# Week 22: The t Test for Two Dependent Samples

Today we’re going to talk about our first test involving dependent samples: the t test for two dependent samples!

When Would You Use It?
The t test for two dependent samples is a parametric test used to determine if two dependent samples represent two populations with different mean values.

What Type of Data?
The t test for two dependent samples requires interval or ratio data.

Test Assumptions

• If each sample contains the same subjects (e.g., a setup that involves testing subjects at time A and then again at time B), order effects must be controlled for.
• If a matched subjects design is employed, within each pair of matched subjects, the two subjects must be randomly assigned to one of the two experimental conditions.

Test Process
Step 1: Formulate the null and alternative hypotheses. The null hypothesis claims that the two sample means are equal. The alternative hypothesis claims otherwise (one population mean is greater than the other, less than the other, or that the means are simply not equal).

Step 2: Compute the t-score. The t-score is computed as follows: Step 3: Obtain the p-value associated with the calculated t-score. The p-value indicates the probability of a difference in the two sample means that is equal to or more extreme than the observed difference between the sample means, under the assumption that the null hypothesis is true.

Step 4: Determine the conclusion. If the p-value is larger than the prespecified α-level, fail to reject the null hypothesis (that is, retain the claim that the population means are equal). If the p-value is smaller than the prespecified α-level, reject the null hypothesis in favor of the alternative.

Example
For the data for this example, I decided to compare the age at which the internet thought I would die in 2011 to the age at which the internet thinks I would die in 2016. That is, I took 8 different online “death tests” in 2011, then re-took them this evening. The data are in the following table: I wanted to see if there was a significant difference in the average “age of death” between 2011 and 2016, based on what information I gave these tests. Here, n = 8. Set α = 0.05.

H0: µ2011 = µ2016 (or µ2011 – µ2016 = 0)
Ha: µ2011 ≠ µ2016 (or µ2011 – µ2016 ≠ 0)

Computations: Since our p-value is smaller than our alpha-level, we reject Hand claim that the population means are significantly different (with evidence in favor of the mean being higher in 2011).

Example in R

```dat=read.table('clipboard',header=T) #"dat" is the name of the imported raw data

diffs = y2011 - y2016

n=length(diffs)

D = sum(diffs)

sdev = sd(diffs)

t = D/(sdev/sqrt(n))        #t score

pval = pt(t, n-1)*2         #p-value

#pt calculates the left-hand area

#multiply by two because it is a two-sided test```

(Here’s a list of the tests, by the way.)

# Week 21: The z Test for Two Independent Proportions

Hello, all! Today we’re going to talk about a two sample test involving proportions. Specifically, we’re going to talk about the z test for two independent proportions!

When Would You Use It?
The z test for two independent proportions is a nonparametric test used to determine if, in a 2 x 2 contingency table, the underlying populations represented by the samples have equal proportions of observations in one of the two categories of the dependent variable.

What Type of Data?
The z test for two independent proportions requires categorical or nominal data.

Test Assumptions

• The data represent a random sample of independent observations.

Test Process
Step 1: Formulate the null and alternative hypotheses. The data appropriate for this type of test is usually summarized in a 2 x 2 table (see the example below to get a better understanding of this). The null hypothesis claims that for the category of interest of the dependent variable, the proportion of observations from the first category of the independent variable that belong to the category of interest is equal to the proportion of observations from the second category of the independent variable that belong to the category of interest.

Step 2: Compute the test statistic. The test statistic here is a z-score and is computed as follows: Step 3: Obtain the p-value associated with the calculated z-score. The p-value indicates the probability of observing a difference in proportions as extreme or more extreme than the observed sample difference, under the assumption that the null hypothesis is true.

Step 4: Determine the conclusion. If the p-value is larger than the prespecified α-level, fail to reject the null hypothesis (that is, retain the claim that the proportions are equal in both groups of the independent variable). If the p-value is smaller than the prespecified α-level, reject the null hypothesis in favor of the alternative.

Example
For today’s example, I wanted to see if there was a significant difference in the proportion of gold medals for European countries versus the rest of the world in the 2012 London Summer Olympics. I sampled a total of 55 countries (all countries that won at least one gold medal), then tallied the number of gold medals, the number of non-gold medals, and whether or not the country was in Europe. This data is summarized in the following table: Let’s test the claim that the proportion of gold medals for European and non-European countries is different. Set α = 0.05.

H0: π1 = π2
Ha: π1 ≠ π2

Here, n1 = 353, n2 = 516, p1 = 0.323, and p2 = 0.353. The values of p and z and the resulting p-value are calculated as: Since our p-value is larger than our alpha-level (0.3632 > 0.05), we fail to reject H0 and claim that the proportions are equal in the population.

Example in R
This example assumes that your data are in columns, with one column containing the number of gold medals per country, one column containing the number of total medals per country, and one coded column telling you whether a country belongs to Europe or not.

```dat=read.table('clipboard', header=T) #'dat' is the name of the imported raw data
euro = subset(dat,europe == "y")
non = subset(dat,europe == "n")
a = sum(euro\$gold)
b = sum(euro\$total) - a
c = sum(non\$gold)
d = sum(non\$total) - c
n1 = sum(a + b)
n2 = sum(c + d)
goldsum = sum(dat\$gold)
othersum = sum(total)
p1= a/n1
p2 = c/n2
p = (a + c)/(n1 + n2)
z = (p1 - p2)/(sqrt((p*(1-p))*((1/n1)+(1/n2))))
pval = (pnorm(z))*2          #p-value
#pnorm calculates the left-hand area
#multiply by two because it is a two-sided test```

# Week 20: The Chi-Square Test of Independence

Hello, people! Today we’re going to talk about another chi-square test: the chi-square test of independence!

When Would You Use It?
The chi-square test of independence is a nonparametric test used to determine if the two variables represented in a contingency table are independent of one another.

What Type of Data?
The chi-square test of independence requires categorical or nominal data.

Test Assumptions

• The data represent a random sample of independent observations.
• The expected frequency of each cell in the contingency table is at least 5.

Test Process
Step 1: Formulate the null and alternative hypotheses. The data appropriate for this type of test is usually summarized in an r x c table, where r is the number of rows of the table and c is the number of columns of the table (see the example below to get a better understanding of this). The null hypothesis claims that the in the population from which the sample was drawn, the observed frequency of each cell in the table is equal to the respective expected frequencies of each cell in the table. The alternative hypothesis claims that for at least one cell, the observed and expected frequencies are different.

Step 2: Compute the test statistic. The test statistic here, unsurprisingly, a chi-square value. To compute this value, use the following equation: Eij, the expected cell count for the ijth cell, is calculated as follows: Step 3: Obtain the critical value. The critical value can be obtained using a chi-square table (such as this one here). Find the column corresponding to your specified alpha-level, then find the row corresponding to your degrees of freedom. The degrees of freedom is calculated as df = (r – 1)(c – 1), where r is the number of rows in the table and c is the number of columns in the table. Compare your obtained chi-square value to the value at the intersection of your selected alpha-level and degrees of freedom.

Step 4: Determine the conclusion. If your test statistic is equal to or greater than the table value, reject the null hypothesis. If your test statistic is smaller than the table value, fail to reject the null (that is, claim that the observed cell frequencies match those of the expected cell frequencies).

Example
The example I’ll use today involves looking at some Nobel Prize data. Specifically, I want to see if the category of Nobel Prize (chemistry, physics, etc.) is independent of gender. The data come from here. The sample size I used was n = 761; I omitted organizations who had won the award and just looked at individuals. I also chose to omit the “Economics” category, as that had been the most recently added and did not have a lot of observations for either gender yet. Set α = 0.05.

