… overall, more males have had part-time jobs…”</p>
<p>I got a 2 on question 2 and a 2 on question 4. So thats 85.7% on FRQ max. I need a 65% on multiple choice, which is like 12-13 wrong. I think I got more than 13 wrong on MC.</p>
<p>“the maximum/minimum could be an outlier that is not representative of the spread/skewness… the Q1 and Q3 include a good, representative portion of the distribution, so outliers would not change the statistic drastically.”</p>
<p>Outliers are measures of skewness. That is partly why the mean/median ratio works so often.</p>
<p>ok.
lets go through this one at a time see what I did…you can comment if you think its wrong/correct. The ones that I AM NOT COMPLETELY SURE OF HAVE A * noted by the problem letter.
- (a) I drew a segmented bar graph that I dont have the time to recreate. I had frequency on the vertical axis and the categories on the horizontal axis. My male bar was shaded and my female not. I think you get the idea.
(b*) The distribution shows that more females never had a part-time job than males, and that more males have part-time jobs than females during or not during summer. More males proportionally have part-time jobs than females regardless of the time although proportionally more females have part-times jobs in the summer compared with those not during the summer and “never”.
(c)Chi-square for independence should be used with:
Ho: Gender and job experience are independent (no association)
versus
Ha: Gender and job experience are dependent (association exists)</p>
<p>2 (a) 128.4086
(b) .47178
(c).04271</p>
<p>3 (a) Number each student 01-24. Read off of a random number table skipping 00 and ending after all numbers have been read. The first twelve selected will be assigned the computer simulation dissection, and the second twelve the physical dissection.
(b*) The situation clearly describes the possible presence of a lurking variable, which causes ambiguity in deciding whether it is the treatment or the characteristics of the subjects which causes the result. For example, if it is actually the computer skills of those who chose the computer simulation, and they learn better using a computer, the cause is not the software, but the subject’s prior knowledge. (THIS ONE MAY BE WAYY OFF)</p>
<p>4 (a) (-0.3732, 2.3732) We are 95% confident that the difference between the true mean response time of the southern fire station and the true mean response time of the northern fire station is between -0.3732 minutes and 2.3732 minutes.
(b*) Ho: u1-u2=0
Ha: u1-u2/=0
No, it does not support his claim because the 95% interval includes the value of 0.</p>
<p>5(a*) The probability of obtaining a sample statistic as extreme as the one obtained is 0.0761 if the Ho of survival rate of the two treatments are equal is assumed true.
(b) Given that the p-value > a (.0761> .05) we fail to reject the null hypothesis at the 5% significance level. There is not significant evidence to suggest that CC alone produces a higher survival rate.
(c) Type II error is the only error possible since we failed to reject the Ho. One possible consequence is that we are unnecessarily employing the use of MMR.</p>
<p>I think 6 has be covered extensively enough. you guys can talk that out</p>
<p>3b.) Students may decide to pick the treatment that better suits them, so their performance on the pre-test may not fully reflect how effective this treatment is. Students could also just go into a group with their friends, which would also not reflect how effective the treatment was.</p>
<p>^ In short: Voluntary bias.</p>
<p>Or the better students may choose one method over the other. Thats what I put atleast : /</p>
<p>axeback: voluntary bias is where a survey is offered on a “only if you want to” basis and is not present in experiments where a treatment must be placed on subjects :)</p>
<p>On 3b I started to write: “If student chooses which program they, they will likely perform better.” Then I thought I was over-complicating the answer, so I instead wrote something to the effect of: “If the students choose which program they want, the sample is not random; this create bias, and the statistical validity of the experiment is therefore compromised.”</p>
<p>I couldn’t think of the specific kind of bias. Response, nonresponse, voluntary response, convenience bias, wouldn’t work — I don’t think, anyway.</p>
<p>Do you think I’ll still get full credit for that part?</p>