chi-square – dataanalysisclassroom

Lesson 91 – The One-Sample Hypothesis Tests in R

The last time we did something in R is in Lesson 83. Since then, we have been learning the hypothesis testing framework. In Lesson 84, we first made friends with the hypothesis tests — the concept of needing evidence beyond a reasonable doubt. In Lesson 85, we learned the framework — the elements of hypothesis testing that provide us with a systematic way of setting up the test and using data to verify the hypothesis.

In Lessons 86, 87, and 88, we learned the one-sample hypothesis test on proportion, mean, and the variance, respectively. Then, in Lesson 89, we learned the one-sample sign test as a complement to the one-sample hypothesis test on the mean, which can account for outliers as well. Last week (Lesson 90), we dived into the bootstrap-based method where we relaxed the assumptions on the null distribution. The bootstrap-based process provides us the flexibility to run the test on various statistics of interest.

Today, let’s pause and look at how to perform these hypothesis tests in R. Brace yourself for an R drive.

To help us with this lesson today, we will make use of the eight data points on 2003 Ford Focus vehicle mileage that Joe used in Lesson 89.

The Basic Steps

Step1: Get the data

You can get the data from here.

Step 2: Create a new folder on your computer

Let us call this folder “lesson91”.

Step 3: Create a new code in R

Create a new code for this lesson. “File >> New >> R script”. Save the code in the same folder “lesson91” using the “save” button or by using “Ctrl+S”. Use .R as the extension — “lesson91_code.R”

Step 4: Choose your working directory

“lesson91” is the folder where we stored the code. Use “setwd(“path”)” to set the path to this folder. Execute the line by clicking the “Run” button on the top right.

setwd("path to your folder")

Step 5: Read the data into R workspace

Since the sample data is small enough, instead of saving the data in a text file and reading it into R workspace, we can directly input the data into the workspace as follows:

## Data Entry ##
# FORD MPG - reported # 
  x = c(21.7, 29, 28.1, 31.5, 24, 21.5, 28.7, 29)

The Baseline for Null Hypotheses

Let’s run the hypothesis tests on the mean, standard deviation, and the proportion. For this, we have to make some modest assumptions.

For the mean, we will use the EPA specified rating of 26.2 MPG. $\mu = 26.2$ MPG.

For comparing the standard deviation, we are unsure of the baseline standard deviation from the EPA rating. However, they provide us with a range that is between 22 MPG and 32 MPG. If we can assume that the range is covered by six standard deviations, i.e., $32 - 22 = 6 \sigma$ , we can get an estimate for baseline $\sigma$ . In this case, $\sigma = 1.67$ MPG.

Further, let’s also assume that one in eight cars will usually have a mileage less than the minimum rating of 22 MPG. $p = \frac{1}{8}$

# Compare to Baseline
  mu = 26
 
  rangex = c(22,32)
  sigma = diff(range_ford)/6

  n = length(x)
  p = 1/n

Hypothesis Test on the Mean

For the hypothesis test on the mean, we learned three methods, the t-test, the sign-test, and the bootstrap-based test.

Let’s start with the t-test. For the t-test, the null and the alternate hypothesis are:

$H_{0}: \mu \ge 26.2$ MPG

$H_{A}: \mu < 26.2$ MPG

The null distribution is a T-distribution with (n-1) degrees of freedom. The test statistic is $t_{0}=\frac{\bar{x}-\mu}{\frac{s}{\sqrt{n}}}$ . We have to verify how likely it is to see a value as large as $t_{0}$ in the null distribution.

Let’s compute the test statistic.

#T-Test #
   xbar = mean(x)

   s = sd(x)

   to = (xbar-mu)/(s/sqrt(n))

> to
   [1] 0.5173082

The test statistic is 0.517.

For a selected rejection rate of $\alpha$ we have to compute the p-value corresponding to this test statistic, or the critical value on the T-distribution.

   alpha = 0.05
   
   pvalue_ttest = pt(to,df=(n-1))   
   
   tcritical = qt(alpha,df=(n-1))

When you execute these lines, you will see that the p-value is 0.69, and the critical value from the t-distribution is -1.89. The “pt” function computed $P(T \le t_{0})$ from a T-distribution with user-specified degrees of freedom. For our case, df = 7. The “qt” function computes the quantile corresponding to a probability of 5%, i.e., $\alpha=0.05$ from a T-distribution with user-specified degrees of freedom.

