Tag: hypothesis test

Lesson 89 – The One-Sample Sign Test

Hypothesis Tests – Part IV

H_{0}: P(X > \mu) = 0.5

H_{A}: P(X > \mu) > 0.5

H_{A}: P(X > \mu) < 0.5

H_{A}: P(X > \mu) \neq 0.5

Joe is ready to buy his first (used) car as anyone does when they are new to the whole driving experience.

After a careful search on craigslist, he narrowed down to check out a 2003 Ford Focus and a 2011 Hyundai Sonata Hybrid. He liked their specs and wanted to verify the all-important gas mileage.

While he is taking the words of the owners at face value, he also wants to check the mileage reported by the general public to the EPA.

In case you are unaware, the United States Environmental Protection Agency (USEPA) maintains a website www.fueleconomy.gov where they provide information on the fuel economy for different vehicles.
They give their rating based on standardized tests to reflect typical driving conditions and driver behavior. The public also shares their miles per gallon (MPG) estimates.

At any rate, Joe navigated through this interesting website and checked out the data that people shared for these two varieties.

Here is the data shared by eight people for the 2003 Ford Focus.

As you can see, EPA rates an average of 26 MPG for this variety.

Joe wants to check if the average mileage for this car as derived from a sample data conforms to the EPA rating or not.

Joe already knows how to do this using the one-sample t-test. So he goes about establishing the null and the alternative hypothesis, deciding on the acceptable rejection rate, computing the test statistic and its corresponding p-value.

Here is a summary of his test.

From the sample of eight vehicles, he computed \bar{x} as 26.69 MPG, which he compares with EPA’s rating — \mu of 26 MPG. The test statistic t_{0} is 0.52.

His alternate hypothesis is \mu < 26 MPG against the null hypothesis of \mu \ge 26 MPG. So, the critical value from the T-distribution with 7 degrees of freedom (n=8) is -1.9.

Since the test statistic is greater than the critical value, he cannot reject the hypothesis.

Until further evidence is available, he will continue to believe that the 2003 Ford Focus will have an average fuel economy of 26 MPG.

Enter the Sign Test

Joe can also perform a One-Sample Sign Test if he needs more confirmation.

The sign test determines whether a given sample X is generally larger or smaller or different than a selected value \mu. It works well when there are outliers in the data.

Joe has to establish the following null hypothesis.

H_{0}: P(X > 26) = 0.5

If H_{0} is true, about half of the values of sample X will be greater than 26 MPG, and about half of them will be less than 26 MPG. In other words, the sample data will be dispersed around \mu with equal likelihood.

To this null hypothesis, he will cast an alternate hypothesis

H_{A}: P(X > 26) < 0.5

If H_{A} is true, more than half of the values of sample X will be less than 26 MPG. In other words, the sample shows low mileages — significantly less than 26 MPG.

Look at this visual. Joe is plotting the sample data on the number line and indicating 26 MPG cutoff line along with whether the sample data is greater than 26 MPG or less than 26 MPG using a plus and a minus sign — the sign test 😉

Three vehicles from the sample had a reported mileage of less than 26 MPG. Out of the eight data points, the number of positives is five, and the number of negative is three.

The number of positives, S^{+}, is the test statistic for the sign test.

In a sample of eight (n = 8), S^{+}=5.

S^{+} could have been 0, 1, 2, 3, 4, 5, 6, 7, or 8.

Under the null hypothesis, S^{+} follows a binomial distribution with a probability of 0.5. We consider this as the null distribution.

Look at this visual.

It shows the possible S^{+} values on the x-axis and P(S^{+}=k) on the y-axis. The null distribution is the distribution of the probability that S^{+} can be 0, 1, 2, …, or 8. For n = 8 and p = 0.5.

P(S^{+}=k) = {8 \choose k} p^{k}(1-p)^{8-k}

For instance, P(S^{+}=0) = {8 \choose 0} p^{0}(1-p)^{8-0} = 0.0039

Here is the full binomial table.

Now, since the sample yields S^{+}=5, p-value for this test is the probability of finding up to five positives — P(S^{+} \le 5).

P(S^{+} \le 5)=\sum_{x=0}^{x=5} {8 \choose x} p^{x}(1-p)^{8-x}=0.8555

Since the p-value is greater than 0.05, the rejection rate Joe has decided upon; he cannot reject the null hypothesis.

If you noticed, the sign test only uses the signs of the data and not the magnitudes. So it is resistant to outliers.

