Data analysis (Mathematics I)

The mean is the total divided by the count: (2+4+6+8+10)/5=30/5=6. For equally spaced data the middle value coincides with the mean. Do not forget to divide by the count, 5.

In increasing order the data are 2, 3, 4, 7, 9. With 5 values, an odd number, the median is the third one, namely 4. The mean is 25/5=5, and in general the median and the mean do not agree. Do not forget to sort first.

In increasing order the data are 2, 3, 5, 7, 8, 9. With 6 values, an even number, the median is the mean of the middle two, the third and fourth: (5+7)/2=6. Taking only one of the middle values and answering 5 or 7 is wrong.

The mode is the value that occurs most often. The value 9 occurs 3 times and every other value at most twice, so the mode is 9. The value 8 is the mean and 8.5 is the median; these three measures of centre are different things.

The range is maximum minus minimum: 10−2=8. Do not answer with the maximum 10 itself, and do not confuse the range with the interquartile range. A single outlier can change the range a great deal.

The median is 4. The lower half 1, 2, 3 has median Q1=2, and the upper half 5, 6, 7 has median Q3=6. The interquartile range is Q3−Q1=6−2=4. Do not confuse it with the range, which is 7−1=6.

With 8 values the lower half is 1, 3, 4, 5 and the upper half is 6, 8, 9, 10. Q3 is the median of the upper half, (8+9)/2=8.5. The value 5.5 is the overall median Q2. Do not forget to average the middle two values of the upper half.

The left end of the box is Q1 and the right end is Q3, so the length of the box is Q3−Q1, the interquartile range. The range runs from the end of one whisker to the end of the other, that is maximum minus minimum. The standard deviation cannot be read off a box plot.

IQR=16−10=6, so the boundary is 16+1.5×6=16+9=25. The value 26 exceeds it and is therefore an outlier. The values 24 and 20 are at most the boundary 25, so they are not outliers. Remember to add 1.5×IQR to Q3.

The total is 98+102+101+99+100=500, and 500/5=100. You can also see it from the differences from the working mean 100, namely −2, +2, +1, −1, 0, whose average is 0, so the mean is 100. Since the total is 500, the mean cannot be 100.5.

The original total was 10×5=50. Correcting the error makes the total 50−8+13=55, so the mean is 55/5=11. You can also divide the increase of 5 by the count 5 and see that the mean rises by 1.

The mean is 3. The deviations are −2, −1, 0, 1, 2, so the variance is (4+1+0+1+4)/5=10/5=2. Do not forget to divide the sum of squared deviations, 10, by the count. The value √2 is the standard deviation.

The mean is 5. The variance is (9+1+1+9)/4=20/4=5, and the standard deviation is its positive square root, √5. Do not report the variance 5 as the standard deviation.

The mean is 4 and the mean of the squares is (1+9+25+49)/4=84/4=21. The variance is 21−4²=21−16=5. The value 21 stops at the mean of the squares and 16 at the square of the mean, both forgetting the subtraction. The defining formula also gives 5.

The standard deviation is √(variance)=√9=3. The value 81 squares the variance again, and √3 takes one square root too many. Keep the relation exact: the standard deviation is the positive square root of the variance.

For y=bx+a the standard deviation is multiplied by |b|, so it is 2×4=8. The +3 merely shifts everything and does not change the spread. The value 11 adds the constant as well, computing 2×4+3, and 16 confuses this with the factor b²=4 for the variance.

For y=bx+a the variance is multiplied by b², so it is 3²×2=18. The value 6 multiplies only by 3, which is the factor for the standard deviation. The −2 has no effect on the variance. Keep the two apart: the standard deviation scales by |b| and the variance by b².

The overall total is 4×5+6×10=20+60=80 points, so the mean is 80/10=8 points. The typical error is to ignore the group sizes and compute (5+10)/2=7.5. Weight by the sizes and work from the total.

The standard deviation is multiplied by |b|, so it is |−2|×2=4. A standard deviation measures spread and can never be negative, so −4 is wrong. Changing the sign leaves the width of the spread the same. The value 8 confuses this with the factor b²=4 for the variance.

Adding the same value to every item raises the mean by that amount, but each deviation from the mean is unchanged, so the variance and standard deviation stay the same. It helps to picture the whole shape of the spread simply sliding along.

From variance = mean of the squares − square of the mean, the mean of the squares is 6+4²=22. Hence the sum of the squares is 22×5=110. The value 50 comes from computing (6+4)×5. Use the formula the other way round: mean of the squares = variance + square of the mean.

Sloping upward to the right, meaning y tends to increase as x increases, is positive correlation, while sloping downward to the right is negative correlation. Note that the shape of a scatter plot alone tells you nothing about causation.

The correlation coefficient is the covariance divided by the product of the standard deviations, and it always lies in −1≦r≦1. The endpoints ±1 do occur, when all the points lie on a single straight line, so the equalities are included. Negative values also occur.

Since r is close to −1 the correlation is strongly negative, so the points gather near a line sloping downward to the right. If it sloped upward, r would be positive. Because r measures a linear relationship, a parabolic arrangement would not give a value such as −0.9.

When all points lie on one straight line the correlation coefficient is ±1, and for a downward slope r=−1. The value r=−1 is the strongest possible negative correlation, 0 means no correlation, and 1 corresponds to points on an upward-sloping line.

r=sxy/(sx·sy)=1.6/(√2×√2)=1.6/2=0.8. Dividing by the product of the variances, 2, instead of the standard deviations, √2, gives the wrong answer 0.4. Do not simply report the covariance 1.6.

Correlation is not evidence of causation. In this example it is natural to think that a third variable, temperature, is raising both. The strength of a correlation, the size of r, is a separate matter from whether there is causation.

Sales of hot coffee would be expected to fall as the temperature rises, which is a downward slope, that is negative correlation. Height and weight, study time and score, and population and number of shops all suggest positive correlation, where one rises as the other rises.

Under the assumption of fairness, the probability of the observed outcome, 1/256≒0.4%, is smaller than the threshold of 5%. Rather than believing that something very rare has happened, it is more natural to doubt the assumption, so we reject the hypothesis and judge that heads is favoured. The comparison is with the threshold of 5%, not with 50%.

A probability of 6% is at or above the threshold of 5%, so the outcome cannot be called rare and the hypothesis cannot be rejected. Failing to reject, however, is not a proof that the hypothesis is correct. Moving the threshold after seeing the result is a mistaken attitude.