Hypergeometric Distribution – dataanalysisclassroom

Lesson 93 – The Two-Sample Hypothesis Test – Part II

On the Difference in Proportions

$H_{0}: p_{1}-p_{2} = 0$

$H_{A}: p_{1}-p_{2} > 0$

$H_{A}: p_{1}-p_{2} < 0$

$H_{A}: p_{1}-p_{2} \neq 0$

Joe and Mumble are interested in getting people’s opinion on the preference for a higher than 55 mph speed limit for New York State.

Joe spoke to **ten** of his rural friends, of which **seven** supported the idea of increasing the speed limit to 65 mph. Mumble spoke to **eighteen** of his urban friends, of which **five** favored a speed limit of 65 mph over the current limit of 55 mph.

Can we say that the sentiment for increasing the speed limit is stronger among rural than among urban residents?

We can use a hypothesis testing framework to address this question.

Last week, we learned how Fisher’s Exact test could be used to verify the difference in proportions. The test-statistic for the two-sample hypothesis test follows a hypergeometric distribution when $H_{0}$ is true.

We also learned that, in more generalized cases where the number of successes is not known apriori, we could assume that the number of successes is fixed at $t=x_{1}+x_{2}$ , and, for a fixed value of $t$ , we reject $H_{0}:p_{1}=p_{2}$ for the alternate hypothesis $H_{A}:p_{1}>p_{2}$ if there are more successes in random variable $X_{1}$ compared to $X_{2}$ .

In short, the p-value can be derived under the assumption that the number of successes $X=k$ in the first sample $X_{1}$ has a hypergeometric distribution when $H_{0}$ is true and conditional on a total number of t successes that can come from any of the two random variables $X_{1}$ and $X_{2}$ .

$P(X=k) = \frac{\binom{t}{k}*\binom{n_{1}+n_{2}-t}{n_{1}-k}}{\binom{n_{1}+n_{2}}{n_{1}}}$

Let’s apply this principle to the two samples that Joe and Mumble collected.

Let $X_{1}$ be the random variable that denotes Joe’s rural sample. He surveyed a total of $n_{1}=10$ people and $x_{1}=7$ favored an increase in the speed limit. So the proportion $p_{1}$ based on the number of successes is 0.7.

Let $X_{2}$ be the random variable that denotes Mumble’s urban sample. He surveyed a total of $n_{2}=18$ people. $x_{2}=5$ out of the 18 favored an increase in the speed limit. So the proportion $p_{2}$ based on the number of successes is 0.2778.

Let the total number of successes in both the samples be $t=x_{1}+x_{2}=7+5=12$ .

Let’s also establish the null and alternate hypotheses.

$H_{0}: p_{1}-p_{2}=0$

$H_{A}: p_{1}-p_{2}>0$

The alternate hypothesis says that the sentiment for increasing the speed limit is stronger among rural ( $p_{1}$ ) than among urban residents ( $p_{2}$ ).

Larger values of $x_{1}$ and smaller values of $x_{2}$ support the alternate hypothesis $H_{A}$ that $p_{1}>p_{2}$ when t is fixed.

For a fixed value of t, we reject $H_{0}$ , if there are more number of successes in $X_{1}$ compared to $X_{2}$ .

Conditional on a total number of t successes from any of the two random variables, the number of successes $X=k$ in the first sample has a hypergeometric distribution when $H_{0}$ is true.

In the rural sample that Joe surveyed, seven favored an increase in the speed limit. So we can compute the p-value as the probability of obtaining more than seven successes in a rural sample of 10 when the total successes t from either urban or rural samples are twelve.

$p-value=P(X \ge k) = P(X \ge 7)$

$P(X=k) = \frac{\binom{t}{k}*\binom{n_{1}+n_{2}-t}{n_{1}-k}}{\binom{n_{1}+n_{2}}{n_{1}}}$

$P(X=7) = \frac{\binom{12}{7}\binom{10+18-12}{10-7}}{\binom{10+18}{10}} =\frac{\binom{12}{7}\binom{16}{3}}{\binom{28}{10}} = 0.0338$

A total of 12 successes exist, out of which the number of ways of choosing 7 is $\binom{12}{7}$ .

A total of 28 – 12 = 16 non-successes exist, out of which the number of ways of choosing 10 – 7 = 3 non-successes is $\binom{16}{3}$ .