What about Hyundai Sonata Hybrid

Joe quickly looks for some data on the 2011 Hyundai Sonata Hybrid. Sure enough, he finds this.

There is a sample of 24 vehicles of this type, and EPA rates this at 36 MPG.

Let’s see how Joe uses the sign test to verify if the sample data conforms to the EPA specified rating.

He first establishes the null and the alternative hypothesis.

H_{0}: P(X > 36) = 0.5

H_{A}: P(X > 36) < 0.5

If H_{0} is true, about half of the 24 values found in the sample will be greater than 36 MPG, and about half of them will be negative.

If H_{A} is true, more than half of the 24 values will be less than 36 MPG. In other words, the sample reveals a significantly lower mileage than the EPA rated 36 MPG.

Joe then plots the data on the number line to visually find the number of positive and negative signs.

He counts them as 7.

There is one number that is exactly 36 MPG. He removes this from the sample and conducts the test with 23 data points.

The test statistic is S^{+}=7 out of 23 values (n = 23). The p-value for the test is then the probability of seeing value for S^{+} of up to 7 in a binomial distribution with n = 23 and p = 0.5.

P(S^{+} \le 7)=\sum_{x=0}^{x=7} {23 \choose x} p^{x}(1-p)^{23-x}=0.046

Since the p-value is less than \alpha, Joe has to reject the null hypothesis that the average mileage for the 2011 Hyundai Sonata Hybrid is 36 MPG.

Why can’t Joe just conduct the t-test?

He can. Here is the result of the t-test.

The test statistic t_{0} falls in the rejection region, i.e., t_{0}<t_{critical}, or the p-value is less than \alpha.

He will reject the null hypothesis test.

What about the outliers?

Suppose there is an error in entering the 21st value. Let’s say it is 48.8 MPG instead of 38.8 MPG.

How will this change the result of the t-test?

\bar{x} and s will be 33.875 MPG and 6.099 MPG respectively.

The test statistic t_{0} will be -1.706.

At 5% rejection rate and 23 degrees of freedom, t_{critical} will be -1.713.

Since t_{0}>t_{critical}, Joe would not be able to reject the null hypothesis.

Whereas, if he uses the sign test for his investigation, as he is doing now, the outlier data point 48.8 MPG will still show up only as a positive sign and the resultant S^{+} will still be seven and the p-value P(S^{+} \le 7) will still be 0.046.

The sign test is resistant to outliers.

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Lesson 88 – The One-Sample Hypothesis Test – Part III

On Variance

H_{0}: \sigma^{2} = \sigma^{2}_{0}

H_{A}: \sigma^{2} > \sigma^{2}_{0}

H_{A}: \sigma^{2} < \sigma^{2}_{0}

H_{A}: \sigma^{2} \neq \sigma^{2}_{0}

Joe is cheery after an intense semester at his college. He is meeting Devine today for a casual conversation. We all know that their casual conversation always turns into something interesting. Are we in for a new concept today?

Devine: So, how did you fare in your exams.

Joe: Hmm, I did okay, but, interestingly, you are asking me about my performance in exams and not what I learned in my classes.

Devine: Well, Joe, these days, the college prepares you to be a good test taker. Learning is a thing of the past. I am glad you are still learning in your classes.

Joe: That is true to a large extent. We have exams after exams after exams, and our minds are compartmentalized to regurgitate one module after the other — no time to sit back and absorb what we see in classes.

By the way, I heard of an intriguing phenomenon from one of my professors. It might be of interest to you.

Devine: What is it.

Joe: In his eons of teaching, he has observed that the standard deviation of his class’s performance is 16.5 points. He told me that over the years, this had fed back into his ways of preparing exams. It seems that he subconsciously designs exams where the students’ grades will have a standard deviation of 16.5.

Devine: That is indeed an interesting phenomenon. Do you want to verify his hypothesis?

Joe: How can we do that?

Devine: Assuming that his test scores are normally distributed, we can conduct a hypothesis test on the variance of the distribution — H_{0}: \sigma^{2} = \sigma^{2}_{0}

Joe: Using a hypothesis testing framework?

Devine: Yes. Let’s first outline a null and alternate hypothesis. Since your professor is claiming that his exams are subconsciously designed for a standard deviation of 16.5, we will establish that this is the null hypothesis.

H_{0}: \sigma^{2} = 16.5^{2}

We can falsify this claim if the standard deviation is greater than or less than 16.5, i.e.,

H_{A}: \sigma^{2} \neq 16.5^{2}

The alternate hypothesis is two-sided. Deviation in either direction (less than or greater than) will reject the null hypothesis.