H0: Nobel Prize category is independent of gender
Ha: Nobel Prize category is not independent of gender

Observed counts are in the following table: The expected cell counts, as calculated by the Eij formula above, are displayed in the following table: Calculating the chi-square value gives us: The degrees of freedom for this test is df = (5 – 1)(2 – 1) = 4, which gives us a critical chi-square value of 9.488 by the table. Since our calculated chi-square value, 32.894, is larger than the table value, this suggests that we reject the null and claim that prize category and gender are not independent.

# Week 19: The Chi-Square Test for Homogeneity

What’s up, y’all? Today we’re going to talk about the chi-square test for homogeneity!

When Would You Use It?
The chi-square test for homogeneity is a nonparametric test used to determine whether or not r independent samples, categorized on a single dimension, are homogeneous with respect to the proportion of observations in each of the c categories.

What Type of Data?
The chi-square test for homogeneity requires categorical or nominal data.

Test Assumptions

• The data represent a random sample of independent observations.
• The expected frequency of each cell in the contingency table is at least 5.

Test Process
Step 1: Formulate the null and alternative hypotheses. The data appropriate for this type of test is usually summarized in an r x c table, where r is the number of rows of the table and c is the number of columns of the table (see the example below to get a better understanding of this). The null hypothesis claims that the in the population from which the sample was drawn, the observed frequency of each cell in the table is equal to the respective expected frequencies of each cell in the table. The alternative hypothesis claims that for at least one cell, the observed and expected frequencies are different.

Step 2: Compute the test statistic. The test statistic here, unsurprisingly, a chi-square value. To compute this value, use the following equation: Eij, the expected cell count for the ijth cell, is calculated as follows: Step 3: Obtain the critical value. The critical value can be obtained using a chi-square table (such as this one here). Find the column corresponding to your specified alpha-level, then find the row corresponding to your degrees of freedom. The degrees of freedom is calculated as df = (r – 1)(c – 1), where r is the number of rows in the table and c is the number of columns in the table. Compare your obtained chi-square value to the value at the intersection of your selected alpha-level and degrees of freedom.

Step 4: Determine the conclusion. If your test statistic is equal to or greater than the table value, reject the null hypothesis. If your test statistic is smaller than the table value, fail to reject the null (that is, claim that the observed cell frequencies match those of the expected cell frequencies).

Example
The example for this test comes from Amazon. Specifically, I want to see if the number of 4+ star ratings was homogeneous across the six different price ranges for laptop computers. I chose a random sample of n = 15 from each of the six price ranges and determined how many of the 15 laptops selected had four or more stars for their average review. The observed counts are in the following table: Set α = 0.05.

H0: The proportion of 4+ star ratings is homogeneous across all price ranges
Ha: The proportion of 4+ star ratings is not homogeneous across all price ranges

The expected cell counts, as calculated by the Eij formula above, are displayed in the following table: Calculating the chi-square value gives us: The degrees of freedom for this test is df = (6 – 1)(2 – 1) = 5, which gives us a critical chi-square value of 11.070 by the table. Since our calculated chi-square value, 3.54, is smaller than the table value, this suggests that we fail to reject the null and claim that the proportion of 4+ star ratings is the same for each price category.

# Week 18: The Siegel-Tukey Test for Equal Variability

Today we’re going to talk about another nonparametric test: the Siegel-Tukey test for equal variability!

When Would You Use It?
The Siegel-Tukey test for equal variability is a nonparametric test used to determine if two independent samples represent two populations with different variances.

What Type of Data?
The Siegel-Tukey test for equal variability requires ordinal data.

Test Assumptions

• Each sample is a simple random sample from the population it represents.
• The two samples are independent.
• The underlying distributions of the samples have equal medians.

Test Process
Step 1: Formulate the null and alternative hypotheses. The null hypothesis claims that the two population variances are equal. The alternative hypothesis claims otherwise (one variance is greater than the other, or that they are simply not equal).

[Note that from here on out, the calculations are exactly the same as for the Mann-Whitney U test. The only thing that differs is how the data are ranked.]

Step 2: Compute the test statistics: U1 and U2. Since this is best done with data, please see the example shown below to see how this is done.

Step 3: Obtain the critical value. Unlike most of the tests we’ve done so far, you don’t get a precise p-value when computing the results here. Rather, you calculate your U values and then compare them to a specific value. This is done using a table (such as the one here). Find the number at the intersection of your sample sizes for both samples at the specified alpha-level. Compare this value with the smaller of your U1 and U2 values.

Step 4: Determine the conclusion. If your test statistic is equal to or less than the table value, reject the null hypothesis. If your test statistic is greater than the table value, fail to reject the null (that is, claim that the variances are equal in the population).

Example
Today’s data come from my 2012 music selection. I wanted to see if the median play counts for two genres—pop and electronic—were the same. I chose these two because I think most of my favorite songs are of one of the two genres. To keep things relatively simple for the example, I sampled n = 8 electronic songs and n = 8 pop songs. Set α = 0.05.

H0: σ2pop = σ2electronic
Ha: σ2pop ≠ σ2electronic

The following table shows several different columns of information. I will explain the columns below. Column 1 is the genre of each song.
Column 2 is the play count for each song, ranked from least to greatest
Column 3 is the rank of each play count. In order to obtain the ranks for this test, start by giving a rank of “1” to the lowest play count value. Then a rank of “2” to the highest play count value, a rank of “3” to the second highest play count value, a rank of “4” to the second lowest play count value, etc. (that is, assign ranks by alternating from one extreme to the other).

To compute U1 and U2, use the following equations: So here, The test statistic itself is the smaller of the above values; in this case, they’re both the same, so we get U = 32. In the table, the critical value for n1 = 8 and n2 = 8 and α = 0.05 for a two-tailed test is 13. Since U > 13, we fail to reject the null and retain the claim that the population variances are equal.

Example in R
No R example this week; most of this is easy enough to do by hand for a small-ish sample.

# AP

Yo, fools!

Awhile back (2011), I decided to try taking an online practice version of the AP Statistics exam. I think it went okay (I unfortunately didn’t record my exact final score on the multiple choice, ‘cause I’m a loser), but I decided to try it again because, you know, I’ve actually got a better background for it now.

Well, not the same practice test. I found a newer one (2010 I think?) on which there were fewer multiple choice questions but the questions were decidedly more difficult. This time I ended up getting 16/18 correct (I shouldn’t have missed one of the ones I did; I just completely blanked on how to do it, even though I’ve taught it in lab like 20 times).

Anyway. If you want to give it a shot, it’s located here!

# Week 17: The Moses Test for Equal Variability

Today we’re going to talk about another nonparametric test: the Moses Test for equal variability!

When Would You Use It?
The Moses Test for equal variability is a nonparametric test used to determine if two independent samples represent two populations with different variances.

What Type of Data?
The Moses Test for equal variability requires ordinal data.

Test Assumptions

• Each sample is a simple random sample from the population it represents.
• The two samples are independent.

Test Process
Step 1: Formulate the null and alternative hypotheses. The null hypothesis claims that the two population variances are equal. The alternative hypothesis claims otherwise (one variance is greater than the other, or that they are simply not equal).

Step 2: Compute the test statistics: U1 and U2. Since this is best done with data, please see the example shown below to see how this is done. [Note that the test statistic calculations are exactly the same as for the Mann-Whitney U test. The only thing that differs is the ranking procedure.]

Step 3: Obtain the critical value. Unlike most of the tests we’ve done so far, you don’t get a precise p-value when computing the results here. Rather, you calculate your U values and then compare them to a specific value. This is done using a table (such as the one here). Find the number at the intersection of your sample sizes for both samples at the specified alpha-level. Compare this value with the smaller of your U1 and U2 values.

Step 4: Determine the conclusion. If your test statistic is equal to or less than the table value, reject the null hypothesis. If your test statistic is greater than the table value, fail to reject the null (that is, claim that the variances are equal in the population).

