Notes for April 1st and 3rd

Two averages from two populations


In the tests to see if the average of some numerical value is significantly different when comparing two populations, we need the averages, standard deviations and sizes of both populations. The score we use is a t-score and the degrees of freedom is the smaller of the two sample sizes minus 1.

Question: Do female Laney students sleep a number of hours each night different from male Laney students?

This uses data sets from a previous class. Here are the numbers for the students who submitted data, with the males listed as group #1. Again, let's assume a two-tailed test, since we don't have any information going in which should be greater, and let's do this test to 90% level of confidence.

With a test like this, we can arbitrary choose which set is the first set and which is the second. Let's do it so  x-bar1 > x-bar2. This way, our t-score will be positive, which is what the table expects.

H0: mu1 = mu2 (average hours of sleep are the same for males and females at Laney)

x-bar1 = 7.54
s1 = 1.47
n1 = 12


x-bar2 = 7.31
s2 = .94
n2 = 26

The degrees of freedom will be 12-1=11, and 10% in two tails gives us the thresholds of +/-1.796. Here is what to type into the calculator.

(7.54-7.31)/sqrt(1.47^2/26+.94^2/12)[enter]

0.4971...

2 tails__0.01_______0.02_____0.05_______0.10______0.20
11_______3.106_____2.718_____2.201_____1.796_____1.383


This number is less than every threshold, and so does not impress us enough to make us reject the null hypothesis. It's possible that larger samples would give us numbers that would show a difference, which if true would mean this example produced a Type II error, but we have no proof of that.

Matched pairs

Was the price of silver in 2007 significantly different than it was in 2008?

Side by side, we have two lists of prices of silver, the highest price in a given month in 2007, followed by the highest price in that same month in 2008. Take the differences in the prices and find the average and standard deviation. The size of the list is 12, so the degrees of freedom are 11. If we assume we did not know which year showed higher prices when we started this experiment, it make sense to make this a two-tailed test. Just for a change of pace, let us use the 90% confidence level.

Mo.___2007___2008
Jan.__13.45__16.23
Feb.__14.49__19.81
Mar.__13.34__20.67
Apr.__14.01__17.74
May___12.90__18.19
Jun.__13.19__17.50
Jul.__12.86__18.84
Aug.__12.02__15.27
Sep.__12.77__12.62
Oct.__14.17__11.16
Nov.__14.69__10.26
Dec.__14.76__10.66

Find the test statistic t, the threshold from Table A-3 and determine if we should reject H0, which in matched pairs tests is always that mu1 = mu2.

Answers in the comments.

Correlation
When we have a data set, sometimes we collect more than one variable of information about the units. For example, in class surveys taken in previous classes, among the numerical variables were the height in inches, the GPA, the opinion about the difficulty of the class, age and average hours of sleep per night.

A question about two variables is if they are related to one another in some simple way. One simple way is correlation, which can be positive or negative. Here is a general definition of each.

Positive correlation between two numerical variables, call them x and y, means that the high values of x tend to be paired with the high values of y, the middle values of x tend to be paired with the middle values of y and the low values of x tend to be paired with the low values of y.

The variables x and y show negative correlation if that the high values of x tend to be paired with the low values of y, the middle values of x tend to be paired with the middle values of y and the low values of x tend to be paired with the high values of y.

If we pick two variables at random, we do not expect to see correlation. We can write this as a null hypothesis, where the test statistic is r², the correlation coefficient. The sign of low correlation is r² = 0. The values of rx,y are always between -1, which means perfect negative correlation, and +1, which means perfect positive correlation. This means 0 <= r² <= 1.

The second orange sheet gives us threshold numbers for the 99% confidence level and 95% confidence level for correlation given the number of points n. For instance, when n = 5, the thresholds are .7709 for 95% confidence and .9197 for 99% confidence. This splits up the numbers from 0 to 1 into three regions.

0 < r² < .7709 We fail to reject the null hypothesis, which means not strong correlation.
.7709 < r² < .9197We reject the null hypothesis with 95% confidence, but not 99% confidence. This is fairly strong correlation.
.9197 < r² We reject the null hypothesis with 99% confidence this is very strong correlation.

Just like with any hypothesis test, we should decide the confidence level before testing. This is a two-tailed test, because whether correlation is positive or negative, the relationships between number sets can often give us vital scientific information.