Would you be able to get some data on his recent exam scores?

Joe: I think I can ask some of my friends and get a sample of up to ten scores. Let me make some calls.

Here is a sample of ten exam scores from our most recent test with him.

60, 41, 70, 61, 69, 95, 33, 34, 82, 82

Devine: Fantastic. We can compute the standard deviation/variance from this sample and verify our hypothesis — whether this data provides evidence for the rejection of the null hypothesis.

Joe: Over the past few weeks, I was learning that we call it a parametric hypothesis test if we know the limiting form of the null distribution. I already know that we are doing a one-sample hypothesis test, but how do we know the type of the null distribution?

Devine: The sample variance (s^{2}) is a random variable that can be described using a probability distribution. Several weeks ago, in lesson 73 where we derived the T-distribution, and in lesson 75 where we derived the confidence interval of the variance, we learned that \frac{(n-1)s^{2}}{\sigma^{2}} follows a Chi-square distribution with (n-1) degrees of freedom.

Since it was more than ten lessons ago, let’s go through the derivation once again. Ofttimes, repetition helps reinforce the ideas.

Joe: I think I remember it vaguely. Let me take a shot at the derivation 🙂

I will start with the equation of the sample variance s^{2}.

s^{2} = \frac{1}{n-1} \sum(x_{i}-\bar{x})^{2}

I will move the n-1 term over to the left-hand side and do some algebra.

(n-1)s^{2} = \sum(x_{i}-\bar{x})^{2}

(n-1)s^{2} = \sum(x_{i} - \mu -\bar{x} + \mu)^{2}

(n-1)s^{2} = \sum((x_{i} - \mu) -(\bar{x} - \mu))^{2}

(n-1)s^{2} = \sum[(x_{i} - \mu)^{2} + (\bar{x} - \mu)^{2} -2(x_{i} - \mu)(\bar{x} - \mu)]

(n-1)s^{2} = \sum(x_{i} - \mu)^{2} + \sum (\bar{x} - \mu)^{2} -2(\bar{x} - \mu)\sum(x_{i} - \mu)

(n-1)s^{2} = \sum(x_{i} - \mu)^{2} + n (\bar{x} - \mu)^{2} -2(\bar{x} - \mu)(\sum x_{i} - \sum \mu)

(n-1)s^{2} = \sum(x_{i} - \mu)^{2} + n (\bar{x} - \mu)^{2} -2(\bar{x} - \mu)(n\bar{x} - n \mu)

(n-1)s^{2} = \sum(x_{i} - \mu)^{2} + n (\bar{x} - \mu)^{2} -2n(\bar{x} - \mu)(\bar{x} - \mu)

(n-1)s^{2} = \sum(x_{i} - \mu)^{2} + n (\bar{x} - \mu)^{2} -2n(\bar{x} - \mu)^{2}

(n-1)s^{2} = \sum(x_{i} - \mu)^{2} - n (\bar{x} - \mu)^{2}

Let me divide both sides of the equation by \sigma^{2}.

\frac{(n-1)s^{2}}{\sigma^{2}} = \frac{1}{\sigma^{2}}(\sum(x_{i} - \mu)^{2} - n (\bar{x} - \mu)^{2})

\frac{(n-1)s^{2}}{\sigma^{2}} = \sum(\frac{x_{i} - \mu}{\sigma})^{2} - \frac{n}{\sigma^{2}} (\bar{x} - \mu)^{2}

\frac{(n-1)s^{2}}{\sigma^{2}} = \sum(\frac{x_{i} - \mu}{\sigma})^{2} - (\frac{\bar{x} - \mu}{\sigma/\sqrt{n}})^{2}

The right-hand side now is the sum of squared standard normal distributions — assuming x_{i} are draws from a normal distribution.

\frac{(n-1)s^{2}}{\sigma^{2}} = Z_{1}^{2} + Z_{2}^{2} + Z_{3}^{2} + … + Z_{n}^{2} - Z^{2}

Sum of squares of (n - 1) standard normal random variables.

We learned in lesson 53 that if there are n standard normal random variables, Z_{1}, Z_{2}, …, Z_{n}, their sum of squares is a Chi-square distribution with n degrees of freedom. Its probability density function is f(\chi)=\frac{\frac{1}{2}(\frac{1}{2} \chi)^{\frac{n}{2}-1}e^{-\frac{1}{2}*\chi}}{(\frac{n}{2}-1)!} for \chi^{2} > 0 and 0 otherwise.

