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 5/11: Hypothesis testing for the mean of a population

t-scores and p values

If we have a z-score between -3.5 and +3.5, Table A-2 lets us find the p value associated with that z-score accurate to four decimal places. For example, if z = 1.71, the p value is .9564, which is to say that z-score is higher than 95.64% of data in a normally distributed set.

The t-score table is smaller, and to read a t-score correctly, we also need n, the size of the sample, because that gives us the Degrees of Freedom, which is n-1 in this case.

If n=10, then d.f. = 9, and the t-score table reads as follows.

___________________________Area in One Tail_____________

_______0.005______0.01______0.025______0.05______0.10__________df=9___3.250_____2.821______2.262_____1.833_____1.383__________
If t = 1.71, that value isn't on our table, but because it lies between the values associated with 0.05 and 0.10, that means that score is in the top 10% of scores, but not in the top 5%.

If instead n=30 and d.f. = 29, here are the t-score values.
___________________________Area in One Tail_____________
_______0.005______0.01______0.025______0.05______0.10__________df=29__2.756_____2.462______2.045_____1.699_____1.311__________

Now a t-score of 1.71 lies between 2.045 and 1.699, which means it is in the top 5%, but not the top 2.5%.

Like the z-score table, the t-score table is symmetric about the value t=0. If d.f.=29, t=-1.71 is a score in the bottom 5%, but not the bottom 2.5%.

Hypothesis testing for the mean of a population


Hypothesis testing for the mean of a population assumes we know the population mean from some previously obtained information. Perhaps that mean has changed over time or the previous information wasn't correct to begin with, but the null hypothesis assumes we know that mean, which we call mux. If we take a sample from the population, we will get the values x-bar, sx and n, and using those values and mux, we can get the t-score.

Just like with the hypothesis test for a proportion, the test can be one-tailed high, one-tailed low or two-tailed.

For example, if we were testing people who had studied using a special method and we were checking scores on a standardized test, we would only be impressed if the average went up, so a one-tailed high test would be appropriate.

If the experiment was dealing with a cholesterol drug, we would want to see a lower average reading, and a one-tailed low test would be used.

If we assume the average duration of a pop song on the charts today is the same as the duration of pop songs in the seventies, we can't assume beforehand if the new readings will be higher or lower, and would be surprised if the new average were significantly different in either direction, so a two-tailed test would be appropriate.

Here is some data we went over in class. In most textbooks, the 'normal' human body temperature is listed at 98.6 degrees Fahrenheit, based on the work of Dr. Carl Wunderlich back in the 19th Century. If we do a test, it should be a two-tailed test, since we would be surprised if the normal temperature is significantly higher or significantly lower than this. Since this is a medical experiment, let's use the 99% level of confidence.

The size of the sample was n=103, which means the degrees of freedom are 102. Our table doesn't have a listing for d.f.=102, and the next lowest available value is d.f.=100. Here are the table values for that row of Table A-3.

___________________________Area in Two Tails____________

_________0.01______0.02_______0.05______0.10______0.20__________df=100__2.626_____2.364______1.984_____1.660_____1.290__________
With a two tailed test at the 99% confidence level, this means we want the 0.01 column. The "middle" 99% of data lies between t-scores of -2.616 and 2.616. If the t-score lies in that range, we will fail to reject H0. If it is greater or equal to 2.616 or less than or equal to -2.616, we will reject H0.

The values from this study found that x-bar = 98.2 degrees and the standard deviation sx was 0.62. Plugging into our t-score equation from above, we get

t = (98.2-98.6)/0.62*sqrt(103) = -6.547671977... ~= -6.548.

We don't get an exact p value for a number so far away from zero, but if we look at outlier z-score table, we can roughly approximate that this p value is somewhere around 1 in 1,000,000,000. We can say with 99% confidence that the average body temperature is not 98.6 degrees, but probably close to the sample average of 98.2 degrees. Our p value shows we could qualify for even greater confidence with our statement, but very rarely do tests ask for more than 99% confidence, and changing the criteria after the fact is not recommended. Still, publishing this incredibly tiny p value will convince people who can read a statistical report that the evidence is very strong indeed.

This test also changed the idea of what should constitute a fever. Instead of one temperature of 100.4 degrees Fahrenheit being the absolute gauge, the temperature will fluctuate depending on the time of day, as do the normal temperature readings.

