Hypothesis Testing | One & Two Tailed Tests

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1 Summary

1.1 What is Hypothesis Testing?

1.2 A few types of Hypothesis:

1.3 The four steps of hypothesis testing:

1.3.1 Reference

Summary

Hypothesis testing is just a method for testing a claim or hypothesis about a parameter in a population, using data measured in a sample.
The “Null Hypothesis” denoted as ${ H }_{ 0 }$ , this means testing a claim that already has some established parameters.
The “Alternative Hypothesis” is denoted as ${ H }_{ a }$ , this is known as the research hypothesis. It involves the claim to be tested.
Four steps of hypothesis testing are:
(i) We state the Hypothesis
(ii) Set the criteria for a decision
(iii) Compute the test statistic
(iv) Make a decision
When testing a hypothesis of a proportion, we use the z-test: $z\quad =\quad \frac { \bar { x } \quad -\quad \mu }{ \frac { \sigma }{ \sqrt { n } } }$

What is Hypothesis Testing?

We use samples because it allows us to measure behaviors and to learn more about the behavior in populations that are often too large or inaccessible.

Hypothesis testing is just a method for testing a claim or hypothesis about a parameter in a population, using data measured in a sample. In this method, we test some hypothesis by determining the likelihood that a sample statistic could have been selected, if the hypothesis regarding the population parameter were true.

Hypothesis in simpler words is basically a claim that we want to test or investigate. Which means doing sampling and getting the information and then testing the hypothesis. It could just be any idea that we want to test. The goal of hypothesis testing is to determine the likelihood that a population parameter, such as the mean ( $\mu$ ), is likely to be true.

A few types of Hypothesis:

The “Null Hypothesis” denoted as ${ H }_{ 0 }$ , this means testing a claim that already has some established parameters. The null hypothesis is always the accepted fact. It is a starting point. We test whether the value stated in the null hypothesis is likely to be true: ( ${ H }_{ 0 }\quad :\quad \mu \quad =\quad { \mu }_{ 0 }$ )

Some of the examples of null hypotheses that are generally accepted as being true are:

(i) DNA is shaped like a double helix.
(ii) There are 8 planets in the solar system (excluding Pluto).

The “Alternative Hypothesis” is denoted as ${ H }_{ a }$ , this is known as the research hypothesis. It involves the claim to be tested. An alternative hypothesis ( ${ H }_{ a }$ ) is a statement that directly contradicts a null hypothesis by stating that that the actual value of a population parameter is less than, greater than, or not equal to the value stated in the null hypothesis. The alternate hypothesis is formulated depending on whether a one-tail or two-tail test is required:

One-tail test $\Rightarrow \quad { H }_{ a }\quad :\quad \mu \quad <\quad { \mu }_{ 0 }\quad or\quad \mu \quad >\quad { \mu }_{ 0 }$

Two-tail test ${ H }_{ a }\quad :\quad \mu \quad \neq \quad { \mu }_{ 0 }$

The four steps of hypothesis testing:

Step 1: We state the Hypothesis. We start by assuming that the hypothesis or claim we are testing is true. This is stated in the null hypothesis.

Step 2: Set the criteria for a decision. To set the criteria for a decision, we state the level of significance for a test. Level of significance, refers to a criterion of judgment upon which a decision is made regarding the value stated in a null hypothesis. The likelihood or level of significance ( $\alpha$ ) is typically set at 5% in behavioral research studies. The alternative hypothesis establishes where to place the level of significance.

Step 3: Compute the test statistic. The test statistic is a mathematical formula that allows researchers to determine the likelihood of obtaining sample outcomes if the null hypothesis were true. The value of the test statistic is used to make a decision regarding the null hypothesis.

Step 4: Make a decision. We use the value of the test statistic to make a decision about the null hypothesis. The decision is based on the probability of obtaining a sample mean, given that the value stated in the null hypothesis is true. There are two decisions a researcher can make; either reject the null hypothesis or retain the null hypothesis.

When testing a hypothesis of a proportion, we use the z-test and the formula for this is:

$z\quad =\quad \frac { \bar { x } \quad -\quad \mu }{ \frac { \sigma }{ \sqrt { n } } }$

Example #1

Q. The average score of all sixth graders in school District A on a math aptitude exam is 75 with a standard deviation of 8.1. A random sample of 100 students in one school was taken. The mean score of these 100 students was 71. Does this indicate that the students of this school are significantly less skilled in their mathematical abilities than the average student in the district? (Use a 5% level of significance.)

Solution:

Firstly we write down the data provided to us in the question:

We know the mean for the population µ = 75 and standard deviation for the population σ = 8.1

Thus, we are testing the sample mean against the population mean with a population standard deviation which is known to us.

We will use the z-test here.

Now we carry out the above steps in order to come to a conclusion.

Step 1: We state the null hypothesis and the alternate hypothesis:

${ H }_{ 0 }\quad :\quad \mu \quad \ge \quad { 75 }$ and ${ H }_{ a }\quad :\quad \mu \quad <\quad { 75 }$

Step 2: We select the level of significance which is stated in the problem as 5% or α = 0.05

Step 3: Compute the test statistics. We first identify the test to be used. In this case we are using the z-test because is known and the sample is n=100 is a large sample.

$z\quad =\quad \frac { \bar { x } \quad -\quad \mu }{ \frac { \sigma }{ \sqrt { n } } }$

$z\quad =\quad \frac { 71\quad -\quad 75 }{ \frac { 8.1 }{ \sqrt { 100 } } } \quad =\quad -4.938$

Step 4: Make a decision. Since the alternate hypothesis states µ < 75, this is a one-tailed test to the left.

For α = 0.05, z in the normal curve table that gives a probability of 0.05 to the left of z.

Hence, the critical value after looking at the table gives a value of 0.5 – 0.05 = 0.45 or z = -1.645. That is P(z < -1.645) = 0.05.

Because 0.4500 is exactly halfway between 0.4495 and 0.4505, we get half way between 1.640 and 1.650 to get z = 1.645. Since 71 is to the left of 75, we have z = -1.645.

That is P(z < -1.645) = 0.05.

Since the computed z = -4.938 < -1.645 (critical z value), we reject the null hypothesis that the students in the school are not less skilled in mathematical ability.