Example
Today’s data come from my 2012 music selection (I’ll use this data next week, too!). I wanted to see if the median play counts for two genres—pop and electronic—were the same. I chose these two because I think most of my favorite songs are of one of the two genres. To keep things relatively simple for the example, I sampled n = 8 electronic songs and n = 8 pop songs. Set α = 0.05.

H0: σ2pop = σ2electronic
Ha: σ2pop ≠ σ2electronic

Here are the raw data: The following tables show several different columns of information. I will explain the columns below. The “Subsample” column: to obtain the rankings for this test, first divide the n1 scores in sample 1 into m1 subsamples (m1 > 1), with each subsample comprised of k scores. Then divide the n2 scores of sample 2 into m2 subsamples (m2 >2), with each subsample comprised of k scores. To form the subsamples, employ sampling without replacement within each of the samples. Ideally, m1, m2, and k should be chosen such that (m1)(k) = n1 and (m2)(k) = n2 if at all possible. Here, m1 = m2 = 4, and k = 2. The “subsample” columns list the four subsamples of play counts for each genre.

The second column contains the average of the k values for a given subset. It’s just the average of the values in the “subsample” column.

The third column contains the differences between each subsample value (X) and the subsample’s mean.

The fourth column is just the third column’s values squared.

The fifth column contains the sum of the values in the fourth column.

The sixth column contains the rank of the value in the fifth column over both genres. The smallest values is ranked as 1 and the largest is ranked as 8 (in this case).

To compute U1 and U2, use the following equations: So here, The test statistic itself is the smaller of the above values,  so we get U = 81. In the table, the critical value for n1 = 8 and n2 = 8 and α = 0.05 for a two-tailed test is 13. Since U > 13, we fail to reject the null and retain the claim that the population variances are equal.

```Example in R
attach(x)
elects=subset(x,genre=="Electronic")
pops=subset(x,genre=="Pop")
group1=matrix(rep(NaN,8),nrow=4)
group2=matrix(rep(NaN,8),nrow=4)

group1=as.matrix(sample(elects[,2],8,replace=F)) #subgroups for sample 1
sub1.1=group1[1:2,]
sub1.2=group1[3:4,]
sub1.3=group1[5:6,]
sub1.4=group1[7:8,]

group2=as.matrix(sample(pops[,2],8,replace=F))   #subgroups for sample 1
sub2.1=group2[1:2,]
sub2.2=group2[3:4,]
sub2.3=group2[5:6,]
sub2.4=group2[7:8,]
samps=rbind(sub1.1,sub1.2,sub1.3,sub1.4,sub2.1,sub2.2,sub2.3,sub2.4)

xbars=rep(NaN,8)                                 #column 2
for(i in 1:8){
xbars[i]=mean(samps[i,])}
xbars=as.matrix(xbars)

diffs1=rep(NaN,8)                                #column 3
for(i in 1:8){
diffs1[i]=(samps[i,1]-xbars[i])}
diffs2=rep(NaN,8)
for(i in 1:8){
diffs2[i]=(samps[i,2]-xbars[i])}

diffs=cbind(diffs1,diffs2)                       #column 4

diffs2=diffs^2                                   #column 5

sumdif=rep(NaN,8)
for(i in 1:8){
sumdif[i]=diffs2[i,1]+diffs2[i,2]}               #column 6```

# Week 16: The Kolmogorov-Smirnov Test for Two Independent Samples

[This is coming out on a Monday ’cause I was super busy yesterday and had no time to make this/post it.]

Today’s test is a non-parametric test for two samples: the Kolmogorov-Smirnov test for two independent samples!

When Would You Use It?
The Kolmogorov-Smirnov test for two independent samples is a nonparametric test used to determine if two independent samples represent two different populations.

What Type of Data?
The Kolmogorov-Smirnov test for two independent samples requires ordinal data.

Test Assumptions

• All of the observations in the samples are randomly selected and independent of one another.
• The scale of the measurement is ordinal.

Test Process
Step 1: Formulate the null and alternative hypotheses. The null hypothesis claims that the the distribution underlying the population for one sample is the same as the distribution underlying the population for the other sample. The alternative claims that the distributions are not the same.

Step 2: Compute the test statistic. The test statistic, in the case of this test, is defined by the point that represents the greatest vertical distance at any point between the cumulative probability distribution constructed from the first sample and the cumulative probability distribution constructed from the second sample. I will refer you to the example shown below to show how these calculations are done in a specific testing situation.

Step 3: Obtain the critical value. Unlike most of the tests we’ve done so far, you don’t get a precise p-value when computing the results here. Rather, you calculate your test statistic and then compare it to a specific value. This is done using a table. Find the number at the intersection of your sample sizes for your specified alpha-level. Compare this value with your test statistic.

Step 4: Determine the conclusion. If your test statistic is equal to or larger than the table value, reject the null hypothesis (that is, claim that the distribution of the data is inconsistent with the hypothesized population distribution). If your test statistic is less than the table value, fail to reject the null.

Example
For this test’s example, I want to use some of my music data from 2012. I know that I tend to listen to music from the “electronic” genre and from the “dance” genre fairly equally, so I want to determine, based on play count, if I can say that the population distributions for these genres are similar. To keep things simple, I will use nelectronic = 6 and ndance = 6.

H0: Felectronic(X) = Fdance(X) for all values of X
Ha: Felectronic(X) ≠ Fdance(X) for at least one value of X

Computations:
For the computations section of this test, I will display a table of values for the data and describe what the values are and how the test statistic is obtained. Column A and Column C, together, show the ranked values of the play counts for electronic (Column A) and dance (Column C).
Column B represents the cumulative proportion in the sample for each play count in Column A. For example, for the play count = 7, the cumulative proportion of that value is just 1/6, since there is no smaller value in Column A.
Column D represents the same thing as column B, except for Column C.
Column E is Column B – Column D.

The test statistic is obtained by determining the largest value from Column E. Here, the test statistic is .5. This value is compared to the critical value at α = 0.05, n1 = 6, n2 = 6, which is .667. Since our test statistic is not larger than our critical value, we fail to reject the null and claim that the distributions of play counts for electronic and dance are similar.

Example in R
No R example this week, as this is pretty easy to do by hand, especially with having to rank things.

# Week 15: The Mann-Whitney U Test

Today we’re going to talk about another nonparametric test: the Mann-Whitney U test!

When Would You Use It?
The Mann-Whitney U test is a nonparametric test used to determine if two independent samples represent two populations with different medians.

What Type of Data?
The Mann-Whitney U test requires ordinal data.

Test Assumptions

• Each sample is a simple random sample from the population it represents.
• The two samples are independent.
• The original scores obtained are continuous random variables (which are later ranked).
• The underlying distributions of the samples are identical in shape (but do not necessarily have to be normal).

Test Process
Step 1: Formulate the null and alternative hypotheses. The null hypothesis claims that the two population medians are equal. The alternative hypothesis claims otherwise (one median is greater than the other, or that they are simply not equal).

Step 2: Compute the test statistics: U1 and U2. Since this is best done with data, please see the example shown below to see how this is done.

Step 3: Obtain the critical value. Unlike most of the tests we’ve done so far, you don’t get a precise p-value when computing the results here. Rather, you calculate your U values and then compare them to a specific value. This is done using a table (such as the one here). Find the number at the intersection of your sample sizes for both samples at the specified alpha-level. Compare this value with the smaller of your U1 and U2 values.

Step 4: Determine the conclusion. If your test statistic is equal to or less than the table value, reject the null hypothesis. If your test statistic is greater than the table value, fail to reject the null (that is, claim that the medians are equal in the population).

Example
Today’s data come from my 2012 music selection. I wanted to see if the median play counts for two genres—pop and electronic—were the same. I chose these two because I think most of my favorite songs are of one of the two genres. To keep things relatively simple for the example, I sampled n = 8 electronic songs and n = 8 pop songs. Set α = 0.05.

H0: θpop = θelectronic
Ha: θpop ≠ θelectronic

The following table shows several different columns of information. I will explain the columns below. Column 1 is the genre of each song.
Column 2 is the play count for each song.
Column 3 is the overall rank of the play count, regardless of which genre it came from.

If there had been ties, I would have summed the number of ranks that were taken by the ties, and then divide that value by the number of ties.

To compute U1 and U2, use the following equations: So here, The test statistic itself is the smaller of the above values; in this case, we get U = 28. In the table, the critical value for n1 = 8 and n2 = 8 and α = 0.05 for a two-tailed test is 13. Since U > 13, we fail to reject the null and retain the claim that the population medians are equal.

Example in R
No R example this week; most of this is easy enough to do by hand for a small-ish sample.

# Week 14: The Median Absolute Deviation Test for Identifying Outliers

Today we’re going do something a little bit different by talking about the median absolute deviation test for identifying outliers!

When Would You Use It?
The median absolute deviation test for identifying outliers is used to determine whether or not a specific sore in a sample of n observations should be classified as an outlier.

What Type of Data?
The median absolute deviation test for identifying outliers requires interval or ratio data.

Test Assumptions
None listed.

Test Process
The equation employed in this test is as follows:

Step 1: Compute the median M of the dataset.

Step 2: Compute the median absolute deviation, or MAD. To do so:

a) Calculate the absolute values of the difference between each score and the median.
b) Arrange these absolute deviations in order from lowest to highest.
c) Find the median of these absolute deviations; this is the MAD value.

Step 3: Determine the Max value. While the selection of this value is somewhat arbitrary, a recommended value is to set Max = 5. This is because if the data are assumed to come from an approximately normal distribution, this value will be very likely to identify extreme or outlier scores.

Step 4: Plug in each X value into the equation to determine if it is an outlier. X is an outlier if the left-hand side of the equation exceeds the Max value. If doing this test by hand, the best way to go about this step is to start with the X that deviates the most from the median and work down from there, but if using a program, it’s easy enough to just test them all at once.

Example
Today’s data is from my 2013 music. I have the lengths (in seconds) of all n = 365 songs from that year, and I want to determine which values are outliers.

Computations:
M = 226

With Max = 5, I found that the songs with the following lengths are outliers:

891
564
636
516
580
534
597
574
537
595
486

This was done using R; the code is below.

Example in R