Class notes for 3/18

Tests before confidence interval for average of population

We have discussed finding a confidence interval for mux by using x-bar, the average of a sample taken from that population. Because we don't know the average of the population, we can't know the standard deviation of the population, since you need the average to compute the standard deviation. Because of this uncertainty, we use Student's t-scores instead of z-scores for the Confidence Level Multipliers for xx%, or CLMxx% for short.

There are times when we aren't allowed to use this method for approximation. We have two questions, and if we answer no to both of them, this method should not be used and only non-parametric methods are useful when dealing with this data set. (We will learn a few non-parametric methods later in the class, but most of these methods are taught in more advanced stats classes.)

Question #1: Is n > 30? If a sample is big enough, Gossett decided that his t-scores are going to give a reasonable approximation. This is always the first question because it is so easy to answer, and if we get a yes here, we don't even need to ask Question #2.

Question #2: Is the underlying data set normally distributed? It might be that we will be told the answer to this question without having to do any work to test this ourselves, and if we are told "yes" we can proceed. If we aren't told, here are two tests. There are other tests than can be done, but these two are easy and useful, and if the data set passes both these tests, we will assume we can move forward.


The z(mid-range) test. Recall that the mid-range is the average of the highest and lowest values in the data set. If one of these scores is a lot farther away from average than the other, we can consider that the data set may be too skewed to be reliable. What this test does is look to see if one extreme value (the highest or lowest) is more than one standard deviation farther away from average than the other extreme value. Understand that one standard deviation is an arbitrary value. The formula in this inequality is z(mid-range), which is where the test's name comes from.


The mean to median test. If the data is normally distributed, then the mean and median of the underlying population should be equal. If the mean and median of the sample are significantly different, then this test should produce a number greater than 1.353. (This is a simple version of a goodness-of-fit test, which we will be exploring in greater depth later in the semester.)

Once we have the average value x-bar, count how many values are greater than x-bar, and call this value above. We should expect that random samples from a normally distributed set should have about a 50%-50% split of values above and below average, so about n/2 in each group.

Again, the cut-off point 1.353 is arbitrary, and it is derived from the chi squared table. If we get a yes to both these questions, we can answer yes to the question about the normally distributed underlying population and proceed to find the interval for mux, which is the following inequality.

x-bar - CLMxx%*sx/sqrt(n) < mux < x-bar + CLMxx%*sx/sqrt(n)

Confidence interval for standard deviation of population


The confidence interval formula for sigmax is completely different than the formula for mux, and uses values from a new table, Table A-4, known as the chi squared table. Chi is pronounced like the chi in chiropractor, "kai", not "chee" or "chai".

Note that we don't have to test for normal distribution of the underlying data.

Let's give an example. If we have a set where n = 13 and sx = 3.21, we use n-1 as our degrees of freedom. Here is the line from Table A-4 for the row that corresponds to 12.

_____0.995___0.99___0.975___0.95___0.90___0.10____0.05___0.025____0.01___0.005
_12__3.074__3.571___4.404__5.226__6.304__18.549__21.026__23.337__26.217__28.299

The values of chi^2R and chi^2L are taken from the table as follows.

90% confidence: The right value is from the 0.05 column, the left value s from the 0.95 column. (Note that 0.95 - 0.05 = 0.90 or 90%)

95% confidence: The right value is from the 0.025 column, the left value s from the 0.975 column. (Note that 0.975 - 0.025 = 0.95 or 95%)99% confidence: The right value is from the 0.005 column, the left value s from the 0.995 column. (Note that 0.995 - 0.005 = 0.99 or 99%)

In this instance, here are the confidence intervals for each of the standard percentages of confidence.

90% confidence: sqrt(3.21^2*12/21.026) < sigmax < sqrt(3.21^2*12/5.226)95% confidence: sqrt(3.21^2*12/23.337) < sigmax < sqrt(3.21^2*12/4.404)99% confidence: sqrt(3.21^2*12/28.299) < sigmax < sqrt(3.21^2*12/3.074)

For practice, use your calculator to give the answers above, rounded to two places after the decimal. Answers in the comments.

More practice sets will be posted later today.

