Hypothesis Testing Course Online

The null and alternative hypotheses statements corresponding to a testing problem differ from problem to problem.

Usually, the claim made about the parameter is chosen as the alternative hypothesis when dealing with a problem. Consider the following problem:

Problem:

A lightbulb manufacturer packs their bulbs into cartons, each carton containing 100 bulbs. Out of these 100 bulbs, 30 bulbs are picked at random testing. According to the manufacturer, the average lifetime of a bulb is 1,000 hours. Now, a new manufacturing process has been introduced, which is said to increase the average lifetime of the bulbs. Check whether the new approach is effective, assuming that the lifetime of the bulbs follows a normal distribution.

Solution:

In the above problem, it has been provided that the average lifetime of the bulbs using the old method is 1,000 hours.

A claim has been made that the new manufacturing process will increase the lifetime of the bulb, i.e., it will be more than 1,000 hours.

So, we have to test if the new process actually increases the lifetime of the bulbs.

Let denote the lifetime of the bulbs.

As such the testing problem can be written as:

To test: H0 : =1,000 against H1 : >1,000

The level of significance of a test () is the probability of type I error. This is usually provided to the experimenter.

A test statistic is used to decide the rejection criteria for the null hypothesis in a testing problem. Different tests have different test statistics

Test Statistic for the Mean of a Normal Distribution

Let X1,X2, X3, ..., Xn denote a set of random samples that follow a normal distribution independently and identically with mean and variance 2. Here both the mean and variance are unknown parameters.

Let the testing problem be defined as

To test: H0 : =0 against H1 : = 1

Where 10

We define the test statistic as -

T = (X-)SE

Where X is the mean of the population from which the random sample has been sampled

And SE is the standard error

Now, under H0,

T =(X-0)SE ~ tn-1

So the value of the test statistic can be calculated at a particular level of significance from a t-distribution table

The retrieved value of the test statistic can be computed methodically by using the observations.

Test Statistic for the Variance of a Normal Distribution

Let X1,X2, X3, ..., Xn denote a set of random samples that follow a normal distribution independently and identically with mean and variance 2. Here both the mean and variance are unknown parameters.

Let the testing problem be defined as
To test: H0 : 2=02 against H1 :2=12

Where 1202

We define the test statistic as
T=(n-1)s22

Where s2 denotes the sample variance

Under H0,

T ~ n-12

So the value of the test statistic can be calculated at a particular level of significance from a chi-square distribution table.

The observed value of the test statistic is computed methodically by using the observations.

After calculating the value of the test statistic T, denoted by Tobs (say) we need to compare it to the critical value to determine whether H0 gets rejected.

The test statistic’s critical value is obtained from tables provided or by using the software. The critical value is calculated at a particular level of significance ,say.

Suppose the calculated value of the test statistic comes out to be greater than the critical value at significance level. In such a case, the null hypothesis is rejected in favor of the alternate hypothesis.

Correctly reporting the results of an experiment is one of the most crucial tasks of the experimenter. While dealing with the problem of hypothesis testing, a particular syntax is followed by statisticians all around the globe.

After comparing the value of the test statistic to the critical value, either the null hypothesis will get rejected, or it will not get rejected.

Case 1: Null Hypothesis Rejected

As the calculated value of the test statistic is greater than the critical value, we reject H0 in favor of H1.

Case 2: Null Hypothesis Is Not Rejected

As the calculated value of the test statistic is less than the critical value, we do not reject H0 in favor of H1.

It is also preferable to report all the values obtained in a tabular format.

One sample t-test is generally used to determine if a significant difference exists between the means of an unknown population and a particular value. It is used when the standard deviation of the population is unknown.

Assumptions:

1.Data must be continuous

2. The data must follow a normal distribution

3. Sampling should be done using simple random sample techniques such that the probability of selection of each sample is equal

The pre-requisites for performing this test are the population mean, sample size, sample mean, sample standard deviation, and sample size.

Let X1,X2, X3, ..., Xn denote a set of random samples that follow a normal distribution independently and identically with mean and variance 2, where the variance is unknown.

Let the testing problem be denoted as:

To test: H0 : =0 against H1 : 0 (two-tailed test)

H0 : =0 against H1 : > 0 (right-tailed test)

H0 : =0 against H1 : < 0 (left-tailed test)

Where is the value of the hypothesized mean

Now, the standard error of the sample is given by:

SE = sn

Where s is the standard deviation of the random sample

The test statistic is defined as

T = (X-)SE =nX-)s

Under H0,

T = (X-0)SE =n(X-0)s ~ tn-1

I.e., the test statistic follows a t-distribution with degrees of freedom n-1

The critical value if given by: Tctitcal= t; n-1 (for a one tailed test)

t2; n-1 (for a two tailed test)

Where is the level of significance

In a testing problem, z-test is used to check the significant difference between two population means when the standard deviation of the population is known.