```x = read.table('clipboard', header=T) #data
M = median(x)
absdev = abs(x-M)
Max = 5
for (i in 1:length(x)){       # if an x value is an outlier, this loop will
dev = (abs(x[i]-M))/MAD       # print its value
if (dev > 5) { print(x[i]) }}```

# Week 13: F Test for Two Population Variances

Today we’re going to talk about variances. Specifically, the F test for two population variances!

When Would You Use It?
The F test for two population variances is a parametric test used to determine if two independent samples represent two populations with homogeneous (similar) variances.

What Type of Data?
The F test for two population variances requires interval or ratio data.

Test Assumptions
None listed.

Test Process
Step 1: Formulate the null and alternative hypotheses. The null hypothesis claims that the two population variances are equal. The alternative hypothesis claims otherwise (one population variance is greater than the other, less than the other, or that the variances are simply not equal).

Step 2: Compute the test statistic, an F value. The test statistic is computed as follows: Step 3: Obtain the p-value associated with the calculated F value. The p-value indicates the probability of a difference in the two sample variances that is equal to or more extreme than the observed difference between the sample variances, under the assumption that the null hypothesis is true. The two degrees of freedom associated with the F value are df1 = n1-1 and df2 = n2-1, where n1 and n2 are the respective sample sizes of the first and second sample.

Step 4: Determine the conclusion. If the p-value is larger than the prespecified α-level, fail to reject the null hypothesis (that is, retain the claim that the population means are equal). If the p-value is smaller than the prespecified α-level, reject the null hypothesis in favor of the alternative.

Example
The data for this example come from my walking data from 2013 and 2015. I want to see if there is a significant difference in the mileage variance for these two years (in other words, was I less consistent with the length of my walks in one year versus the other? Set α = 0.05.

H0: σ12 = σ22 (or σ12 – σ22 = 0)
Ha: σ12 ≠ σ22 (or σ12 – σ22 ≠ 0)

Computations: Since our p-value is smaller than our alpha-level, we reject H0 and claim that the population variances are significantly different (with evidence in favor of the variance being higher for 2015).

Example in R

```year2013=read.table('clipboard', header=F) #data from 2013
s1=var(year2013)
s2=var(year2015)
df1=length(year2013)-1
df2=length(year2015)-1
F=s2/s1                        #test statistic
pval = (1-pf(F, df2, df1))     #p-value```

# Week 12: The t Test for Two Independent Samples

Today we’re going to talk about our first test involving two samples: the t test for two independent samples!

When Would You Use It?
The t test for two independent samples is a parametric test used to determine if two independent samples represent two populations with different mean values.

What Type of Data?
The t test for two independent samples requires interval or ratio data.

Test Assumptions

• Each sample is a simple random sample from the populations they represent.
• The distributions underlying each of the populations are normal.
• The variances of the underlying populations are equal (homogeneity of variance; a formal test for this will come in a later week).

Test Process
Step 1: Formulate the null and alternative hypotheses. The null hypothesis claims that the two sample means are equal. The alternative hypothesis claims otherwise (one population mean is greater than the other, less than the other, or that the means are simply not equal).

Step 2: Compute the t-score. The t-score is computed as follows: Step 3: Obtain the p-value associated with the calculated t-score. The p-value indicates the probability of a difference in the two sample means that is equal to or more extreme than the observed difference between the sample means, under the assumption that the null hypothesis is true.

Step 4: Determine the conclusion. If the p-value is larger than the prespecified α-level, fail to reject the null hypothesis (that is, retain the claim that the population means are equal). If the p-value is smaller than the prespecified α-level, reject the null hypothesis in favor of the alternative.

Example
The data for this example come from the midterm scores of my lab section for STAT 213. While lab attendance is technically optional, the students’ attendance is recorded for each lab (if they show up to lab, they basically get additional instructional materials unlocked to help them study more).

I wanted to see if there was a significant difference in the average midterm score for students who attended lab at least half the time (sample 1) and students who attended lab less than half the time (sample 2). Specifically, I wanted to test the claim that attending lab more frequently was associated with a higher midterm score. Here, n1 = 17 and n2 = 13. Set α = 0.05.

H0: µ1 = µ2 (or µ1 – µ2 = 0)
Ha: µ1 > µ2 (or µ1 – µ2 > 0)

Computations: Since our p-value is smaller than our alpha-level, we reject H0 and claim that the population means are significantly different (with evidence in favor of the mean being higher for those attending labs more often).

Example in R

```x=read.table('clipboard', header=T)
attach(x)
x1=subset(x,attended==1)[,1]                 #attended lab
x2=subset(x,attended==0)[,1]                 #did not attend lab
n1=length(x1)
n2=length(x2)
xbar1=mean(x1)
xbar2=mean(x2)
s1=((sum(x1^2)-(((sum(x1))^2)/n1))/(n1-1))
s2=((sum(x2^2)-(((sum(x2))^2)/n2))/(n2-1))
t = (xbar1 - xbar2)/sqrt(((((n1-1)*s1)
+((n2-1)*s2))/(n1+n2-2))*((1/n1)+(1/n2))) #test statistic
pval = (1-pt(t, n-1))                         #p-value```

# Week 11: The Single-Sample Runs Test

Today we’re going to look at another nonparametric test: the single-sample runs test.

When Would You Use It?
The runs test is a nonparametric test used in a single sample situation to determine if the distribution of a series of binary events, in the population, is random.

What Type of Data?
The single sample runs test requires categorical (nominal) data.

Test Assumptions
None listed.