Class notes for 3/16

One of the major uses of statistics is to find confidence intervals for parameters from statistics, which means to find the approximate range of a value of the population from taking the associated value from a sample, and assigning a confidence level to that range. For the average of a numerical variable, this means finding a range for mux using x-bar, sx, n and looking up a Confidence Level Multiplier on the Student's t-score table, the fifth page of yellow handout, labeled Table A-3.

x-bar - CLMxx%*sx/sqrt(n) < mux < x-bar + CLMxx%*sx/sqrt(n)

There are some things we need to test before we create the interval. We have two questions to ask about the data, and if we get a yes answer to either question, we can proceed. If we get no answers to both, then we should not use this data set to create a confidence interval.

Question #1: Is n > 30?
Question #2: Is the underlying data set normally distributed?

Question #1 is easy to answer, and if the answer is yes, we don't have to bother with Question #2. It might be that you are simply told by the person who collected the data that the answer is yes. But if that is not the case, here are two tests the data set should pass to get a yes answer to Question #2. Both of these tests are using arbitrary cut-off points, the second test being a preview of tests we are going to do later in the class called goodness of fit.

Test #1: The Mid-range Outlier test. Recall that the mid-range is (high + low)/2. Take the z-score for the mid-range, (mid-range - x-bar)/sx, and if this value is more than 0.5 or less than -0.5, we will answer no to Question #2. If we answer yes, this means there is an outlier either high or low, that either the low value is one full standard deviation farther from the average than the high value, or vice versa. In a small data set, this can skew the statistics and make us less confident about the values being near the parameters from the underlying population.

Test #2: The Goodness of Fit test. We can do this test as long as n is 10 or more. If the data set has less than 10 entries and it failed Test #1, you shouldn't try to do the confidence interval. In this test, count the number of entries that are more than x-bar and call that value above. If the underlying data set is normally distributed, we would expect about half the entries would be above average and half below average, so about n/2 in each group. The Goodness of Fit test for this data set is done by plugging in these values into this formula, and making this check.

Is (above - n/2)^2 < face="courier new">1.2 4.7 13.1 3.2 8.8 18.2 3.5 12.5 4.8 8.2 0.7
17.1 0.2 11.4 3.2 8.6 6.3 17.4 17.1 17.8
1.2 0.2 0.1 8.9 4.0 8.6 0.7 2.8 0.5 0.4

x-bar = 6.85
sx = 6.17
n = 30
mid-range = (18.2 + 0.1)/2 = 9.15

Test #1: z(mid-range) = (9.15-6.85)/6.17 = 0.37

Because this is less than one half, we can answer yes and move on to test #2.

Test #2: number of values above average, which we call above = 13. n/2 = 15, so the test is

(13-15)^2/15 = 4/15, which rounds to 0.267 and is below 1.353, so we can answer yes to this and proceed.

n = 30 so degrees of freedom = 29. Here are the three most commonly used CLM values.

CLM90% = 1.699
CLM95% = 2.045
CLM99% = 2.756

Using the formula. here are the values for each of the confidence levels.

x-bar - CLMxx%*sx/sqrt(n) < mux < x-bar + CLMxx%*sx/sqrt(n)

90% confidence:
6.85 - 1.699*6.17/sqrt(30) < mux < style="font-size:100%;">mux < style="font-weight: bold;">


95% confidence:
6.85 - 2.045*6.17/sqrt(30) < mux 6.85 + 2.045*6.17/sqrt(30)
4.55 < mux < style="font-weight: bold;">99% confidence:
6.85 - 2.756*6.17/sqrt(30) < mux < 6.85 + 2.756*6.17/sqrt(30)
3.75 < mux < face="georgia">The sentence explaining the confidence interval would go as follows.

Given a sample of the carries per game of 30 running backs from the 2008 NFL season, we are 99% confident the true average number of carries per game for all NFL running backs that season is between 3.75 and 9.75.

Practice problem set

Here is the data for average yards per carry for the same 30 randomly selected NFL running backs for the 2008 season.

4.6 2.6 3.9 4.4 4.2 3.8 5.5 3.6 3.1 3.7
6.8 5.5 2.0 5.6 4.4 4.8 5.7 3.6 4.3 3.5
5.0 0.0 -2.0 2.8 3.3 4.4 2.0 6.0 1.3 4.0

Problem #1: Find x-bar and sx, rounded to two places after the decimal point.
Problem #2: Do the z(mid-range) test. Does the set pass or fail?
Problem #3: Do the second test for normal distribution. Does the set pass or fail?
Problem #4: Find the endpoints for the 95% confidence interval.
Problem #5: Write the sentence explaining the interval.