Since we have \frac{(n-1)s^{2}}{\sigma^{2}} = Z_{1}^{2} + Z_{2}^{2} + Z_{3}^{2} + … + Z_{n}^{2} - Z^{2}

\frac{(n-1)s^{2}}{\sigma^{2}} follows a Chi-square distribution with (n-1) degrees of freedom.

\frac{(n-1)s^{2}}{\sigma^{2}} \sim \chi_{n-1} with a probability distribution function f(\frac{(n-1)s^{2}}{\sigma^{2}}) = \frac{\frac{1}{2}(\frac{1}{2} \chi)^{\frac{n-1}{2}-1}e^{-\frac{1}{2}*\chi}}{(\frac{n-1}{2}-1)!}

Depending on the degrees of freedom, the distribution of \frac{(n-1)s^{2}}{\sigma^{2}} looks like this.

Smaller sample sizes imply lower degrees of freedom. The distribution will be highly skewed; asymmetric.

Larger sample sizes or higher degrees of freedom will tend the distribution to symmetry.

Devine: Excellent job, Joe. As you have shown \frac{(n-1)s^{2}}{\sigma^{2}} is our test statistic, \chi^{2}_{0}, which we will verify against a Chi-square distribution with (n-1) degrees of freedom.

Have you already decided on a rejection rate \alpha?

Joe: I will go with a 5% Type I error. If my professor’s assumption is indeed true, I am willing to commit a 5% error in my decision-making as I may get a sample from my friends that drives me to reject his null hypothesis.

Devine: Okay. Let’s then compute the test statistic.

s^{2} = \frac{1}{n-1} \sum(x_{i}-\bar{x})^{2}=452.01

\chi^{2}_{0} = \frac{(n-1)s^{2}}{\sigma^{2}} = \frac{9*452.09}{16.5^{2}} = 14.95

Since we have a sample of ten exam scores, we should consider as null distribution, a Chi-square distribution with nine degrees of freedom.

Under the null hypothesis H_{0}: \sigma^{2} = 16.5^{2}, for a two-sided hypothesis test at the 5% rejection level, \frac{(n-1)s^{2}}{\sigma^{2}} can vary between \chi^{2}_{0.025} and \chi^{2}_{0.975}, the lower and the upper percentiles of the Chi-square distribution.

If our test statistic \chi^{2}_{0} is either less than, or greater than the lower and the upper percentiles respectively, we reject the null hypothesis.

The lower and upper critical values at the 5% rejection rate (or a 95% confidence interval) are 2.70 and 19.03.

In lesson 75, we learned how to read this off the standard Chi-square table.

Joe: Aha. Since our test statistic \chi^{2}_{0} is 14.95, we cannot reject the null hypothesis.

Devine: You are right. Look at this visual.

The rejection region based on the lower and the upper critical values (percentiles \chi^{2}_{0.025} and \chi^{2}_{0.975}) is shown in red triangles. The test statistic lies inside.

It is now easy to say that the p-value, i.e., P(\chi^{2}>\chi^{2}_{0}) or P(\chi^{2} \le \chi^{2}_{0}) is greater than \frac{\alpha}{2}.

Since we have a two-sided test, we compare the p-value with \frac{\alpha}{2}.

Hence we cannot reject the null hypothesis.

Joe: Looks like I cannot refute my professor’s observation that the standard deviation of his test scores is 16.5 points.

Devine: Yes, at the 5% rejection level, and assuming that his test scores are normally distributed.

Joe: Got it. If the test scores are not normally distributed, our assumption that \frac{(n-1)s^{2}}{\sigma^{2}} follows a Chi-square distribution is questionable. How then can we test the hypothesis?

Devine: We can use a non-parametric test using a bootstrap approach.

Joe: How is that done?

Devine: You will have to wait until the non-parametric hypothesis test lessons for that. But let me ask you a question based on today’s lesson. What is the main difference between the hypothesis test on the mean, which you learned in lesson 87, and the hypothesis test on the variance which you learned here?

Joe: 😕 😕 😕

For the hypothesis test on the mean, we looked at the difference between \bar{x} and \mu. For the hypothesis on the variance, we examine the ratio of s^{2} to \sigma^{2} and reject the null hypothesis if this ratio differs too much from what we expect under the null hypothesis, i.e., when H_{0} is true.