Test Process|
Step 1: Formulate the null and alternative hypotheses. The null hypothesis claims that the events in the underlying population (represented by the sample series) are distributed randomly. The alternative hypothesis claims that the events in the underlying population are distributed nonrandomly.

Step 2: Compute the number of times each of the two alternatives appears in the sample series (n1 and n2) and the number of runs, r, in the series. A run is a sequence within the series in which one of the alternatives occurs on consecutive trials.

Step 3: Obtain the critical value. Unlike most of the tests we’ve done so far, you don’t get a precise p-value when computing the runs test results. Rather, you calculate your r and then compare it to an upper and lower limit for your specific n1 and n2 values. This is done using a table (such as the one here*). For the values of n1 and n2 in your sample (labeled on this table as n and m), find the intersecting cell of the two values.

Step 4: Determine the conclusion. If your r is greater than or equal to the larger number or smaller than or equal to the smaller number in that cell, you have a statistically significant result, meaning you reject the null (that is, reject the claim that the distribution of the binary events in the population is nonrandom). If your r is between the smaller and larger number, fail to reject the null.

Example
For this example, I decided to see if the coin flips from the website Just Flip a Coin were, in fact, random. I “flipped” the coin a total of 30 times and recorded my results as “H” for heads and “T” for tails. This series is recorded below.

H0: The distribution of heads and tails in the population is random
Ha: The distribution of heads and tails in the population is nonrandom.

THTTHHHHHHTHTTHHTTHHTHHHTHHHHT

Computations:

T  H  TT  HHHHHH  T H  TT  HH  TT  HH  T  HHH  T  HHHH  T

n2 = 11 (tails)

r = 15

According to the table, my lower bound is 9 and my upper bound is 21. Since my r is in between these two values, I do not have a statistically significant result. I fail to reject H0 and claim that the distribution of heads and tails in the population is indeed random.

*Note that this table, like many others, only has a maximum of 20 for either n or m, and is constructed with α = 0.05 for the two-sided test and α = 0.025 for the one-sided test.

Example in R
No R example this week, since it’s probably more work to do this in R than it is to do it by hand, haha.

# Week 10: The z Test for a Population Proportion

Today it’s time for yet another test: the z test for a population proportion!

When Would You Use It?
The z test for a population proportion is used to determine if, in an underlying population comprised of two categories, the proportion of observations (in a sample) in one of the two categories is equal to a specific value.

What Type of Data?
The z test for a population proportion requires categorical (nominal) data.

Test Assumptions

• Each of the n independent observations is randomly selected from a population, and each observation can be classified into one of two mutually exclusive categories.
• It is recommended that this test is employed when the sample is not too small (n > 11).

Test Process
Step 1: Formulate the null and alternative hypotheses. The null hypothesis claims that the likelihood an observation will fall into Category 1 in the population is equal to a certain probability. The alternative hypothesis claims otherwise (that the population proportion for Category 1 is not equal to the value stated in the null).

Step 2: Compute the test statistic, a z-score. The test statistic is computed as follows: Step 3: Obtain the p-value associated with the calculated z-score. The p-value indicates the probability of observing a sample proportion as extreme or more extreme than the observed sample proportion, under the assumption that the null hypothesis is true.

Step 4: Determine the conclusion. If the p-value is larger than the prespecified α-level, fail to reject the null hypothesis (that is, retain the claim that the probability of falling into Category 1 in the population is equal to the value specified in the null hypothesis). If the p-value is smaller than the prespecified α-level, reject the null hypothesis in favor of the alternative.

Example
The data for this example come from my n = 30 students from one of my STAT 213 labs. I want to test the hypothesis, based on my lab, that the proportion of students who come to lab for ALL the 213 labs (for this particular instructor) is 40%. We take attendance, so we know who’s there and who’s not for a given week (note, however, that I’m going to randomly select an attendance sheet from one of the weeks of lab, AND I actually never pay attention to what the proportion of students who show up is, so I’m really just guessing about 40%). So let’s do it! Set α = 0.05 and let π denote the proportion of individuals to come to lab.

H0: π = .4
Ha: π ≠ .4

Computations: Since our p-value is (only slightly) larger than our alpha-level, we fail to reject H0 and claim that the population proportion of students who attend labs is, in fact, .4.

Example in R

```dat = read.table('clipboard',header=F) #'dat' is the name of the imported raw data
#'dat' coded such that 0 = did not attend, 1 = attended
n = nrow(dat)
X = sum(dat)
p1 =  X/n
pi1 = .4
z = (p1-pi1)/(sqrt((pi1*(1-pi1))/n))  #z-score
pval = (1-pnorm(z))*2                 #p-value
#pnorm calculates the left-hand area
#multiply by two because it is a two-sided test```

# Week 9: The Binomial Sign Test for a Single Sample

Today we’re going to look at another nonparametric test: the binomial sign test for a single sample!

When Would You Use It?
The binomial sign test is used in a single sample situation to determine, in a population comprised of two categories, if the proportion of observations in one of the two categories is equal to a specific value.

What Type of Data?
The binomial sign test requires categorical (nominal) data.

Test Assumptions
None listed.

Test Process
Step 1: Formulate the null and alternative hypotheses. The null hypothesis claims that the true proportion of observations in one of the two categories, in the population, is equal to a specific value. The alternative hypothesis claims otherwise (the proportion is either greater than, less than, or not equal to the value claimed in the null hypothesis.

Step 2: Compute the test statistic, a probability value. The test statistic is calculated as follows: That is, your test statistic is the probability of attaining ≥ r observations from “category 1” in a sample of size n, where r is the number of observations from “category 1” in your original sample.

While you can calculate this value by hand, it may be easier to either use a table or an online calculator.

Step 3: Obtain the p-value. If your alternative hypothesis is non-directional (that is, uses “≠”), your p-value is equal to α/2. If your alternative hypothesis is directional (uses “>” or “<”, your p-value is simply equal to α.

Step 4: Determine your conclusion. This depends on your alternative hypothesis. Let p1 denote the sample proportion of observations falling into “category 1”, and let P denote the test statistic value, as calculated above.

If it is nondirectional (≠) reject H0 if P < α/2.
If the alternative hypothesis is directional (>), reject H0 if p1 > π1 and P < α.
If the alternative hypothesis is directional (<), reject H0 if p1 < π1 and P < α.

Example
For this example, I decided to see if the coin flips from the website Just Flip a Coin were, binomially distributed with π1 = π2 = 0.5. I “flipped” the coin a total of 30 times and recorded my results for “category 1” (heads) and “category 2” (tails). The outcomes are displayed in the table below.

H0: π1 = 0.5
Ha: π1 ≠ 0.5

Set α = 0.05.

Computations:  Since P = 0.1002 > 0.025 (α/2 = 0.05/2 = 0.025), we fail to reject H0, the claim that proportion of the number of heads is equal to 0.5 in the population.

Example in R

```x = read.table('clipboard', header=F)
pi1= 0.5                                 #hypothesized probability for "heads"
n = length(as.matrix(x))
tab = as.data.frame(table(x))            #observed frequencies
p1 = tab[1,2]
P = 1 - pbinom(p1, size = n, prob = pi1) #test statistic```

# Week 8: The Chi-Square Goodness-of-Fit Test

Today we’re continuing the theme of nonparametric tests with the chi-square goodness-of-fit test!

When Would You Use It?
The chi-square goodness-of-fit test is a nonparametric test used in a single sample situation to determine if a sample originates from a population for which the observed cell (category) frequencies different from the expected cell (category) frequencies. Basically, this test is used when a particular “distribution” is expected of the categories of a variable and a researcher wants to know if that distribution fits the data.

What Type of Data?
The chi-square goodness-of-fit test requires categorical (nominal) data.

Test Assumptions

• Categorical or nominal data are used in the analysis (the data should represent frequencies for mutually exclusive categories).
• The data consist of a random sample of n observations.
• The expected frequency (as calculated below) of each cell is 5 or greater.