Answers in the comments.

Class notes for 3/11

The positive and negative z-score tables connect z-scores to proportions, numbers between 0 and 1 written to four decimal places of accuracy. For example, z = 1.23 corresponds to .8907, which means that 89.07% of data in a normally distributed set should have a z-score of 1.23 or less, while (100-89.07)% = 10.93% of data will have a z-score of 1.23 or more. Moreover, the normal curve is symmetric around z = 0. This means when z = -1.23, that is the cut-off point between the low 10.93% of the data and the high 89.07% of the data, just the opposite of the percentages that correspond to z = 1.23.



Critical values, or Confidence Level Multipliers (CLMxx%)

Sometimes, instead of being interested in the highest n% or the lowest, we will need to deal with the middle n% for what is known as a confidence interval. In the bottom right hand corner of the Positive z Scores table (first page of yellow handout), there is a small table labeled Common Critical Values. On tests and homework, I will call these Confidence Level Multipliers, or CLMxx%, where xx% is the percentage of confidence associated with the z-scores. The table gives the z-scores I will be calling CLM90%, CLM95% and CLM99%.

CLM90%: The end points are -1.645 and +1.645
CLM95%: The end points are -1.96 and +1.96
CLM99%: The end points are -2.575 and +2.575

For example, what this means for the case of CLM90% is that the low 5% of the data is below z=-1.645, the middle 90% is between z=-1.645 and z=+1.645, and the high 5% is above z=+1.645.


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, t = (x - x-bar)/sigmax, but we use the different look-up table to decide what these numbers mean. 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.

As the degrees of freedom get larger, the numbers in the columns get smaller. The last row has the label Large and reads as follows.

Large____2.576_____2.326_____1.960_____1.645_____1.282

These values exactly correspond to the z-score table. As the data set size gets larger, the differences between the z-distribution and the t-distribution shrink down to nothing.

When can we use t-scores?

Because we don't know sigmax, we are prohibited from using the z-score tables. But there are cases when we shouldn't use the t-score tables either. Here is the decision method.

Step 1: Is n at least 30? If yes, we are good. If no, go to Step 2.

Step 2: Is the sample normally distributed or can we assume the underlying data set is normally distributed? If yes, we can continue. If no, we would only be able to use non-parametric statistical techniques, which are not covered in this course.

For example, the cotinine data sets we have on the handout sheet have one set that looks normally distributed, the smokers data, and two that do not look normally distributed, the exposed and unexposed non-smoker data. Because n=40 for all the sets, we can use the t-score method because we answered yes to Step 1. If the data sets had less than 30 subjects, we would not be able to use the non-smoker data because we would have answered no to the questions from both Step 1 and Step 2.


The formula for a confidence interval for the mean of a population given the mean of a sample.

The confidence interval for the mean of a population given the mean of the sample is a formula that gives us two endpoints as follows

x-bar - CLMxx%*sx/sqrt(n) < mux < x-bar + CLMxx%*sx/sqrt(n)

Let's take an example. We have the heights of males from Data Set #1. The statistics from that set are as follows.

n = 18
x-bar = 71.17
sx = 2.57

The sample size is 18, which is less than 30, so since we answer no in Step 1, we have to move on to Step 2. Here we can answer yes, because we can assume that human height is a normally distributed set. Let's now move on to finding the 95% confidence interval for the average male height given this sample.

Since n = 18, degrees of freedom = 17, and the CLM95% = 2.110. This is because the area in two tails of 5% is the same as the middle region having 95%. Here is our formula with these numbers plugged in.

71.17 - 2.110*2.57/sqrt(18) < mux < style="font-style: italic;">mux < 72.4

From census information, we know the average height of males in the United States is 69.5 inches, so this interval does not contain the true value. This semester, both the data sets had averages for male heights well above average, largely because of how many football and baseball players are enrolled in the classes, as well as a few other tall males who are not on the sports teams. This is a good example that a confidence interval is NOT a promise of a correct answer, and that statistical methods include confidence intervals exactly for this reason. (note: If we changed the interval to the 99% confidence interval, our confidence level multiplier would be 2.898 instead of 2.110, and the 99% confidence interval would contain the correct answer.)