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Lesson 87 – The One-Sample Hypothesis Test – Part II

On Mean

H_{0}: \mu = \mu_{0}

H_{A}: \mu > \mu_{0}

H_{A}: \mu < \mu_{0}

H_{A}: \mu \neq \mu_{0}

Tom grew up in the City of Ruritania. He went to high school there, met his wife there, and has a happy home. Ruritania, the land of natural springs, is known for its pristine water. Tom’s house is alongside the west branch of Mohawk, one such pristine river. Every once in a while, Tom and his family go to the nature park on the banks of Mohawk. It is customary for Tom and his little one to take a swim.

Lately, he has been sensing a decline in the quality of the water. It is a scary feeling as the consequences are slow. Tom starts associating the cause of the poor water quality to this new factory in his neighborhood constructed just upstream of Mohawk. 

Whether or not the addition of this new factory in Tom’s neighborhood reduced the quality of water compared to EPA standards is to be seen.

He immediately checked the EPA specifications for dissolved oxygen concentration in the river, and it is required by the EPA to have a minimum average concentration of 2 mg/L. Over the next ten days, Tom collected ten water samples from the west branch and got his friend Ron to measure the dissolved oxygen in his lab. In mg/L, the data reads like this.

1.8, 2, 2.1, 1.7, 1.2, 2.3, 2.5, 2.9, 1.9, 2.2

Tom wants to test if the average dissolved oxygen he sees from the samples significantly deviates from the one specified by EPA.

Does \bar{x} deviate from 2 mg/L? Is that deviation large enough to prompt caution?

He does this investigation using the hypothesis testing framework.

Since the investigation is regarding \bar{x} , the sample mean, and whether it is different from a selected value, it is reasonable to say that Tom is conducting a one-sample hypothesis test. 

Tom knows this, and so do you and me — the sample mean (\bar{x} ) is a random variable, and it can be described using a probability distribution. If Tom gets more data samples, he will get a slightly different sample mean. The value of the estimate changes with the change of sample and this uncertainty can be represented using a normal distribution by the Central Limit Theorem.

The sample mean is an unbiased estimate of the true mean, so the expected value of the sample mean is equal to the truth. E[\bar{x}]=\mu . Go down the memory lane and find why in Lesson 67.

The variance of the sample mean is V[\bar{x}]=\frac{\sigma^{2}}{n} . Variance tells us how widely the estimate is distributed around the center of the distribution. We know this from Lesson 68.

When we put these two together,

\bar{x} \sim N(\mu, \frac{\sigma^{2}}{n})

or,

\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}} \sim N(0,1)

Now, if the sample size (n) is large enough, it would be reasonable to substitute sample standard deviation (s) in place of \sigma.

When we substitute s for \sigma, we cannot just assume that \bar{x} will tend to a normal distribution.

W. S. Gosset (aka “Student”) taught us that \frac{\bar{x}-\mu}{\frac{s}{\sqrt{n}}} follows a T-distribution with (n-1) degrees of freedom. 

\frac{\bar{x}-\mu}{\frac{s}{\sqrt{n}}} \sim t_{df=n-1}

Anyhow, all this is to confirm that Tom is conducting a parametric hypothesis test. 

CHOOSE THE APPROPRIATE TEST; ONE-SAMPLE OR TWO-SAMPLE AND PARAMETRIC OR NONPARAMETRIC — check.

Tom establishes the null and alternate hypotheses. He assumes that the inception of the factory does not affect the water quality downstream of the Mohawk River. Hence,

H_{0}: \mu \ge 2 mg/L

H_{A}: \mu < 2 mg/L

The alternate hypothesis is one-sided. A significant deviation in one direction (less than) needs to be seen to reject the null hypothesis.

Notice that his null hypothesis is \mu \ge 2 mg/L since it is required by the EPA to have a minimum average concentration of 2 mg/L.

ESTABLISH THE NULL AND ALTERNATE HYPOTHESIS — check.

Tom is taking a 5% risk of rejecting the null hypothesis; \alpha =0.05 . His Type I error is 5%.

Suppose the factory does not affect the water quality, but, the ten samples he collected showed a sample mean much smaller than the EPA prescription of 2 mg/L, he should reject the null hypothesis — so he is committing an error (Type I error) in his decision making.

There is a certain level of subjectivity in the choice of \alpha . If Tom wants to see that the water quality is lower than 2 mg/L, he would perhaps choose to commit a greater error, i.e., select a larger value for \alpha .

If he wants to see that the water quality has not deteriorated, he will choose a smaller value for \alpha .

So, the decision to reject or not to reject the null hypothesis is based on \alpha .