Test Process
Step 1: Formulate the null and alternative hypotheses. The null hypothesis claims that the observed frequency of each cell is equal to the expected frequency of that cell, while the alternative hypothesis claims that at least one cell has an observed frequency that differs from the expected frequency.

Step 2: Compute the test statistic, a chi-square value. The calculations are as follows: Step 3: Obtain the p-value associated with the calculated chi-square. The p-value indicates the probability of observing deviations from the expected values that are larger than those in the observed sample, under the assumption that the null hypothesis is true.

Step 4: Determine the conclusion. If the p-value is larger than the prespecified α-level, fail to reject the null hypothesis (that is, retain the claim that the observed frequencies in the cells are equal to the expected frequencies of the cells). If the p-value is smaller than the prespecified α-level, reject the null hypothesis in favor of the alternative.

Example
For this test’s example, I wanted to determine if the dice on my iPod Yahtzee app were fair. That is, I wanted to see if there was an equal probability for all six sides to come up on any given roll. So I rolled the five dice 115 times (for a total of n = 575 individual die rolls) and simply recorded the faces showing (note that I did not actually “play” Yahtzee while doing this, which is to say that I just kept rolling all five dice no matter what happened and I didn’t “hold” any of them at any point). I’m going to claim in my null hypothesis that all six side values have an equal probability of showing on a given die. That is,

H0: p1 = p2 = p3 = p4 = p5 = p6 = 1/6
Ha: The probability for at least one value is not 1/6

Set α = 0.05.

Computations: Since our p-value is larger than our alpha level of .05, we fail to reject H0 and claim that the observed values are equal to the expected values and the dice are fair.

Example in R

```x=read.table('clipboard', header=F)
n=length(x)
tab=as.data.frame(table(x))           #observed frequencies
p=1/length(tab\$x)
Chi=rep(NaN,length(tab\$x))
for (i in 1:length(tab\$x)){
Chi[i]=((tab\$Freq[i]-(p*n))^2)/(p*n)
}
Chisquare=sum(Chi)
1-pchisq(Chisquare,length(tab\$x)-1)   #p-value```

# Week 7: The Kolmogorov-Smirnov Goodness-of-Fit Test for a Single Sample

Today we’re going to do our first test of goodness-of-fit with the Kolmogorov-Smirnov goodness-of-fit test for a single sample.

When Would You Use It?
The Kolmogorov-Smirnov goodness-of-fit test is a nonparametric test used in a single sample situation to determine if the distribution of a sample of values conforms to a specific population (or probability) distribution.

What Type of Data?
The Kolmogorov-Smirnov goodness-of-fit test requires ordinal data.

Test Assumptions
None listed.

Test Process
Step 1: Formulate the null and alternative hypotheses. The null hypothesis claims that the distribution of the data in the sample is consistent with the hypothesized theoretical population distribution. The alternative claims that the distribution of the data in the sample is inconsistent with the hypothesized theoretical population distribution.

Step 2: Compute the test statistic. The test statistic, in the case of this test, is defined by the point that represents the greatest vertical distance at any point between the cumulative probability distribution constructed from the sample and the cumulative probability distribution constructed under the hypothesized population distribution. Since the specifics of the cumulative probability distribution calculations depend on which distributions are used, I will refer you to the example shown below to show how these calculations are done in a specific testing situation.

Step 3: Obtain the critical value. Unlike most of the tests we’ve done so far, you don’t get a precise p-value when computing the results here. Rather, you calculate your test statistic and then compare it to a specific value. This is done using a table (such as the one here). Find the number at the intersection of your sample size n and the specified alpha-level. Compare this value with your test statistic.

Step 4: Determine the conclusion. If your test statistic is equal to or larger than the table value, reject the null hypothesis (that is, claim that the distribution of the data is inconsistent with the hypothesized population distribution). If your test statistic is less than the table value, fail to reject the null.

Example
For this test’s example, I wanted to determine, from a sample of n = 59 IQ scores, if scores in the population follow a normal distribution with a mean µ = 100 and standard deviation σ = 15. Set α = 0.05.

H0: IQ scores in the population follow a normal distribution with a mean of 100 and a standard deviation of 15.
Ha: IQ scores in the population deviate from a normal distribution with a mean of 100 and a standard deviation of 15.

Computations:

For the computations section of this test, I will display a table of values for the first three and the last of the IQ scores (sorted from smallest to largest) and describe what the values are and how the test statistic is obtained. Column A represents the IQ scores of the sample, ranked from lowest to highest.

Column B represents the z-scores of the IQ tests, calculated by taking the difference of the score and the mean (100), then dividing by the standard deviation (15).

Column B is not necessary, but is used to make the calculation of Column C easier. Column C is the proportion of cases between the z-score (Column B) and the hypothesized mean of the population’s distribution (100 in this case). For example, for an IQ of 81, the proportion of scores falling between 82 and 100 is .385.

Column D represents the percentile rank of a given IQ score in the hypothesized population distribution. An IQ of 82, for example, is the 11.5th percentile.

Column E represents the cumulative proportion, in the sample, for each IQ. For an IQ of 82, the cumulative proportion is just 1/59, while the cumulative proportion for the highest value, 145, is 59/59.

Column F is the absolute difference between the ith values in Column D and Column E. This represents the differences between the proportions in the sample population and the proportions expected under the hypothesized population distribution.

Finally, Column G is the absolute difference between the value of Column D for a given row and the value of Column E for the preceeding row. For example, for an IQ of 89, Column G is calculated by taking |0.232 – 0.017|.

The test statistic is obtained by determining the largest value from either Column F or Column G. That is, whichever column has the largest value, then that largest value becomes the test statistic. When these values are computed for the whole dataset, the largest value is 0.438. This value is compared to the critical value at α = 0.05, n > 35, which ends up being: Since our test statistic is larger than our critical value, we reject H0 and claim that IQ scores in the population deviate from a normal distribution with mean 100 and standard deviation 15.

Example in R

```x=read.table('clipboard', header=F)
x=as.matrix(x)
x=sort(x)                                                   #column A
mu=100
sd=15
B=(x-mu)/sd                                                 #column B
pmu=.5
pz=pnorm(abs(z), mean = 0, sd = 1, lower.tail = TRUE)
C=abs(pmu-pz) #column C
D=pnorm(z, mean = 0, sd = 1, lower.tail = TRUE)             #column D
E=rep(NaN,length(x))                                        #column E
for (i in 1:length(x)){
e[i]=i/length(x)
}
F=abs(e-dz)                                                 #column F
ee=c(0,e[1:(length(x)-1)])
G=abs(ee-dz)                                                #column G
if(max(G)>max(F)){Tstat=max(G)}else{Tstat=max(F)}
Tstat                                                       #test statistic```

# Week 6: The Wilcoxon Signed-Ranks Test

Today we’re going to talk about our first nonparametric test: the Wilcoxon signed-ranks test!

When Would You Use It?
The Wilcoxon signed-ranks test is a nonparametric test used in a single sample situation to determine if the sample originates from a population with a specific median θ.

What Type of Data?
The Wilcoxon signed-ranks test requires ordinal data.

Test Assumptions

• The sample is a simple random sample from the population of interest.
• The original scores obtained for each of the individuals in the sample are in the format of interval or ratio data.
• The underlying population distribution is symmetrical.

Test Process
Step 1: Formulate the null and alternative hypotheses. The null hypothesis claims that the median in the population is equal to a specific value; the alternative hypothesis claims otherwise (the population median is greater than, less than, or not equal to the value specified in the null hypothesis).

Step 2: Compute the test statistic. Since this is best done with data, please see the example shown below to see how this is done.