> pvalue_ttest
   [1] 0.6895594

> tcritical
   [1] -1.894579

Since the p-value is greater than 0.05, or since $t_{0}>t_{critical}$ , we cannot reject the null hypothesis.

In R, there is a function to run this test — “t.test“

We could have used the function directly on the sample data x.

# using t-Test function in R #
   t.test(x,alternative = "less", mu = mu, conf.level = 0.95)

One Sample t-test
 data:  x
 t = 0.51731, df = 7, p-value = 0.6896
 alternative hypothesis: true mean is less than 26
 95 percent confidence interval:
      -Inf 29.20539
 sample estimates:
 mean of x 
   26.6875

It provides $t_{0}$ and the p-value as the outputs from which we can decide on the null hypothesis.

Next, let’s look at how to conduct the sign-test in R. The null and the alternative hypothesis for the sign-test are

$H_{0}: P(X > 26.2) = 0.5$

$H_{A}: P(X > 26.2) < 0.5$

If $H_{0}$ is true, about half of the values of sample X will be greater than 26.2 MPG, and about half of them will be negative. If $H_{A}$ is true, more than half of the values of sample X will be less than 26.2 MPG, i.e., the sample shows low mileages — significantly less than 26.2 MPG.

To run this test, we should compute $s^{+}$ , the test statistic that determines the number of values exceeding 26.2 MPG.

#Sign-Test
 splus = length(which(x>mu))

 > splus
   [1] 5

$s^{+}$ is 5.

Under the null hypothesis, $s^{+}$ follows a binomial distribution with a probability of 0.5. Using this assumption, we compute the p-value using the “binomial.test” function in R.

# Using Binom Test function in R 
   binom.test(splus,n,p=0.5,alternative = "less")

Exact binomial test
 data:  splus and n
 number of successes = 5, number of trials = 8, p-value = 0.8555
 alternative hypothesis: true probability of success is less than 0.5
 95 percent confidence interval:
  0.0000000 0.8888873
 sample estimates:
 probability of success 
                  0.625

The test provides the p-value, i.e., $P(S^{+} \le 5)$ from a binomial distribution with n = 8 and p = 0.5.

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

Since the p-value is greater than 0.05, the rejection rate, we cannot reject the null hypothesis.

Lastly, we will check out the bootstrap-based test on the mean. For the bootstrap-based one-sided test, the null hypothesis and the alternate hypothesis are

$H_{0}: P(\bar{x} > 26.2) = 0.5$

$H_{A}: P(\bar{x} > 26.2) < 0.5$

Use the following lines to set up and run the test.

#Bootstrap Test
 N = 10000

 xmean_null = matrix(NA,nrow=N,ncol=1)
  for (i in 1:N)
   {
     xboot = sample(x,n,replace=T)
     xmean_null[i,1] = mean(xboot)
   }
 
hist(xmean_null,col="pink",xlab="Mean of the Distribution",font=2,font.lab=2,main="Null Distribution Assuming Ho is True")

points(mu,10,pch=15,cex=2,col="blue")

abline(v=c(quantile(xmean_null,0.95)),col="black",lwd=2,lty=2)

pvalue_bootstrap = length(which(xmean_null>mu))/N

> pvalue_bootstrap
   [1] 0.7159

In the loop, we are executing the “sample” function to draw a bootstrap-replicate from the original data. Then, from this bootstrap-replicate, we compute the mean. Repeated such sampling and computation of the bootstrap-replicated mean statistic forms the null distribution. From this null distribution, we calculate the p-value, i.e., the proportion of the null distribution that exceeds the baseline of 26.2 MPG.

Since the p-value (0.716) is greater than a 5% rejection rate, we cannot reject the null hypothesis.

We can also observe that the code provides a way to visualize the null distribution and the basis value of 26.2 MPG. If the basis value $\mu=26.2$ is so far out on the null distribution of $\bar{x}$ that less than 5% of the bootstrap-replicated test statistics are greater than $\mu$ , we would have rejected the null hypothesis.

