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SKIDMORE PS 217 - PS 217 Exam 1

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Exam 1 PS217, Fall 19961. Below is a sample. For the data from this sample, compute the mean, median, mode, varianceand standard deviation. [10 pts]146535253SX = 34SX2 = 150 X =XÂn=349= 3.78SS = X2-( X )2ÂnÂ= 150 -11569= 21.56s2=SSn -1=21.568= 2.695s = s2= 2.695 = 1.64The median would be 4 (4 scores above and 4 scores below). The mode would be 5 (mostfrequently occurring score). Thus, the distribution would be a bit negatively skewed.2. Test the hypothesis that the sample you see in the first question was drawn from a normalpopulation with m = 5 and s = 2. [10 pts]Because s is known, the appropriate statistic would be a z-score.H0: m = 5H1: m ≠ 5Decision Rule: If |zObt| ≥ 1.96, reject H0. sX =sn=29= .67z =X -mX sX =3.78 - 5.67= -1.82Because |zObt| < 1.96, we would retain H0. Thus, it is plausible that the sample was drawnfrom a population with m = 5. OTOH, you should realize that you may be making a TypeII error, because with only 9 scores in your sample, you may have insufficient power.3. You all remember the OJ Simpson trial, right? OK, tell me in words what a Type I and a Type IIerror would be for that trial, which tested the H0 that Simpson was not guilty. In this context, whicherror would be more serious? Why? In typical psychological research, we set a = .05. What leveldo you think a is set to for most criminal trials in this country? Why? [10 pts]A Type I Error would be concluding that OJ was guilty, when in fact he was innocent. AType II Error would be concluding that OJ was innocent, when in fact he was guilty.Because we generally think that it’s a very serious error to falsely imprison (or execute)an innocent person, a Type I Error would be thought of as more serious. On that basis,courtroom decisions are likely to have a probability of Type I Error set at a “probability”that is much smaller than .05 (one would hope .001 or so), because we would not want totolerate falsely convicting 5 people out of 100 on trial.4. The average age for registered voters in the county is m = 39.7 years with s = 11.8. Thedistribution of ages is approximately normal. During a recent jury trial in the county courthouse, astatistician noted that the average age for the 12 jurors was X = 50.4 years. [10 pts]a. How likely is it to obtain a jury this old or older by chance? sX =sn=11.812= 3.41z =X -mX sX =50.4 - 39.73.41= 3.14The probability of a z of 3.14 or greater is .0008.b. Is it reasonable to conclude that this jury is not a random sample of registered voters?Yes, it is extremely unlikely to draw a sample of 12 from a population with m = 39.7 andhave the sample mean turn out to be 50.4,. (Of course, it is possible to get such a samplefrom that population, just very unlikely! So, you could be making a Type I Error.)5. As we saw in class, gestation periods are normally distributed with m = 268 days and s = 16days. Suppose that 16 women became pregnant. How likely is it that the mean gestation period forthis sample would fall between 276 and 284 days? [5 pts] sX =sn=1616= 4z =X -mX sX =276 - 2684= 2z =X -mX sX =284 - 2684= 4The proportion of scores greater than or equal to z=2 would be .0228. The proportion ofscores greater than or equal to z=4 would be .00003. Thus, the proportion between z=2and z=4 would be .02277, or roughly 2.3%.6. A normal distribution has a mean of 120 and a standard deviation of 20. For this distribution,[5 pts]a. What score separates the top 40% (highest scores) from the rest?.40 in Column C yields a z = .25, which would mean a score = 125.b. Scores between 60 and 100 make up what percentage of the distribution? z =X -ms=60 -12020= -3z =X -ms=100 -12020= -1The proportion of scores less than z = -1 would be .1587. The proportion of scores lessthan z = -3 would be .0013. Thus, the proportion of scores between 60 and 100 would be.1587 - .0013 = .1574.c. What range of scores would form the middle 60% of this distribution?The middle 60% of the distribution means that you’re eliminating the extreme 20% of thescores in each end, so you’d look up .20 in Column C. That yields z = 0.84, so –0.84 and+0.84 would cut off the lower and upper 20% of the distribution (leaving the middle60%). Thus, the values that would cut off the middle 60% of the distribution would be103.2 and


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