Assumptions:

1. Data should be continuous

2. The data should follow a normal distribution

3. The sample should be generated from the population using simple random sampling techniques, such that the probabilities of selecting the samples are equal.

4. The population standard deviation should be known.

Let X1,X2, X3, ..., Xn denote a set of random samples that follow a normal distribution independently and identically with mean and variance 2, where the variance is known.

Let the testing problem be denoted as:

To test: H0 : =0 against H1 : 0 (two-tailed test)

H0 : =0 against H1 : > 0 (right-tailed test)

H0 : =0 against H1 : < 0 (left-tailed test)

Where is the value of the hypothesized mean

The test statistic is defined as

Z=X-/n

Where X=i=1nXin

Under H0

Z = (X-0)/n ~ N(0,1)

I.e, the test statistic follows a standard normal distribution

The critical value is given by: Zctitcal= z; n-1 (for a one-tailed test)

z2; n-1 (for a two-tailed test)

Where is the level of significance

The use of t-test can be extended beyond one sample, i.e., it can also be used to check for a significant difference between the means of two different independent populations.

Assumptions:

1. Data must be continuous

2. Random sampling techniques from the population should generate the data.

3. The data should follow a normal distribution

4. The variances of the two independent groups should be equal

Let X1,X2, X3, ..., XnX denote the first random sample and Y1,Y2, Y3, ..., YnY denote the second random sample such that they are independent of each other.

Let the first sample follow a normal distribution with mean X and variance sX2.

Let the second sample follow a normal distribution with mean Y and variance sY2.

To test: H0 : X=Y against H1 : XY (two-tailed test)

H0 : X=Y against H1 : X> Y (right-tailed test)

H0 : X=Y against H1 : X< Y (left-tailed test)

We define the test statistic as

T=X-YSE(1nX+1nY)

Where SE is the pooled standard deviation is given by

SE ={(nx-1)sX2}+{nY-1)sY2}nX+nY-2

Under H0, T ~ tnX+nY-2

The critical value is given by: Tctitcal=t; nX+nY-2 (for a one-tailed test)

t2; nX+nY-2 (for a two-tailed test )

Where is the level of significance

A paired t-test is used to check for the presence of any significant difference between two variables under the same subject. Usually, the two variables are separated by time.

Example:

An experimenter may want to find if there is any significant difference between deaths due to COVID-19 in May 2020 as compared to June 2020.

So, a paired t-test is used to check whether the mean difference between the pairs of observations differs significantly.

Assumptions:

1. The samples under study must be independent, i.e., any measurements made on the first sample should not affect the second sample.

2. Each sample pair must be obtained from the same subject, e,g., the weights of patients before and after undergoing a diet.

3. Each sample pair must follow a normal distribution.

Let X1,X2, X3, ..., Xn denote the first random sample and Y1,Y2, Y3, ..., Yn denote the second random sample such that they are independent of each other. Let both of them be normally distributed.

Let Z be a new random variable denoting the difference between the two samples, i.e.,

Z=X-Y

Let Z denote the mean of the differences and sZ2 denote the variance of the difference.

To test: H0 : Z=0 against H1 : Z0 (two-tailed test)

H0 : Z=0 against H1 : Z>0 (right-tailed test)

H0 : Z=0 against H1 : Z<0 (left-tailed test)

The test statistic is given by

T=ZsZ/n

Under H0, T ~ tn-1

The critical value if given by: Tctitcal= t; n-1 (for a one tailed test)

t2; n-1 (for a two tailed test)

Where is the level of significance

If observations are taken from a population with a given mean, it is not necessary that they will be identical. Due to the presence of random observation error, the observations fluctuate around the mean. This is a natural, inevitable variation. On top of this, another source of variation or sources of variation is deliberately introduced or suspected to enter due to circumstances beyond our control.

Hence, observations are heterogeneous or not homogeneous concerning the source or sources of variation.
Example:
An experimenter wishes to assess the effect of a sleeping drug on the average amount of sleep of patients.

A deliberately introduced source of variation, for example, a sleeping drug, is called “treatment” or “factor”. Thus certain patients who do not receive the “treatment” form one group, and the other groups are formed by changing the “dose” of the drug. Besides the drug, the patients can be classified according to other factors such as age or gender.

The effect of these sources of variation; that is, treatment can be assessed by analyzing the total variation and spilling it into components corresponding to these sources of variation.