Step 3: Obtain the critical value. Unlike most of the tests we’ve done so far, you don’t get a precise p-value when computing the results here. Rather, you calculate your test statistic and then compare it to a specific value. This is done using a table (such as the one here). Find the number at the intersection of your sample size n and the specified alpha-level. Compare this value with your test statistic.

Step 4: Determine the conclusion. If your test statistic is equal to or less than the table value, reject the null hypothesis. If your test statistic is greater than the table value, fail to reject the null (that is, claim that the median in the population is in fact equal to the value specified in the null hypothesis).

Example
The data for this example come from a little analysis of my Facebook friends’ birthdays I did awhile ago. In that analysis, I recorded the birth months for the n = 97 friends who had their birthdays visible. All I did at the time was see how many birthdays there were per month. Now, however, I want to see if the median number of birthdays per month is equal to a certain value—say, 8. Set α = 0.05.

H0: θ = 8
Ha: θ ≠ 8

The following table shows several different columns of information. I will explain the columns below. Column 1 is just the name of the month.
Column 2 is the number of observed birthdays in my sample for that corresponding month.
Column 3 is calculated by taking the number of observed birthdays minus the hypothesized median, which is θ = 8 in this case.
Column 4 is the absolute value of the difference in Column 3.
Column 5 is the rank of the value in Column 4. Ranking is done as follows: rank the values of Column 4 from smallest to largest. If there are ties, sum the number of ranks that are taken by the ties, and then divide that value by the number of ties. For example, there are five months that have a |D| = 3. That means that this tied value takes up the rank places of the 1st, 2nd, 3rd, 4th, and 5th observations, had they not been all tied at 1. Thus, I sum these rank places and divide by 5 to get (1+2+3+4+5)/5 = 3, then assign all these five months the rank of 3.
Column 6 contains the same values as Column 5, but signs them depending on the sign in Column 3.

The next step is to sum all the positive ranks in Column 6 and sum all the negative ranks in Column 6. Doing so, we get: The test statistic itself is the absolute value of the smaller of the above values; in this case, we get T = 37.5. In the table, the critical value for n = 12 and α = 0.05 for a two-tailed test is 13. Since T > 13, we fail to reject the null and retain the claim that the population median is, in fact, 8.

Example in R

No R example this week; most of this is easy enough to do by hand for a small-ish sample.

# Week 5: The Single-Sample Test for Evaluating Population Kurtosis

Last week we did a test for population skew, which represents the third moment about the mean. Now we’re going to move onto the fourth moment by doing a single-sample test to evaluate population kurtosis!

When Would You Use It?
The test of population kurtosis test is a parametric test used in a single sample situation to assess if a sample originates from a population that is mesokurtic (as opposed to leptokurtic or platykurtic).

What Type of Data?
The test for kurtosis requires interval or ratio data.

Test Assumptions
None listed.

Test Process
Step 1: Formulate the null and alternative hypotheses. The null hypothesis claims that the kurtosis parameter γ2 in the population is equal to 0, which corresponds to a mesokurtic distribution; the alternative hypothesis claims otherwise (the population kurtosis parameter is greater than, less than, or not equal to the value specified in the null hypothesis, suggesting a leptokurtic or platykurtic distribution).

Step 2: Compute the test statistic value, a z-score. The test statistic requires several calculations to be obtained. The calculations are as follows: Step 3: Obtain the p-value associated with the calculated chi-square. The p-value indicates the probability of observing a test statistic as extreme or more extreme than the observed test statistic, under the assumption that the null hypothesis is true.

Step 4: Determine the conclusion. If the p-value is larger than the prespecified α-level, fail to reject the null hypothesis (that is, retain the claim that the population distribution is mesokurtic). If the p-value is smaller than the prespecified α-level, reject the null hypothesis in favor of the alternative.

Example
The data for this example come from my n = 365 song downloads from 2010. I want to create a hypothesis test regarding the kurtosis of the distribution of song lengths (in seconds).

H0: γ2 = 0
Ha: γ2  0

Set α = 0.05.

Computations: Since our p-value is basically zero, it is smaller than our alpha-level, and we reject H0 and claim that the population is not mesokurtic (γ2 ≠ 0).

Example in R

```dat = read.table('clipboard',header=T) #'dat' is the name of the imported raw data
n = length(dat)
xbar = mean(dat)
k = ((((sum((dat-xbar)^4))*n*(n+1))/(n-1))-(3*((sum((dat-xbar)^2))^2)))/((n-2)*(n-3))
s = sqrt(((sum(dat^2))-(((sum(dat))^2)/n))/(n-1))
g2 = k/(s^4)
A = (24*n*(n-2)*(n-3))/(((n+1)^2)*(n+3)*(n+5))
B = (g2*(n-2)*(n-3))/((n-1)*(n+1)*sqrt(A))
C = ((6*((n^2)-(5*n)+2))/((n+7)*(n+9)))*sqrt((6*(n+3)*(n+5))/(n*(n-2)*(n-3)))
D = 6+(8/C)*((2/C)+sqrt(1+(4/(C^2))))
E = (1-(2/D))/(1+(B*sqrt(2/(D-4))))
z = (1-(2/(9*D))-((E)^(1/3)))/(sqrt(2/(9*D)))      #test statistic
pval = (1-pnorm(z))*2                              #p-value```

# Data Dump

Holy crapples, I just found the best place for big datasets from online personality tests.

Sample sizes in the ten thousands? WHAT IS THIS NONSENSE

I’m not too concerned about the accuracy, really; it’s data that would be useful for my weekly stats examples. And just screwing around with R.

‘Cause I like that.

Anyway.

# Week 4: The Single-Sample Test for Evaluating Population Skewness

Last week we did a test for population variance, which represents the second moment about the mean. Today we’re going to go one moment further and do a single-sample test to evaluate population skewness (which represents the third moment about the mean)!

When Would You Use It?
The test of population skewness test is a parametric test used in a single sample situation to determine if a sample originates from a population that is symmetrical (that is, not skewed).

What Type of Data?
The test for skewness requires interval or ratio data.

Test Assumptions
None listed.

Test Process
Step 1: Formulate the null and alternative hypotheses. The null hypothesis claims that the skewness parameter γ in the population is equal to 0, which corresponds to symmetry; the alternative hypothesis claims otherwise (the population skewness parameter is greater than, less than, or not equal to the value specified in the null hypothesis, suggesting there is some skew).

Step 2: Compute the test statistic value, a z-score. The test statistic requires several calculations to be obtained. The calculations are as follows: Step 3: Obtain the p-value associated with the calculated chi-square. The p-value indicates the probability of observing a skew as extreme or more extreme than the observed sample skew, under the assumption that the null hypothesis is true.

Step 4: Determine the conclusion. If the p-value is larger than the prespecified α-level, fail to reject the null hypothesis (that is, retain the claim that there is symmetry (no skew) in the population). If the p-value is smaller than the prespecified α-level, reject the null hypothesis in favor of the alternative.

Example
As in the last test, the data for this example come from my n = 365 song downloads from 2010. I want to create a hypothesis test regarding the skew of the distribution of song lengths (in seconds). Based on the following histogram, I’m going to say that this distribution has a right skew. Thus,

H0: γ = 0
Ha: γ > 0

Set α = 0.05.

Computations: Since our p-value is basically zero, it is smaller than our alpha-level, and we reject H0 and claim that the population is indeed positively skewed (γ > 0)

Example in R

```dat = read.table('clipboard',header=T) #'dat' is the name of the imported raw data
hist(dat)                              #creates histogram of data
n = 365
m3 =(n*sum((dat-mean(dat))^3))/((n-1)*(n-2))
s3 = sqrt((sum(dat^2)-(((sum(dat))^2)/n))/(n-1))
g1 = m3/(s3)^3
b1 = ((n-2)*g1)/(sqrt(n*(n-1)))
A = b1*sqrt(((n+1)*(n+3))/(6*(n-2)))
B = (3*((n^2)+(27*n)-70)*(n+1)*(n+3))/((n-2)*(n+5)*(n+7)*(n+9))
C = sqrt(2*(B-1))-1
D = sqrt(C)
E = 1/sqrt((log(D)))
F = A/(sqrt(2/(C-1)))
z = E*log(F+sqrt((F^2)+1))            #test statistic
pval = (1-pnorm(z))                   #p-value```

# Week 3: The Single-Sample Chi-Square Test for a Population Variance

Today we’re going to move away from testing for means and do the single-sample chi-square test for a population variance!