There is an important warning: Correlation is not causation. Just because two number sets have a relation, it doesn't mean that x causes y or y causes x. Sometimes there is a hidden third factor that is the cause of both of the things we are looking at. Sometimes, it's random chance and there is no causative agent at all.


Here is a set of five points, listed as (x,y) in each case.

(1,1)
(2,2)
(3,4)
(4,4)
(6,5)

As we can see, the points are ordered from low to high in both coordinates, so we expect some correlation. If we input the points into our calculator, we get a value for r (which is the same as rx,y) of .933338696...,  and r² = .8712, which is strong positive correlation, but not very strong positive correlation. Assuming the 95% confidence level is good enough for us, we can use the a and b variables from out calculator to give us the equation of the line

yp = .797x + .649

This is called the predictor line (that's where the p comes from) or the line of regression or the line of least squares or the trendline. Any such line for a given data set has two important criteria it meets. It passes through the centroid (x-bar, y-bar), the center point of all the data, and it minimizes the sum of the absolute values of the residuals, which is |y - yp| for all points.

Let's find the absolute values of the residuals for each of the five points, using the rounded values of a and b.

Point (1,1): |1 - .797*1 - .649| = 0.446
Point (2,2): |2 - .797*2 - .649| = 0.243
Point (3,4): |4 - .797*3 - .649| = 0.96
Point (4,4): |4 - .797*4 - .649| = 0.163
Point (6,5): |5 - .797*6 - .649| = 0.431

As we can see, the point (3,4) is farthest from the line, while the point (4,4) is the closest. The centroid (3.2, 3.2) is exactly on the line if you use the un-rounded values of a and b, and even using the rounded values, the centroid only misses the line by .0006.

In class, we used five points, but the last point was (5,6) instead of (6,5). This changes the numbers. rx,y goes up to .973328527..., which is above the 99% confidence threshold. The formula for the new predictor line is

yp = 1.2x - .2

Where we see the difference in these two different examples is in the residuals.

Point (1,1): |1 - 1.2*1 + .2| = 0
Point (2,2): |2 - 1.2*2 + .2| = 0.2
Point (3,4): |4 - 1.2*3 + .2| = 0.6
Point (4,4): |4 - 1.2*4 + .2| = 0.6
Point (5,6): |6 - 1.2*5 + .2| = 0.2

The closest point is now exactly on the line, which is a rarity, but even the farthest away point is only .6 units away, closer than the farthest away on the line with the lower correlation coefficient.

As we get more points in our data set, we lower our threshold that shows correlation strength. This way, a few points that are outliers do not completely ruin the chances of the data showing correlation, though sometimes strong outliers can mess up the data set and the correlation coefficient gets so close to zero that we cannot reject the null hypothesis that the two variables are not simply related.

Notes for March 25 and 27

The first hypothesis tests we studied were checking to see if an experimental sample produced a value that was significantly different from some known value produced either by math or by earlier experiments.

For example, in the lady tasting tea, since she has two choices each time a mixture is given to her, the math would say that her chance of getting it right by just guessing is 50% or H0: p = .5. In testing psychic abilities, there are five different symbols on the cards, so random guessing should get the right answer 1 out of 5 times, or 20%, so H0: p = .2.

In a test for average human body temperature, the assumption of 98.6 degrees Fahrenheit being the average came from an experiment performed in the 19th Century.

We can also do tests by taking samples from two different populations. The null hypothesis, as always, is an equality, the assumption that the parameters from the two different populations are the same. As always, we need convincing evidence that the difference is significant to reject the null hypothesis, and we can choose just how convincing that evidence must be by setting the confidence level, which is usually either 90% or 95% or 99%.

Two proportions from two populations


Like with the one proportion test, the test statistic is a z-score. We have the proportions from the two samples, p-hat1 = f1/n1 and p-hat2 = f2/n2, but we also need to create the pooled proportion p-bar = (f1 + f2)/(n1 + n2).

Here's an example from the polling data from last year.

Question: Was John McCain's popularity in 2008 Iowa significantly different from his popularity in Pennsylvania?

Let's assume we don't know either way, so it will be a two tailed test. Polling data traditionally uses the 95% confidence level, so that means the z-score will have to be either greater than or equal to 1.96 or less than or equal to -1.96 for us to reject the null hypothesis. Here are our numbers, with Iowa as the first data set.

f1 = 263
n1 = 658
p-hat1 = .400

f2 = 283
n2 = 657
p-hat2 = .430

p-bar = (263+283)/(658+657) = .415 (q-bar = .585)

Type this into your calculator.