A total sample of 10 + 18 = 28 exists, out of which the number of ways of choosing ten samples is $\binom{28}{10}$ .

When we put them together, we can derive the probability $P(X=7)$ for the hypergeometric distribution when $H_{0}$ is true.

$P(X=7) = \frac{\binom{12}{7}\binom{10+18-12}{10-7}}{\binom{18+18}{10}} =\frac{\binom{12}{7}\binom{16}{3}}{\binom{28}{10}} = 0.0338$

Applying the same logic for k = 8, 9, and 10, we can derive their respective probabilities.

$P(X=8) = \frac{\binom{12}{8}\binom{10+18-12}{10-8}}{\binom{10+18}{10}} =\frac{\binom{12}{8}\binom{16}{2}}{\binom{28}{10}} = 0.0045$

$P(X=9) = \frac{\binom{12}{9}\binom{10+18-12}{10-9}}{\binom{10+18}{10}} =\frac{\binom{12}{9}\binom{16}{1}}{\binom{28}{10}} = 0.0003$

$P(X=10) = \frac{\binom{12}{10}\binom{10+18-12}{10-10}}{\binom{10+18}{10}} =\frac{\binom{12}{10}\binom{16}{0}}{\binom{28}{10}} = 5.029296*10^{-6}$

The p-value can be computed as the sum of these probabilities.

$p-value=P(X \ge k) = P(X = 7)+P(X = 8)+P(X = 9)+P(X = 10)=0.0386$

Visually, the null distribution will look like this.

The x-axis shows the number of possible successes in $X_{1}$ . They range from k = 0 to k = 10. The vertical bars are showing $P(X=k)$ as derived from the hypergeometric distribution. The area highlighted in red is the *p-value*, the probability of finding $\ge$ seven successes in a rural sample of 10 people.

The p-value is the probability of obtaining the computed test statistic under the null hypothesis.

The smaller the p-value, the less likely the observed statistic under the null hypothesis – and stronger evidence of rejecting the null.

Suppose we select a rate of error $\alpha$ of 5%.

Since the p-value (0.0386) is smaller than our selected rate of error (0.05), we reject the null hypothesis for the alternate view that the sentiment for increasing the speed limit is stronger among rural ( $p_{1}$ ) than among urban residents ( $p_{2}$ ).

Let me remind you that this decision is based on the assumption that the null hypothesis is correct. Under this assumption, since we selected , we will reject the true null hypothesis 5% of the time. At the same time, we will fail to reject the null hypothesis 95% of the time. In other words, 95% of the time, our decision to not reject the null hypothesis will be correct.

What if Joe and Mumble surveyed many more people?

You must be wondering that Joe and Mumble surveyed just a few people, which is not enough to derive any decent conclusion for a question like this. Perhaps they just called up their friends!

Let’s do a thought experiment. How would the null distribution look like if Joe and Mumble had double the sample size and the successes also increase in the same proportion? Would the p-value change?

Say Joe had surveyed 20 people, and 14 had favored an increase in the speed limit. $n_{1} = 20; x_{1} = 14; p_{1} = 0.7$ .

Say Mumble had surveyed 36 people, and 10 had favored an increase in the speed limit. $n_{2} = 36; x_{2} = 10; p_{2} = 0.2778$ .

p-value will then be $P(X \ge 14)$ when there are 24 total successes.

The null distribution will look like this.

Notice that the null distribution is much more symmetric and looks like a bell curve (normal distribution) with an increase in the sample size. The p-value is 0.0026. More substantial evidence for rejecting the null hypothesis.

Is there a limiting distribution for the difference in proportion? If there is one, can we use it as the null distribution for the hypothesis test on the difference in proportion when the sample sizes are large.

While we embark on this derivation, let’s ask Joe and Mumble to survey many more people. When they are back, we will use new data to test the hypothesis.

But first, what is the limiting distribution for the difference in proportion?

We have two samples $X_{1}$ and $X_{2}$ of sizes $n_{1}$ and $n_{2}$ .

We might observe $x_{1}$ and $x_{2}$ successes in each of these samples. Hence, the proportions $p_{1}, p_{2}$ can be estimated using $\hat{p_{1}} = \frac{x_{1}}{n_{1}}$ and $\hat{p_{2}} = \frac{x_{2}}{n_{2}}$ .

See, we are using $\hat{p_{1}}, \hat{p_{2}}$ as the estimates of the true proportions $p_{1}, p_{2}$ .