Hypothesis Test on the Variance

For the hypothesis test on the variance, we learned two methods, the Chi-square test, and the bootstrap-based test.

Let’s first look at the Chi-square test. The null and alternate hypothesis is as follows:

$\sigma^{2}=1.67^{2}$

$\sigma^{2} \neq 1.67^{2}$

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

If you can recall from Lesson 88, $\frac{(n-1)s^{2}}{\sigma^{2}}$ is our test statistic, $\chi^{2}_{0}$ , which we will verify against a Chi-square distribution with $(n-1)$ degrees of freedom.

# Chi-square Test
chi0 = ((n-1)*s^2)/sigma^2

pvalue_chi0 = pchisq(chi0,df=(n-1))

chilimit_right = qchisq(0.975,df=(n-1))

chilimit_left = qchisq(0.025,df=(n-1))

The “pchisq” function computes $P(\chi^{2} \le \chi^{2}_{0})$ from a Chi-square distribution with user-specified degrees of freedom, df = 7. The “qchisq” function computes the quantile corresponding to a specified rate of rejection. Since we are conducting a two-sided test, we calculate the limiting value on the left tail and the right tail.

 > pvalue_chi0
   [1] 0.9999914

The p-value is 0.999. Since it is greater than 0.975, we reject the null hypothesis based on the two-sided test.

> chi0
   [1] 35.60715

> chilimit_left
   [1] 1.689869

> chilimit_right
   [1] 16.01276

The lower and the upper bound from the null distribution are 1.69 and 16.01, respectively. Whereas, the test statistic $\chi^{2}_{0}$ is 35.60, well beyond the upper bound acceptable value. Hence we reject the null hypothesis.

Now, let’s look at how to run a bootstrap-based test for the standard deviation. For the bootstrap-based two-sided test, the null hypothesis and the alternate hypothesis are

$H_{0}: P(\sigma > 1.67) = 0.5$

$H_{A}: P(\sigma > 1.67) \neq 0.5$

Use the following lines to set up and run the test.

# Bootstrap Test for Standard Deviation
 N = 10000
 
xsd_null = matrix(NA,nrow=N,ncol=1)
 for (i in 1:N)
   {
     xboot = sample(x,n,replace=T)
     xsd_null[i,1] = sd(xboot)
   }

hist(xsd_null,col="pink",xlab="Standard Deviation of the Distribution",font=2,font.lab=2,main="Null Distribution Assuming Ho is True")

points(sigma,10,pch=15,cex=2,col="blue")
   abline(v=c(quantile(xsd_null,0.025),quantile(xsd_null,0.975)),col="black",lwd=2,lty=2)

pvalue_bootstrap = length(which(xsd_null>sigma))/N

> pvalue_bootstrap
   [1] 0.9747

As before, in the loop, we are executing the “sample” function to draw a bootstrap-replicate from the original data. Then, from this bootstrap-replicate, we compute the standard deviation. Repeated such sampling and computation of the bootstrap-replicated standard deviation statistic forms the null distribution for $\sigma$ . From this null distribution, we calculate the p-value, i.e., the proportion of the null distribution that exceeds the baseline of 1.67 MPG.

Since the p-value (0.975) is greater than or equal to 0.975, ( $\frac{\alpha}{2}$ rejection rate), we reject the null hypothesis.

The code provides a way to visualize the null distribution and the basis value of 1.67 MPG. The basis value is far left on the null distribution. Almost 97.5% of the bootstrap-replicated test statistics are greater than $\sigma = 1.67$ — we reject the null hypothesis.

Hypothesis Test on the Proportion

For the hypothesis test on the proportion, we can employ the binomial distribution (parametric) as the null distribution, or use the bootstrap-based method (non-parametric) to generate the null distribution. For the parametric approach, the null and the alternative hypothesis are

$H_{0}: p \le \frac{1}{8}$

$H_{A}: p > \frac{1}{8}$

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

From a sample of eight cars, the null distribution is the probability distribution of observing any number of cars having a mileage of less than 22 MPG. In other words, out of the eight cars, we could have 0, 1, 2, 3, …, 8 cars to have a mileage less than 22 MPG, the lower bound specified by EPA.