(.400-.430)/sqrt(.415x.585/658+.415x.585/657[enter]

The answer is -1.103..., which rounds to -1.10. This would say the difference we see in the two samples is not enough to convince us of a significant difference in popularity for McCain between the two states, so we would fail to reject the null hypothesis. In the actual election, McCain had 45.2% of the vote in Pennsylvania and 44.8% of the vote in Iowa, which are fairly close to equal.

Student's t-scores

Confidence intervals are used over and over again in statistics, most especially in trying to find out what value the parameter of the population has, when all we can effectively gather is a statistic from a sample. For numerical data, we aren't allowed to use the normal distribution table for this process, because the standard deviation sx of a sample isn't a very precise estimator of sigmax of the underlying population. To deal with this extra level of uncertainty, a statistician named William Gossett came up with the t-score distribution, also known as Student's t-score because Gossett published all his work under the pseudonym Student. He used this fake name for publishing to get around a ban on publishing in journals established by his superiors at the Guinness Brewing Company where he worked.

The critical t-score values are published on table A-3. The values depend on the Degrees of Freedom, which in the case of a single sample set of data is equal to n-1. For every degree of freedom, we could have another positive and negative t-score table two pages long, just like the z-score table, but that would take up way too much room, so statistics textbooks have reverted instead to publishing just the highlights. There are five columns on the table, each column labeled with a number from "Area in One Tail" and "Area in Two Tails". Let's look at the degrees of freedom row for the number 13.

1 tail___0.005______0.01_____0.025______0.05______0.10
2 tails__0.01_______0.02_____0.05_______0.10______0.20
13_______3.012_____2.650_____2.160_____1.771_____1.333


What this means is that if we have a sample of size 14, then the degrees of freedom are 13 and we can use these numbers to find the cut-off points for certain percentages. The formula for t-scores looks like the formula for z-scores, where z = (x - mux)/sigmax and t = z = (x - x-bar)/sx. Because we don't know sigmax, we use the t-score table. For example, the second column in row 13 is the number 2.650. This means that in a sample of 14, a Student's t-score of -2.650 is the cutoff for the bottom 1%, the t-score of +2.650 is the cutoff for the top 1% and the middle 98% is between t-scores of -2.650 and +2.650.

Using the t-score table

Let's say we have a t-score of 2.53 and n = 25, which means the degrees of freedom are 25-1 = 24. Here is the line of the t-score table that corresponds to d.f. = 24.

1 tail___0.005______0.01_____0.025______0.05______0.10
2 tails__0.01_______0.02_____0.05_______0.10______0.20
24_______2.797_____2.492_____2.064_____1.711_____1.318


What does this mean for our t-score of 2.53. If it was a z-score, the look-up table would give us an answer to four digits, .9943, which is a score that would be beyond the 99% confidence threshold for one tail (.9943 > .9500) but not beyond the confidence interval for 99% confidence and one tail because those thresholds are .9950 high and .0050 low. On the t-score table, all we can say is 2.53 is between 2.797 and 2.492, the closest scores on our line. In an two tailed test, it is beyond the 0.02 threshold (which would be 98% confidence, a number we don't use much) but not beyond the 99% threshold. In a one tailed (high) test our t-score is between 0.005 and 0.01, which means it passes the 99% threshold. Unlike the z-score table, t-scores only work with positive values, so if we get a negative t-score test, we follow these rules.

1. You have a negative t-score and the test is two tailed. Take the absolute value of the t-score and work with it.
2. You have a negative t-score and the test is one tailed low. Again, the absolute value will work.
3. You have a positive t-score and the test is one tailed low. This would be a problem, since only a negative t-score is useful in a one-tailed low test. You should fail to reject H0.

In the example below, we have yet another choice which always lets a one-tailed test be a one-tailed high test.

Two averages from two populations


In the tests to see if the average of some numerical value is significantly different when comparing two populations, we need the averages, standard deviations and sizes of both populations. The score we use is a t-score and the degrees of freedom is the smaller of the two sample sizes minus 1.

Question: Do female Laney students sleep a number of hours each night different from male Laney students?

This uses data sets from a previous class. Here are the numbers for the students who submitted data, with the males listed as group #1. Again, let's assume a two-tailed test, since we don't have any information going in which should be greater, and let's do this test to 90% level of confidence.