Take $X_{1}$ . If the probability of success (proportion) is $p_{1}$ , in a sample of $n_{1}$ , we could observe $x_{1}=0, 1, 2, 3, \cdots, n_{1}$ successes with a probabilty $P(X=x_{1})$ that is governed by a binomial distribution. In other words,

$x_{1} \sim Bin(n_{1},p_{1})$

Same logic applies to $X_{2}$ .

$x_{2} \sim Bin(n_{2},p_{2})$

A binomial distribution tends to a normal distribution for large sample sizes; it can be estimated very accurately using the normal density function. We learned this in Lesson 48.

If you are curious as to how a binomial distribution function $f(x)=\frac{n!}{(n-x)!x!}p^{x}(1-p)^{n-x}$ can approximated to a normal density function $f(x)=\frac{1}{\sqrt{2 \pi \sigma^{2}}} e^{\frac{-1}{2}(\frac{x-\mu}{\sigma})^{2}}$ , look at this link.

But what is the limiting distribution for  and ?

$x_{1}$ is the sum of $n_{1}$ independent Bernoulli random variables (yes or no responses from the people). For a large enough sample size $n_{1}$ , the distribution function of $x_{1}$ , which is a binomial distribution, can be well-approximated by the normal distribution. Since $\hat{p_{1}}$ is a linear function of $x_{1}$ , the random variable $\hat{p_{1}}$ can also be assumed to be normally distributed.

When both $\hat{p_{1}}$ and $\hat{p_{2}}$ are normally distributed, and when they are independent of each other, their sum or difference will also be normally distributed. We can derive it using the convolution of $\hat{p_{1}}$ and $\hat{p_{2}}$ .

Let $Y = \hat{p_{1}}-\hat{p_{2}}$

$Y \sim N(E[Y], V[Y])$ since both $\hat{p_{1}}, \hat{p_{2}} \sim N()$

If $Y \sim N(E[Y], V[Y])$ , we can standardize it to a standard normal variable as

$Z = \frac{Y-E[Y]}{\sqrt{V[Y]}} \sim N(0, 1)$

We should now derive the expected value $E[Y]$ and the variance $V[Y]$ of Y.

$Y = \hat{p_{1}}-\hat{p_{2}}$

$E[Y] = E[\hat{p_{1}}-\hat{p_{2}}] = E[\hat{p_{1}}] - E[\hat{p_{2}}]$

$V[Y] = V[\hat{p_{1}}-\hat{p_{2}}] = V[\hat{p_{1}}] + V[\hat{p_{2}}]$

Since they are independent, the co-variability term which carries the negative sign is zero.

We know that $E[\hat{p_{1}}] = p_{1}$ and $V[\hat{p_{1}}]=\frac{p_{1}(1-p_{1})}{n_{1}}$ . Recall Lesson 76.

When we put them together,

$E[Y] = p_{1} - p_{2}$

$V[Y] = \frac{p_{1}(1-p_{1})}{n_{1}} + \frac{p_{2}(1-p_{2})}{n_{2}}$

and finally since $Z = \frac{Y-E[Y]}{\sqrt{V[Y]}} \sim N(0, 1)$ ,

$Z = \frac{\hat{p_{1}} - \hat{p_{2}} - (p_{1} - p_{2})}{\sqrt{\frac{p_{1}(1-p_{1})}{n_{1}} + \frac{p_{2}(1-p_{2})}{n_{2}}}} \sim N(0, 1)$

A few more steps and we are done. Joe and Mumble must be waiting for us.

The null hypothesis is $H_{0}: p_{1}-p_{2}=0$ . Or, $p_{1}=p_{2}$ .

We need the distribution under the null hypothesis — the null distribution.

Under the null hypothesis, let’s assume that $p_{1}=p_{2}$ is $p$ , a common value for the two population proportions.

Then, the expected value of Y, $E[Y]=p_{1}-p_{2}=p-p = 0$ and the variance $V[Y] = \frac{p(1-p)}{n_{1}} + \frac{p(1-p)}{n_{2}}}$

$V[Y] = p(1-p)*(\frac{1}{n_{1}}+\frac{1}{n_{2}})$

This shared value $p$ for the two population proportions can be estimated by pooling the samples together into one sample of size $n_{1}+n_{2}$ where there are $x_{1}$ and $x_{2}$ total successes.