The null distribution is the distribution of the probability of observing these outcomes. In a Binomial null distribution with n = 8 and p = 1/8, what is the probability of getting 0, 1, 2, …, 8? and if more than an acceptable number (as seen in the null distribution) is seen, we reject the null hypothesis.

We can generate and plot that null distribution in R using the following lines.

## Hypothesis Test on the Proportion
 
ncars = length(which(x<22))

plot(0:n,dbinom(0:n,n,prob=p),type="o",xlab="Number of cars with MPG less than the range",ylab="P(X=x)",font=2,font.lab=2)

x_at_alpha = qbinom(0.95,n,prop) # approx quantile for 5% rejection 

cord.x <- c(x_at_alpha,seq(x_at_alpha,n),x_at_alpha) 

cord.y <- c(0, dbinom(x_at_alpha:n,n,prob=p), 0) 

polygon(cord.x,cord.y,col="pink")

points(ncars,0.002,pch=15,col="blue",cex=2)

pval = sum(dbinom(ncars:n,n,prob=p))

In the first line, we are computing the number of cars that have a mileage of less than 22 MPG, which is the test statistic. We observe this to be two cars.

> ncars
   [1] 2

In the next few lines, we are plotting the null distribution as computed from a binomial distribution with n = 8 and p = 1/8, and visually showing the region of rejection using the “polygon” function.

Then, we compute the p-value as the probability of observing two or more cars having a mileage less than 22 MPG.

$P(Y \ge 2) = \sum_{k=2}^{k=8} {8 \choose k} p^{k}(1-p)^{8-k}=0.2636$

> pval
   [1] 0.2636952

Since the p-value is greater than 0.05, we cannot reject the null hypothesis (based on the limited sample) that one in eight cars (2003 Ford Focus) will have a mileage of less than 22 MPG.

Let’s wrap up the day by running the bootstrap test on the proportion. For the bootstrap-based hypothesis test on the proportion, the null and the alternate hypothesis are

$H_{0}: P(p > \frac{1}{8}) = 0.5$

$H_{A}: P(p > \frac{1}{8}) > 0.5$

If we get a large proportion of the null distribution to be greater than 1/8, we will reject the null hypothesis.

Use the following lines to execute the test.

# Bootstrap Test on Proportion 
 N = 10000

xprop_null = matrix(NA,nrow=N,ncol=1)
 for (i in 1:N)
   {
     xboot = sample(x,n,replace=T)
     xprop_null[i,1] = length(which(xboot<22))/n
   }

hist(xprop_null,col="pink",xlab="Proportion of the Distribution",font=2,font.lab=2,main="Null Distribution Assuming Ho is True")

points(prop,10,pch=15,cex=2,col="blue")

abline(v=c(quantile(xprop_null,0.95)),col="black",lwd=2,lty=2)

pval=length(which(xprop_null>prop))/N

> pval
   [1] 0.631

We are executing the “sample” function N = 10,000 times to draw bootstrap-replicates from the original data. Then, for each bootstrap-replicate, we compute the proportion that is less than 22 MPG. The 10,000 bootstrap-replicated proportions forms the null distribution for $p$ . From this null distribution, we calculate the p-value, i.e., the proportion of the null distribution that exceeds the baseline of $\frac{1}{8}$

Since the p-value (0.631) is less than 0.95, ( $1-\alpha$ rejection rate), we cannot reject the null hypothesis. 63% of the bootstrap-replicated proportion is greater than the baseline. If more than 95% of the bootstrap-replicated proportion is greater than the baseline, we would have rejected the null hypothesis. This scenario would have unfolded if the original sample had several cars with less than a mileage of 22 MPG. The bootstrap replicates would show them, in which case the proportion would be much greater than 1/8 in many replicates, and overall, the p-value would exceed 0.95.

Take the next two weeks to digest all the one-sample hypothesis tests, including how to execute the tests in R. In two weeks, we will move on to the two-sample hypothesis tests.

Here is the full code for today’s lesson.

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You can also follow me on Twitter @realDevineni for updates on new lessons.

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.