DECIDE ON AN ACCEPTABLE RATE OF ERROR OR REJECTION RATE — check.

Since \frac{\bar{x}-\mu}{\frac{s}{\sqrt{n}}} \sim t_{df=n-1}, the null distribution is a T-distribution with (n-1) degrees of freedom.

The test statistics is then t_{0} = \frac{\bar{x}-\mu}{\frac{s}{\sqrt{n}}}, and Tom has to verify how likely it is to see a value as large as t_{0} in the null distribution.

Look at his visual.

The distribution is a T-distribution with 9 degrees of freedom. Tom had collected ten samples for this test. 

Since he opted for a rejection level of 5%, there is a cutoff on the distribution at -1.83.

-1.83 is the quantile corresponding to a 5% probability (rate of rejection) for a T-distribution with nine degrees of freedom.

If the test statistic (t_{0}) is less than t_{critical} which is -1.83, he will reject the null hypothesis. 

This decision is equivalent to rejecting the null hypothesis if P(T \le t_{0}) (the p-value) is less than \alpha.

From his data, the sample mean (\bar{x}) is 2.06 and the sample standard deviation (s) is 0.46. 

t_{0} = \frac{\bar{x}-\mu}{\frac{s}{\sqrt{n}}}=\frac{2.06-2}{\frac{0.46}{\sqrt{10}}}=0.41.

t_{critical} can be read off from the standard T-Table, or P(T \le 0.41) can be computed from the distribution. 

At df = 9, and \alpha=5\%, t_{critical}=-1.83 and P(T \le 0.41)=0.66.

COMPUTE THE TEST STATISTIC AND ITS CORRESPONDING P-VALUE FROM THE OBSERVED DATA — check.

Since the test-statistics t_{0} is not in the rejection region, or since p-value > \alpha, Tom cannot reject the null hypothesis that \mu \ge 2 mg/L. 

MAKE THE DECISION; REJECT THE NULL HYPOTHESIS IF THE P-VALUE IS LESS THAN THE ACCEPTABLE RATE OF ERROR — check.

Tom could easily have checked the confidence interval of the true mean to make this decision. Recall that the confidence interval is the range or an interval where the true value will be. So, based on the T-distribution with df = 9, Tom could develop the 90% confidence interval (why 90%?) and check if \mu = 2 mg/L is within that confidence interval.

Look at this visual. Tom just did that.

The confidence interval is from 1.8 mg/L to 2.32 mg/L and the null hypothesis that \mu = 2 mg/L is within the interval.

Hence, he cannot reject the null hypothesis.

While looking at the confidence interval gives us a visual intuition on what decision to make, it is always better to compute the p-value and compare it to the rejection rate.

Together, the p-value and \alpha provide the risk levels associated with decisions.

In this journey through the hypothesis framework, the next time we meet, we will unfold the knots of the test on the variance. Till then, meditate on this.

For a hypothesis test, just reporting the p-value in itself is more informative. Once the p-value is known, any person who understands the context of the problem can decide for themselves whether or not to reject the null hypothesis. In other words, they can set their level of rejection and compare the p-value to it.

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Lesson 86 – The One-Sample Hypothesis Test – Part I

On Proportion

H_{0}: p = p_{0}

H_{A}: p > p_{0}

H_{A}: p < p_{0}

H_{A}: p \neq p_{0}

Our journey to the abyss of hypothesis tests begins today. In lesson 85, Joe and Devine, in their casual conversation about the success rate of a memory-boosting mocha, introduced us to the elements of hypothesis testing. Their conversation presented a statistical hypothesis test on proportion — whether the percentage of people who would benefit from the memory-booster coffee is higher than the percentage who would claim benefit randomly.

In this lesson, using a similar example on proportion, we will dig deeper into the elements of hypothesis testing.

To reiterate the central concept, we wish to test our assumption (null hypothesis H_{0}) against an alternate assumption (alternate hypothesis H_{A}). The purpose of a hypothesis test, then, is to verify whether empirical data supports the rejection of the null hypothesis.

Let’s assume that there is a vacation planner company in Palm Beach, Florida. They are finishing up their new Paradise Resorts and advertised that this Paradise Resorts’ main attraction is it’s five out of seven bright and sunny days!

.

.

.

I know what you are thinking. It’s Florida, and five out of seven bright and sunny? What about the muggy thunderstorms and summer hurricanes?

Let’s keep that skepticism and consider their claim as a proposition, a hypothesis.