$p = \frac{x_{1}+x_{2}}{n_{1}+n_{2}}$

Look at this estimate carefully. Can you see that the pooled estimate $p$ is a weighted average of the two proportions ( $p_{1}$ and $p_{2}$ )?

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Okay, tell me what $x_{1}$ and $x_{2}$ are? Aren’t they $n_{1}\hat{p_{1}}$ and $n_{2}\hat{p_{2}}$ for the given two samples?

So $p = \frac{n_{1}\hat{p_{1}}+n_{2}\hat{p_{2}}}{n_{1}+n_{2}}=\frac{n_{1}}{n_{1}+n_{2}}\hat{p_{1}}+ \frac{n_{2}}{n_{1}+n_{2}}\hat{p_{2}}$

or, $p = w_{1}\hat{p_{1}}+ w_{2}\hat{p_{2}}$

At any rate,

$E[Y]= 0$

$V[Y] = p(1-p)*(\frac{1}{n_{1}}+\frac{1}{n_{2}})$

$p=\frac{x_{1}+x_{2}}{n_{1}+n_{2}}$

To summarize, when the null hypothesis is

$H_{0}:p_{1}-p_{2}=0$

for large sample sizes, the test-statistic $z = \frac{\hat{p_{1}}-\hat{p_{2}}}{\sqrt{p(1-p)*(\frac{1}{n_{1}}+\frac{1}{n_{2}})}} \sim N(0,1)$

If the alternate hypothesis $H_{A}$ is $p_{1}-p_{2}>0$ , we reject the null hypothesis when the p-value $P(Z \ge z)$ is less than the rate of rejection $\alpha$ . We can also say that when $z > z_{\alpha}$ , we reject the null hypothesis.

If the alternate hypothesis $H_{A}$ is $p_{1}-p_{2}<0$ , we reject the null hypothesis when the p-value $P(Z \le z)$ is less than the rate of rejection $\alpha$ . Or when $z < -z_{\alpha}$ , we reject the null hypothesis.

If the alternate hypothesis $H_{A}$ is $p_{1}-p_{2} \neq 0$ , we reject the null hypothesis when the p-value $P(Z \le z)$ or $P(Z \ge z)$ is less than the rate of rejection $\frac{\alpha}{2}$ . Or when $z < -z_{\frac{\alpha}{2}}$ or $z > z_{\frac{\alpha}{2}}$ , we reject the null hypothesis.

Okay, we are done. Let’s see what Joe and Mumble have.

The rural sample $X_{1}$ has $n_{1}=190$ and $x_{1}=70$ .

The urban sample $X_{2}$ has $n_{2}=310$ and $x_{2}=65$ .

Let’s first compute the estimates for the respective proportions — $p_{1}$ and $p_{2}$ .

$\hat{p_{1}}=\frac{x_{1}}{n_{1}}=\frac{70}{190} = 0.3684$

$\hat{p_{2}}=\frac{x_{2}}{n_{2}}=\frac{65}{310} = 0.2097$

Then, let’s compute the pooled estimate $p$ for the population proportions.

$p = \frac{x_{1}+x_{2}}{n_{1}+n_{2}}=\frac{70+65}{190+310}=\frac{135}{500}=0.27$

Next, let’s compute the test-statistics under the large-sample assumption. 190 and 310 are pretty large samples.

$z = \frac{\hat{p_{1}}-\hat{p_{2}}}{\sqrt{p(1-p)*(\frac{1}{n_{1}}+\frac{1}{n_{2}})}}$

$z = \frac{0.3684-0.2097}{\sqrt{0.27(0.73)*(\frac{1}{190}+\frac{1}{310})}}=3.8798$

Since our alternate hypothesis $H_{A}$ is $p_{1}-p_{2}>0$ , we compute the p-value as,
$p-value=P(Z \ge 3.8798) = 5.227119*10^{-5} \approx 0$

Since the p-value (~0) is smaller than our selected rate of error (0.05), we reject the null hypothesis for the alternate view that the sentiment for increasing the speed limit is stronger among rural () than among urban residents ().

Remember that the test-statistic is computed for the null hypothesis that $p_{1}-p_{2}=0$ . What if the null hypothesis is not that the difference in proportions is zero but is equal to some value? $p_{1}-p_{2}=0.25$

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Lesson 92 – The Two-Sample Hypothesis Test – Part I

Fisher’s Exact Test

You may remember this from Lesson 38, where we derived the hypergeometric distribution from first principles.