Now, this analysis can be done in several ways, Analysis of Variance or ANOVA being one such method. The analysis of variance is a body of statistical methods of analyzing observations assumed to be of the structure

Yi= b1xi1+b2xi2+...+bpxip+ei , i=1(1)n j=1(1)p

, where the coefficients {xij} are the values of “counter variables” or “indicator variables’ which refer to the presence or absence of the effects {bj} in the conditions under which the observations are taken as: xij is the number of times bj occurs in the ith observation and this is usually 0 or 1. In general, in the analysis of variance, all factors are treated qualitatively.

Now the experimenter may also be interested to know if the effect of any of the treatments in an ANOVA setup differs significantly concerning the other treatments.

Let the data be modeled as

Yi=+i+ei , i=1(1)n

Where is the process mean

i denotes the effect due to the ith treatment

ei is the random error associated with the process

To test: H0: 1=2=3= ... = n=0 against H1: not H0

Now there may be two cases that the experimenter may face.

Case I: Null Hypothesis Is Not Rejected

In this case, since H0 is not rejected in favor of H1, no significant difference exists between the effect of the treatments.

Case II: Null Hypothesis Rejected

If the null hypothesis gets rejected in favor of the alternate hypothesis, then the experimenter can claim that the effects due to one or more treatments are different.

Pairwise testing is used on all treatment pairs to determine which treatments are responsible for the difference. This process is known as post hoc analysis.

Many courses are available today that provide quality education on hypothesis testing. These courses are especially beneficial because they will save you a lot of time and energy.

The main advantage of opting for an online course is that you can learn at your own pace. In offline courses, once a topic is covered, it will be up to you to learn it because the professor may move on to the next topic without waiting for you to finish. This does not happen in online courses. Online courses follow your pace of learning and thus offer better learning opportunities.

Another significant advantage of online courses is that you can attend classes from the comfort of your home, significantly reducing travel expenses.

When you opt for online courses, you will be provided with a choice of instructors and can select someone who suits your needs best. This will allow you to learn much more effectively than offline courses, where your choices remain limited.

Online courses also have excellent doubt-clearing facilities that offline courses lack.

So, in light of the given data, an online hypothesis testing course is better than an offline one.

The syllabus for a hypothesis testing course coverse.

1. Test of a statistical hypothesis and critical region

2. Type I and type II errors

3. Level of significance and power of test

4. Optimum tests in different situations

5. Unbiased tests

6. Neyman-Pearson lemma

7. Construction of most powerful (MP) and uniformly most powerful (UMP) critical regions

8. MP and UMP regions in random sampling from a normal distribution

9. Construction of type A regions

10. Construction of type A1 regions

11. Optimum regions and sufficient statistics

12. Randomized tests

13. Composite hypotheses and similar regions

14. Similar regions and complete sufficient statistics

15.Construction of most powerful similar regions

16. Test to derive the mean of a normal distribution

17. Test for the variance of a normal distribution

18. Monotonicity of power function

19. Consistency

20. Invariance

21. Likelyhood-ratio tests

22. Comparing the means of k normal distributions with common variance

23. Properties of likelihood-ratio tests

The complex process of Hypothesis testing is being broadly leveraged industry-wide to make well-informed, data-driven decisions towards assured results. The power of Hypothesis testing enables professionals to test their theories before putting them into action, which can significantly benefit organisations to reap value while cutting risks of potential repercussions.

Its active implementation in business, as well as investment opportunities, is helping experts perform statistical analysis against containing datasets and receive decisive predictions towards a winning strategy. As Hypothetical testing is strengthening its statistical methods to enhance accuracy, more and more businesses are incorporating it to test their theories before committing resources to it, leading to a thriving future projection in the coming days.

Today, all major jobs are in the data science field. A data science course may just land you your dream job.

Hypothesis testing is one of the most pivotal concepts in statistics. This concept is used in all industries. As a result, there has been a huge demand for hypothesis testing courses in India.

These courses offer all the knowledge that any data scientist may possess, allowing you to apply for your dream job no matter your educational background.

Hypothesis testing is a part of statistics. So, solving these problems generally falls on the data scientists or data analysts who deal with statistics as a whole.

The median salary of data scientists in India is Rs. 46,953 per annum.

The entry-level salary for an analyst with an experience of less than a year is Rs. 3,67,000. For an experienced data analyst with more than 20 years of experience, the salary is Rs. 2 million.

Different factors affect the job of a data analyst. A base-level data analyst should know basic statistics and software like python, SQL, and R. Apart from these, they must also possess project management and organizational skills.

Data analysts should also have an analytical mind that allows them to work seamlessly with large unstructured data sets.

Other factors that determine their salary are the company they work at, its size and reputation, their position, work experience, and geographic location.