When Would You Use It?
The chi-square test is a parametric test used in a single sample situation to determine if a sample originates from a population with a specific variance σ2.

What Type of Data?
The chi-square test for variance requires interval or ratio data.

Test Assumptions

• The sample is a simple random sample from the population of interest.
• The distribution underlying the data is normal.

Test Process

Step 1: Formulate the null and alternative hypotheses. The null hypothesis claims that the variance in the population is equal to a specific value; the alternative hypothesis claims otherwise (the population variance is greater than, less than, or not equal to the value specified in the null hypothesis.

Step 2: Compute the chi-square value. The chi-square value is computed as follows: Step 3: Obtain the p-value associated with the calculated chi-square. The p-value indicates the probability of observing a sample variance as extreme or more extreme than the observed sample variance, under the assumption that the null hypothesis is true.

Step 4: Determine the conclusion. If the p-value is larger than the prespecified α-level, fail to reject the null hypothesis (that is, retain the claim that the variance in the population is equal to the value specified in the null hypothesis). If the p-value is smaller than the prespecified α-level, reject the null hypothesis in favor of the alternative.

Example
The data for this example come from my n = 365 song downloads from 2010. I want to create a hypothesis test regarding the variance of the song lengths (in seconds). I have no idea what the variance is, but I’m going to say that I suspect the variance to be (120)2, or two minutes squared. Set α = 0.05

H0: σ2 = 14,400 seconds
Ha: σ2 ≠ 14,440 seconds

The sample variance is calculated to be 12182.44.

Computations: Since our p-value is smaller than our alpha-level, we reject H0 and claim that the population variance is greater than (120)2 seconds.

Example in R

```dat=read.table('clipboard',header=T) #'dat' is the name of the imported raw data
sigma = 120^2
s = var(dat)
n = 365
chisq = ((n-1)*s)/(sigma)  #chi-square value
pval = (pchisq(chisq, n-1))*2    #p-value
#n-1 is the degrees of freedom```

# Week 2: The Single-Sample t Test

Today we’ll be discussing another commonly used statistical test—one that is highly related to last week’s test: the single-sample t test!

When Would You Use It?
The single-sample t test is a parametric test used in a single sample situation to determine if the sample originates from a population with a specific mean µ. This test is used when the population standard deviation, σ, is not known (it must be estimated with the sample standard deviation, s).

What Type of Data?
The single-sample t test requires interval or ratio data.

Test Assumptions

• The sample is a simple random sample from the population of interest.
• The distribution underlying the data is normal.

Test Process
Step 1: Formulate the null and alternative hypotheses. The null hypothesis claims that the mean in the population is equal to a specific value; the alternative hypothesis claims otherwise (the population mean is greater than, less than, or not equal to the value specified in the null hypothesis.

Step 2: Compute the t-score. The t-score is computed as follows: Step 3: Obtain the p-value associated with the calculated t-score. The p-value indicates the probability of observing a sample mean as extreme or more extreme than the observed sample mean, under the assumption that the null hypothesis is true.

Step 4: Determine the conclusion. If the p-value is larger than the prespecified α-level, fail to reject the null hypothesis (that is, retain the claim that the mean in the population is equal to the value specified in the null hypothesis). If the p-value is smaller than the prespecified α-level, reject the null hypothesis in favor of the alternative.

Example
The data for this example are the recorded mileages of my n = 306 walks from 2015. Mileage is recorded to the second decimal place. Since there is no real way I can determine what the population standard deviation should be, I will estimate it with the sample standard deviation, and thus must use a t test for a test of the mean value. I’m going to guess that my average walk is greater than 7 miles, ‘cause I honestly can’t remember what the actual average was, but I’m pretty sure it was more than 7. Set α = 0.05.

H0: µ = 7 miles
Ha: µ > 7 miles

The sample mean is calculated to be 8.246 and the sample standard deviation is calculated to be 4.429

Computations: Since our p-value is much smaller than our alpha-level, we reject H0 and claim that the population mean is greater than 7 miles.

Example in R

```dat=read.table('clipboard',header=T) #'dat' is the name of the imported raw data
mu = 7
s = sd(dat)
n = 306
xbar = mean(dat)
t = (xbar-mu)/(s/sqrt(n))  #t-score
pval = (1-pt(t, n-1))*2          #p-value
#n-1 is the degrees of freedom```

# Week 1: The Single-Sample z Test

Today we’ll be discussing one of the most commonly used statistical tests and one of the first ones taught in introductory stats: the single-sample z test!

When Would You Use It?
The single-sample z test is a parametric test used in a single sample situation to determine if the sample originates from a population with a specific mean µ. This test is used when the population standard deviation, σ, is known.

What Type of Data?
The single-sample z test requires interval or ratio data.

Test Assumptions

• The sample is a simple random sample from the population of interest.
• The distribution underlying the data is normal.

Test Process
Step 1: Formulate the null and alternative hypotheses. The null hypothesis claims that the mean in the population is equal to a specific value; the alternative hypothesis claims otherwise (the population mean is greater than, less than, or not equal to the value specified in the null hypothesis.

Step 2: Compute the z-score. The z-score is computed as follows: Step 3: Obtain the p-value associated with the calculated z-score. The p-value indicates the probability of observing a sample mean as extreme or more extreme than the observed sample mean, under the assumption that the null hypothesis is true.

Step 4: Determine the conclusion. If the p-value is larger than the prespecified α-level, fail to reject the null hypothesis (that is, retain the claim that the mean in the population is equal to the value specified in the null hypothesis). If the p-value is smaller than the prespecified α-level, reject the null hypothesis in favor of the alternative.

Example
The data for this example are n = 400 IQ scores from random sample of individuals ages 18 to 87. The test used to obtain the scores was constructed similar to the Stanford-Binet IQ test, meaning that we can assume that the average IQ score in the population should be 100 and the population standard deviation σ is known and is equal to 15. Let’s test the claim that the population mean is actually different than 100. Set α = 0.05.

H0: µ = 100
Ha: µ ≠ 100

The sample mean is calculated to be 101.864

Computations: Since our p-value is smaller than our alpha-level (0.012 > 0.05), we reject H0 and claim that the population mean is different from 100.

Example in R

```dat=read.table('clipboard',header=T) #'dat' is the name of the imported raw data
mu = 100
sigma = 15
n = 400
xbar = mean(dat)
z = (xbar-mu)/(sigma/sqrt(n))  #z-score
pval = (1-pnorm(z))*2          #p-value
#pnorm calculates the left-hand area
#multiply by two because it is a two-sided test```

# Statistics Sunday: An Introductory Post

Every Sunday of this year, I plan on focusing on one of the statistical tests featured in the Handbook of Parametric and Nonparametric Statistical Procedures (5th edition) by David J. Sheskin. I will include the following information in each post:

• When would you use the test? What type of research question might the test help to answer?
• For what type of data is the test appropriate? Do you need interval/ratio data, categorical data, etc.?
• Assumptions. What assumptions must be met in order for the test to be accurately employed?
• Process. The steps (and equations) of the test.
• Example. The test carried out with real data.
• Example in R. The R code for the above example.

I’ll start this tomorrow, and while I’ll probably put a little menu button up at the top of my blog homepage to link to all the tests (a thing like a “Statistics Sundays” button or whatnot), I figured I should explain it here, too.

YAY!