If there are R Pepsi cans in a total of N cans (N-R Cokes) and we are asked to identify them correctly, in our choice selection of R Pepsis, we can get k = 0, 1, 2, … R Pepsis. The probability of correctly selecting k Pepsis is

$P(X=k) = \frac{\binom{R}{k}\binom{N-R}{R-k}}{\binom{N}{R}}$

X, the number of correct guesses (0, 1, 2, …, R) assumes a hypergeometric distribution. The control parameters of the hypergeometric distribution are N and R.

For example, if there are five cans in total, out of which three are Pepsi cans, picking exactly two Pepsi cans can be done in $\binom{3}{2}*\binom{2}{1}$ ways. Two Pepsi cans selected from three in $\binom{3}{2}$ ways; one Coke can be selected from two Coke cans in $\binom{2}{1}$ .

The overall possibilities of selecting three cans from a total of five cans are $\binom{5}{3}$ .

Hence, $P(X=2)=\frac{\binom{3}{2}*\binom{2}{1}}{\binom{5}{3}}=\frac{6}{10}$

Now, suppose there are eight cans out of which four are Pepsi, and four are Coke, and we are testing John’s ability to identify Pepsi.

Since John has a better taste for Pepsi, he claims that he has a greater propensity to identify Pepsi from the hidden cans.

Of course, we don’t believe it, and we think his ability to identify Pepsi is no different than his ability to identify Coke.

Suppose his ability (probability) to identify Pepsi is $p_{1}$ and his ability to identify Coke is $p_{2}$ . We think $p_{1}=p_{2}$ and John thinks $p_{1} > p_{2}$ .

The null hypothesis that we establish is
$H_{0}: p_{1} = p_{2}$

John has an alternate hypothesis
$H_{A}: p_{1} > p_{2}$

Pepsi and Coke cans can be considered as two samples of four each.

Since there are two samples (Pepsi and Coke) and two outcomes (identifying or not identifying Pepsi), we can create a 2×2 contingency table like this.

John now identifies four cans as Pepsi out of the eight cans whose identity is hidden as in the fun experiment.

It turns out that the result of the experiment is as follows.

John correctly identified three Pepsi cans out of the four.

The probability that he will identify three correctly while sampling from a total of eight cans is

$P(X=3)=\frac{\binom{4}{3}*\binom{4}{1}}{\binom{8}{4}}=\frac{\frac{4!}{1!3!}\frac{4!}{3!1!}}{\frac{8!}{4!4!}}=\frac{16}{70}=0.2286$

If you recall from the prior hypothesis test lessons, you will ask for the null distribution. The null distribution is the probability distribution of observing any number of Pepsi cans while selecting from a total of eight cans (out of which four are known to be Pepsi). This will be the distribution that shows P(X=0), P(X=1), P(X=2), P(X=3), and P(X=4). Let’s compute these and present them visually.

$P(X=0)=\frac{\binom{4}{0}*\binom{4}{4}}{\binom{8}{4}}==\frac{1}{70}=0.0143$

$P(X=1)=\frac{\binom{4}{1}*\binom{4}{3}}{\binom{8}{4}}==\frac{16}{70}=0.2286$

$P(X=2)=\frac{\binom{4}{2}*\binom{4}{2}}{\binom{8}{4}}==\frac{36}{70}=0.5143$

$P(X=3)=\frac{\binom{4}{3}*\binom{4}{1}}{\binom{8}{4}}==\frac{16}{70}=0.2286$

$P(X=4)=\frac{\binom{4}{4}*\binom{4}{0}}{\binom{8}{4}}==\frac{1}{70}=0.0143$

In a hypergeometric null distribution with N = 8 and R = 4, what is the probability of getting a larger value than 3? If this has a sufficiently low probability, we cannot say that it may occur by chance.

This probability is the p-value. It is the probability of obtaining the computed test statistic under the null hypothesis. The smaller the p-value, the less likely the observed statistic under the null hypothesis – and stronger evidence of rejecting the null.

$P(X \ge 3)=P(X=3) + P(X=4) = 0.2286+0.0143=0.2429$

Let us select a rate of error $\alpha$ of 10%.

Since the p-value (0.2429) is greater than our selected rate of error (0.1), we cannot reject the null hypothesis that the probability of choosing Pepsi and the probability of choosing Coke are the same.

