Exercises: With the following three sets of values, perform each of the three exercises (R1, R2, R3) below:
- X=9,10,99,1000,124,5,6,7
- X=-2,-1,0,1,2
- X=66312,9934,1000,4432,40522,224402,44556,1034,455
- X=889,23,33,4,5,5,6,778,12,5,4
R1: What is the average of these values (use R like a calculator to show your work, then check it with the
mean()function)?R2: Do the same with median…
R3: Do the same with standard deviation…
R4: If the mean and the median are the same, what does this say?
R5: If the mean is much larger than the median what does this say?
R6: If I have two samples, one with 10 observations and a mean of 20, one with 20 observations and a mean of 17, and I combine my samples together, what is the overall mean?
R7: A test has a normal distribution of scores. The mean is 300, the variance is 150. What is the z-score that corresponds to a score of 330? What is the z-score the corresponds to a score of 280? What percentile is 330 at? What percentile is 280 at?
R8: What score corresponds to the 42nd percentile (i.e. p=0.42)? What score corresponds to the 55th percentile? (i.e. p=0.55)?
R9: What percentage of scores are between 295 and 335? What percentage of scores are between 310 and 312?
SOLUTIONS:
R4: The data are likely to have a fairly even or symmetrical distribution.
R5: That there is probably a positive skew in your data.
R6: (10*20+20*17)/(10+20) = 18
R7: z=(330-300)/sqrt(150) = 2.45 p=.9929
z=(280-300)/sqrt(150)= - 1.63 p: 1-.9484 = 0.0516
R8: p=.42 is a negative z-score. So, calc. z for 1-.42 = .58.
z=-.20. X=-.2*sqrt(150)+300=297.55
p=.55, z=.12, X=.12*sqrt(150)+300 = 301.5
R9: Calc p for 240 and 400; calc p for 310 and 410
z1=(295-300)/sqrt(150) z=-.41, p=1-.6591=.3409
z2=(335-300)/sqrt(150) z=2.86, p=.9979
Percent between these two values = 99.79-34.09=65.7%
z1=(310-300)/sqrt(150) = .82, p=.7939
z2=(312-300)/sqrt(150) = .98, p=.8365
Total between = .8365-.7939=0.043 or 4.3 percent of the observations are between 310 and 312.