With a test like this, we can arbitrary choose which set is the first set and which is the second. Let's do it so  x-bar1 > x-bar2. This way, our t-score will be positive, which is what the table expects.

H0: mu1 = mu2 (average hours of sleep are the same for males and females at Laney)

x-bar1 = 7.54
s1 = 1.47
n1 = 12


x-bar2 = 7.31
s2 = .94
n2 = 26

The degrees of freedom will be 12-1=11, and 10% in two tails gives us the thresholds of +/-1.796. Here is what to type into the calculator.

(7.54-7.31)/sqrt(1.47^2/26+.94^2/12)[enter]

0.4971...

2 tails__0.01_______0.02_____0.05_______0.10______0.20
11_______3.106_____2.718_____2.201_____1.796_____1.383


This number is less than every threshold, and so does not impress us enough to make us reject the null hypothesis. It's possible that larger samples would give us numbers that would show a difference, which if true would mean this example produced a Type II error, but we have no proof of that.

Class notes for 5/13: Two population tests for proportions and averages

The first hypothesis tests we studied were checking to see if an experimental sample produced a value that was significantly different from some known value produced either by math or by earlier experiments.

For example, in the lady tasting tea, since she has two choices each time a mixture is given to her, the math would say that her chance of getting it right by just guessing is 50% or H0: p = .5. In testing psychic abilities, there are five different symbols on the cards, so random guessing should get the right answer 1 out of 5 times, or 20%, so H0: p = .2.

In a test for average human body temperature, the assumption of 98.6 degrees Fahrenheit being the average came from an experiment performed in the 19th Century.

We can also do tests by taking samples from two different populations. The null hypothesis, as always, is an equality, the assumption that the parameters from the two different populations are the same. As always, we need convincing evidence that the difference is significant to reject the null hypothesis, and we can choose just how convincing that evidence must be by setting the confidence level, which is usually either 90% or 95% or 99%.

Two proportions from two populations


Like with the one proportion test, the test statistic is a z-score. We have the proportions from the two samples, p-hat1 = f1/n1 and p-hat2 = f2/n2, but we also need to create the pooled proportion p-bar = (f1 + f2)/(n1 + n2).

Here's an example from the polling data from last year.

Question: Was John McCain's popularity in Iowa significantly different from his popularity in Pennsylvania?

Let's assume we don't know either way, so it will be a two tailed test. Polling data traditionally uses the 95% confidence level, so that means the z-score will have to be either greater than or equal to 1.96 or less than or equal to -1.96 for us to reject the null hypothesis. Here are our numbers, with Iowa as the first data set.

f1 = 263
n1 = 658
p-hat1 = .400

f2 = 283
n2 = 657
p-hat2 = .430

p-bar = (263+283)/(658+657) = .415 (q-bar = .585)

Type this into your calculator.

(.400-.430)/sqrt(.415x.585/658+.415x.585/657[enter]

The answer is -1.103..., which rounds to -1.10. This would say the difference we see in the two samples is not enough to convince us of a significant difference in popularity for McCain between the two states, so we would fail to reject the null hypothesis. In the actual election, McCain had 45.2% of the vote in Pennsylvania and 44.8% of the vote in Iowa, which are fairly close to equal.


Two averages from two populations


In the tests to see if the average of some numerical value is significantly different when comparing two populations, we need the averages, standard deviations and sizes of both populations. The score we use is a t-score and the degrees of freedom is the smaller of the two sample sizes minus 1.

Question: Do female Laney students sleep more hours each night than male Laney students?

We will take our data from the larger of the two class surveys, Data Set #2. Here are the numbers for the students who submitted data, with the females listed as group #1. Again, let's assume a two-tailed test, since we don't have any information going in which should be greater, and let's do this test to 90% level of confidence.

H0: mu1 = mu2 (average hours of sleep are the same for males and females at Laney)

x-bar1 = 7.31
s1 = .94
n1 = 26

x-bar2 = 7.54
s2 = 1.47
n2 = 12

The degrees of freedom will be 12-1=11, and 10% in two tails gives us the thresholds of +/-1.796. Here is what to type into the calculator.

(7.31-7.54)/sqrt(.94^2/26+1.47^2/12)[enter]

-0.4971...

This number is between the thresholds, and so does not impress us enough to make us reject the null hypothesis. It's possible that larger samples would give us numbers that would show a difference, which if true would mean this example produced a Type II error, but we have no proof of that.