Since they claim that their resorts will have five out of seven bright and sunny days, we can assume a null hypothesis (H_{0}) that p = \frac{5}{7}. We can pit this against an alternate hypothesis (H_{A}) that p < \frac{5}{7} and use observational (experimental or empirical) data to verify whether H_{0} can be rejected.

We can go down to Palm Beach and observe the weather for a few days. Or, we may have been to Palm Beach enough number of times that we can bring that empirical data out of our memory. Suppose we observe or remember that seven out of 15 days, we had bright and sunny days.

With this information, we are ready to investigate Paradise Resorts’ claim.

Let’s refresh our memory on the essential steps for any hypothesis test. 

1. Choose the appropriate test; one-sample or two-sample and parametric or nonparametric. 

2. Establish the null and alternate hypothesis. 

3. Decide on an acceptable rate of error or rejection rate (\alpha).

4. Compute the test statistic and its corresponding p-value from the observed data. 

5. Make the decision; Reject the null hypothesis if the p-value is less than the acceptable rate of error, \alpha.
Choose the appropriate test; one-sample or two-sample and parametric or nonparametric

We are verifying a statement about the parameter (proportion, p) of the population — whether or not p = \frac{5}{7}. So it is a one-sample hypothesis test. Since we are testing for proportion, we can assume a binomial distribution to derive the probabilities. So it is a parametric hypothesis test. 

Establish the null and alternate hypothesis

Paradise Resorts’ claim is the null hypothesis — five out of seven bright and sunny days. The alternate hypothesis is that the proportion is less than what they claim.

H_{0}: p = \frac{5}{7}

H_{A}: p < \frac{5}{7}

We are considering a one-sided alternative since departures in one direction (less than) are sufficient to reject the null hypothesis.

Decide on an acceptable rate of error or rejection rate \alpha

Our decision on the acceptable rate of rejection is the risk we take for rejecting the truth. If we select 10% for \alpha, it implies that we are rejecting the null hypothesis 10% of the times. If the null hypothesis is true, by rejecting it, we are committing an error — Type I error.

A simple thought exercise will make this concept more clear. Suppose Paradise Resorts’ claim is true — the proportion of bright and sunny days is \frac{5}{7}. But, our observation provided a sample out of the population where we ended up seeing very few bright and sunny days. In this case, we have to reject the null hypothesis. We committed an error in our decision. By selecting \alpha, we are choosing the acceptable rate of error. We are accepting that we might reject the null hypothesis (when it is true), \alpha\% of the time.

The next step is to create the null distribution.

At the beginning of the test, we agreed that we observed seven out of 15 days to be bright and sunny. We collected a sample of 15 days out of which seven days were bright and sunny. The null distribution is the probability distribution of observing any number of days being bright and sunny, i.e., out of the 15 days, we could have had 0, 1, 2, 3, …, 14, 15 days to be bright and sunny. The null distribution is the distribution of the probability of observing these outcomes. In a Binomial null distribution with n=15 and p = 5/7, what is the probability of getting 0, 1, 2, …, 15?

P(X=0)={15 \choose 0} p^{0}(1-p)^{15-0}

P(X=1)={15 \choose 1} p^{1}(1-p)^{15-1}

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P(X=15)={15 \choose 15} p^{15}(1-p)^{15-15}

It will look like this.

On this null distribution, you also see the region of rejection as defined by the selected rejection rate \alpha. Here, \alpha=10\%. In this null distribution, the quantile corresponding to \alpha=10\% is 8 days. Hence, if we observe more than eight bright and sunny days, we are not in the rejection region, and, if we observe eight or less bright and sunny days, we are in the rejection region.

Compute the test statistic and its corresponding p-value from the observed data

Next, the question we ask is this.

In a Binomial null distribution with n = 15 and p = 5/7, what is the probability of getting a value that is as large as 7? If the value has a sufficiently low probability, we cannot say that it may occur by chance.

This probability is called 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 \le 7)=\sum_{x=0}^{x=7}{15 \choose x} p^{x}(1-p)^{15-x}=0.04

You can see this probability in the figure below. The grey shade within the pink shade is the p-value.

Make the decision; Reject the null hypothesis if the p-value is less than the acceptable rate of error

It is evident at this point. Since the p-value (0.04) is less than our selected rate of error (0.1), we reject the null hypothesis, i.e., we reject Paradise Resorts’ claim that there will be five out of seven bright and sunny days.

