Given the following data: 100, 95, 95, 90, 85, 75, 65, 60, 55. Find the median, mean, and mode. Is there a most appropriate measure?

median = 85, mean = 80, mode = 95; The mode is certainly inappropriate. Beyond that, the data set is too small to know whether the mean or median is more appropriate.Make a sketch of the following, indicating the approximate locations for the mean, median and mode:

- a normal distribution
- a skewed distribution
- a rectangular distribution

See notes.Given the data set: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, ''x''. Find the smallest positive integer value for ''x'' such that ''x'' is an outlier. Find the value for ''both'' definitions.

Experiment to find these values.Given the following set of golf scores: 67, 70, 72, 74, 76, 76, 78, 80, 82, 85. Find the median, mean, mode, and standard deviation. What percentage of scores are in the interval of one standard deviation from the mean?

median = 76, mean = 76, mode = 76, standard deviation = 5.52; 6 out of 10 scores are between 70.48 (76 - 5.52) and 81.52 (76 + 5.52), giving 60% of scores for this data set within one standard deviation of the meanGive at least five uses and at least five misuses of statistics.

See notes on Uses and Abuses of StatisticsGive the four levels or categories of data and give an example of each.

See notes on Variables and Scales of MeasurementGiven the following grades on a test: 86, 92, 100, 93, 89, 95, 79, 98, 68, 62, 71, 75, 88, 86, 93, 81, 100, 86, 96, 52

A stem-and-leaf plot is a quick way to construct a histogram by hand, whereby one lumps together all values that have common leading digits up to some level of precision (i.e., the "stem") using the remaining digit(s) of each data value to form the "bars" (i.e., the "leaves"), as shown below for the data set $13,15,17,17,19,24,25,25,25,26,29,30,31,33,46,62$:

6 | 2 5 | 4 | 6 3 | 0 1 3 2 | 4 5 5 5 6 9 1 | 3 5 7 7 9

Make a stem-and-leaf plot that represents the test grade data.10 | 0 0 9 | 2 3 3 5 6 8 8 | 1 6 6 6 8 9 7 | 1 5 9 6 | 8 5 | 2

Find the mode, median, mean, range, standard deviation, and interquartile range

mode = 86, median = 87, mean = 84.5, range = 52 to 100, standard deviation = 13.17, interquartile range (IQR) = 77 to 94`In R: grades = c(86, 92, 100, 93, 89, 95, 79, 98, 68, 62, 71, 75, 88, 86, 93, 81, 100, 86, 96, 52) median(data) mean(data) range(data) sd(data) IQR(data) # Sadly, the mode() function in R does something else. # So you can write your own function, or if the data # set is small enough (as it is here), you can use # sort(data) to make it easier to count duplicate entries TI-83: Enter data in L1 with [STAT] : EDIT : Edit... Then [STAT] : CALC : 1-Var Stats This returns a scrollable list with: * median given by "Med" * mean given by x * range given by minX and maxX (technically, their difference) * standard deviation given by Sx (presuming the data is a sample) * interquartile range given by Q3 - Q1`

Are there any outliers? Explain clearly.

For a score to be an outlier, the score must be outside the interval of three standard deviations from the mean (44.99 to 124.01). There is not an outlier.

What is an experimental design and why is it important? Describe a completely randomized experimental design and a rigorously controlled design.

See the notes.Given this sample of freshman GPA scores:

2.2, 2.9, 3.5, 4.0, 3.9, 3.5, 2.9, 2.8, 3.1, 3.5, 3.8, 4.0, 3.8, 2.4, 3.9, 3.4, 2.8, 2.4, 1.8, 3.6, 3.1, 2.9, 3.8, 4.0

Is there an outlier? (Check both tests and explain)

1.8 is NOT an outlier, as 1) it is not below 1.35 (i.e., the mean minus three standard deviations); and 2) It is not below 1.425 (Q1 minus 1.5 times 0.95).Draw a frequency histogram using 5 to 6 categories. Be consistent with the rules for making histograms.

One set of boundaries includes: 1.65 to 2.05; 2.05 to 2.45; 2.45 to 2.85; 2.85 to 3.25; 3.25 to 3.65; 3.65 to 4.05, as shown below -- although a variety of answers will work.

In R: data = c(2.2, 2.9, 3.5, 4.0, 3.9, 3.5, 2.9, 2.8, 3.1, 3.5, 3.8, 4.0, 3.8, 2.4, 3.9, 3.4, 2.8, 2.4, 1.8, 3.6, 3.1, 2.9, 3.8, 4.0) hist(data, breaks=seq(from=1.65,to=4.05,by=0.40), col="gray", xlab="GPA", main="GPA Scores")

Is the distribution significantly skewed?

Pearson's Index gives a value of $-0.95$, so it is not significantly skewed (although the distribution looks skewed from the histogram)