John should have selected all four Pepsi cans for us to be able to reject the null hypothesis ( $H_{0}: p_{1} = p_{2}$ ) in favor of the alternative hypothesis ( $H_{A}: p_{1} > p_{2}$ ) conclusively.

The Famous Fisher Test

We just saw a variant of the famous test conducted by Ronald Fisher in 1919 when he devised an offhand test of a lady’s ability to differentiate between tea prepared in two different ways.

One afternoon, at tea-time in Rothamsted Field Station in England, a lady proclaimed that she preferred her tea with the milk poured into the cup after the tea, rather than poured into the cup before the tea. Fisher challenged the lady and presented her with eight cups of tea; four made the way she preferred, and four made the other way. She was told that there were four of each kind and asked to determine which four were prepared properly. Fisher subsequently used this experiment to illustrate the basic issues in experimentation.
sourced from Chapter 5 of “Teaching Statistics, a bag of tricks” by Andrew Gelman and Deborah Nolan

This test, now popular as Fisher’s Exact Test, is the basis for the two-sample hypothesis test to verify the difference in proportions. Just like how the proportion (p) for the one-sample test followed a binomial null distribution, the test-statistic for the two-sample test follows a hypergeometric distribution when $H_{0}$ is true.

Here, where we know the exact number of correct Pepsi cans, the true distribution of the test-statistic (number of correct Pepsi cans) is hypergeometric. In more generalized cases where the number of successes is not known apriori, we need to make some assumptions.

Say there are two samples represented by random variables $X_{1}$ and $X_{2}$ with sample sizes $n_{1}$ and $n_{2}$ . The proportion $p_{1}$ is based on the number of successes ( $x_{1}$ ) in $X_{1}$ , and the proportion $p_{2}$ is based on the number of successes ( $x_{2}$ ) in $X_{2}$ . Let the total number of successes in both the samples be $t=x_{1}+x_{2}$ .

If the null hypothesis is $H_{0}: p_{1} = p_{2}$ , then, large values of $x_{1}$ and small values of $x_{2}$ support the alternate hypothesis that $H_{A}: p_{1} > p_{2}$ when t is fixed.

In other words, for a fixed value of $t=x_{1}+x_{2}$ , we reject $H_{0}: p_{1} = p_{2}$ , if there are more successes in $X_{1}$ compared to $X_{2}$ .

So the question is: what is the probability distribution of $x_{1}$ when the total successes are fixed at $t$ , and we have a total of $n_{1}+n_{2}$ samples.

When the number of successes is t, and when $H_{0}: p_{1} = p_{2}$ is true, these successes can come from any of the two random variables with equal likelihood.

A total sample of $n_{1}+n_{2}$ exists out of which the number of ways of choosing $n_{1}$ samples is $\binom{n_{1}+n_{2}}{n_{1}}$ .

A total of t successes exist, out of which the number of ways of choosing k is $\binom{t}{k}$ .

A total of $n_{1}+n_{2}-t$ non-successes exist, out of which the number of ways of choosing $n_{1}-k$ is $\binom{n_{1}+n_{2}-t}{n_{1}-k}$ .

When we put them together, we can derive the probability $P(X=k)$ for the hypergeometric distribution when $H_{0}$ is true.

$P(X=k) = \frac{\binom{t}{k}*\binom{n_{1}+n_{2}-t}{n_{1}-k}}{\binom{n_{1}+n_{2}}{n_{1}}}$

Conditional on a total number of t successes that can come from any of the two random variables, the number of successes $X=k$ in the first sample has a hypergeometric distribution when $H_{0}$ is true.
The p-value can thus be derived.

We begin diving into the two-sample tests. Fisher’s Exact Test and its generalization (with assumptions) for the two-sample hypothesis test on the proportion is the starting point. It is a direct extension of the one-sample hypothesis test on proportion — albeit with some assumptions. Assumptions are crucial for the two-sample hypothesis tests. As we study the difference in the parameters of the two populations, we will delve into them more closely.
STAY TUNED!

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Tag: Hypergeometric Distribution

Lesson 93 – The Two-Sample Hypothesis Test – Part II

On the Difference in Proportions

What if Joe and Mumble surveyed many more people?

Okay, we are done. Let’s see what Joe and Mumble have.

Lesson 92 – The Two-Sample Hypothesis Test – Part I

Fisher’s Exact Test

The Famous Fisher Test

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