This decision is based on the assumption that the null hypothesis is correct. Under this assumption, since we selected \alpha=10\%, we will reject the true null hypothesis 10% of the time. At the same time, we will fail to reject the null hypothesis 90% of the time. In other words, 90% of the time, our decision to not reject the null hypothesis will be correct.

Now, suppose Paradise Resorts’ hypothesis is false, i.e., they mistakenly think that there are five out of the seven bright and sunny days. However, it is not five in seven, but four in seven. What would be the consequence of their false null hypothesis?

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Let’s think this through again.
Our testing framework is based on the assumption that

H_{0}: p = \frac{5}{7}

H_{A}: p < \frac{5}{7}

For this test, we select \alpha=10\% and make decisions based on the observed outcomes.

Accordingly, if we observe eight or less bright and sunny days, we will reject the hypothesis, and, if we see more than eight bright and sunny days, we will fail to reject the null hypothesis. Based on \alpha=10\% and the assumed hypothesis that p = \frac{5}{7}, we fix eight as our cutoff point.

Paradise also thinks that p = \frac{5}{7}. If they are under a false assumption and we tested it based on this assumption, we might also commit an error — not rejecting the null hypothesis when it is false. This error is called Type II error or the lack of power in the test.

Look at this image. It has the null hypothesis under our original assumption and the selected \alpha=10\% and its corresponding quantile — 8 days. In the same image, we also see the null distribution if p = \frac{4}{7}. On this null distribution, there is a grey shaded region, which is the probability of not rejecting it based on \alpha=10\% and quantile — 8 days. We assign a symbol \beta for this probability.

What is more interesting is its complement, 1-\beta, which is the probability of rejecting the null hypothesis when it is false. Based on our original assumption (which is false), we selected eight days or less as our rejection region. At this cutoff, if there was another null distribution, 1-\beta is the probability of rejecting it. The key is the choice of \alpha or its corresponding quantile. At a chosen \alpha, 1-\beta measures the ability of the test to reject a false hypothesis. 1-\beta is called the power of the test.

In this example, if the original hypothesis is true, i.e., if H_{0}: p = \frac{5}{7} is true, we will reject it 10% of the time and will not reject it 90% of the time. However, if the hypothesis is false (and p = \frac{4}{7}), we will reject it 48% of the time and will not reject it 52% of the time.

For smaller p, the power of the test increases. In other words, if the proportion of bright and sunny days is smaller compared to the original assumption of 5/7, the probability of rejecting it increases.

Keep in mind that we will not know the actual value of p.

It is a thought that as the difference becomes larger, the original hypothesis is more and more false, and power (1-\beta) is a measure of the probability of rejecting this false hypothesis due to our choice of \alpha.

Look at this summary table. It provides a summary of our discussion of the error framework.

Type I and Type II errors are inversely related.

If we decrease \alpha, and if the null hypothesis is false, the probability of not rejecting it (\beta) will increase.

You can intuitively see that from the image that has the original (false) null distribution and possible true null distribution. If we move the quantile to the left (lower the rejection rate \alpha), the grey shaded region (probability of not rejecting a false null hypothesis, (\beta) increases.

At this point, you must be wondering that all of this is only for a sample of 15 days. What if we had more or fewer samples from the population?

The easiest way to understand the effect of sample size is to run the analysis for different n and different falsities (i.e., the difference from original p) and visualize it.

Here is one such analysis for three different sample sizes. The \alpha level that will be fixed based on the original hypothesis also varies by the sample size.

What we are seeing is the power function on the y-axis and the degree of falsity on the x-axis.

A higher degree of falsity implies that the null hypothesis is false by a greater magnitude. The first point on the x-axis is the fact that the null hypothesis is true. You can see that at this point, the power, i.e., the probability of rejecting the hypothesis, is 10%. At this point, we are just looking at \alpha, Type I error. As the degree of falsity increases, for that \alpha level, the power, 1-\beta (i.e., the probability of rejecting a false hypothesis) increases.

For a smaller sample size, the power increases slowly. For larger sample sizes, the power increases rapidly.

Of course, selecting the optimal sample size for the experiment based on low Type I and Type II errors is doable.

I am sure there are plenty of concepts here that will need some time to process, especially Type I and Type II errors. This week, we focused our energy on the hypothesis test for proportion. The next time we meet, we will unfold the knots of the hypothesis test on the mean.

Till then, happy learning.

If you are still unsure about Type I and Type II errors, this analogy will help.

If the null hypothesis for a judicial system is that the defendant is innocent, Type I error occurs when the jury convicts an innocent person; Type II error occurs when the jury sets a guilty person free.

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