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2.11: If Then Statements

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Hypothesis followed by a conclusion in a conditional statement.

Conditional Statements

A conditional statement (also called an if-then statement ) is a statement with a hypothesis followed by a conclusion . The hypothesis is the first, or “if,” part of a conditional statement. The conclusion is the second, or “then,” part of a conditional statement. The conclusion is the result of a hypothesis.

If-then statements might not always be written in the “if-then” form. Here are some examples of conditional statements:

• Statement 1: If you work overtime, then you’ll be paid time-and-a-half.
• Statement 2: I’ll wash the car if the weather is nice.
• Statement 3: If 2 divides evenly into \(x\), then \(x\) is an even number.
• Statement 4: I’ll be a millionaire when I win the lottery.
• Statement 5: All equiangular triangles are equilateral.

Statements 1 and 3 are written in the “if-then” form. The hypothesis of Statement 1 is “you work overtime.” The conclusion is “you’ll be paid time-and-a-half.” Statement 2 has the hypothesis after the conclusion. If the word “if” is in the middle of the statement, then the hypothesis is after it. The statement can be rewritten: If the weather is nice, then I will wash the car. Statement 4 uses the word “when” instead of “if” and is like Statement 2. It can be written: If I win the lottery, then I will be a millionaire. Statement 5 “if” and “then” are not there. It can be rewritten: If a triangle is equiangular, then it is equilateral.

What if you were given a statement like "All squares are rectangles"? How could you determine the hypothesis and conclusion of this statement?

Example \(\PageIndex{1}\)

Determine the hypothesis and conclusion: I'll bring an umbrella if it rains.

Hypothesis: "It rains." Conclusion: "I'll bring an umbrella."

Example \(\PageIndex{2}\)

Determine the hypothesis and conclusion: All right angles are \(90^{\circ}\).

Hypothesis: "An angle is right." Conclusion: "It is \(90^{\circ}\)."

Example \(\PageIndex{3}\)

Use the statement: I will graduate when I pass Calculus.

Rewrite in if-then form and determine the hypothesis and conclusion.

This statement can be rewritten as If I pass Calculus, then I will graduate. The hypothesis is “I pass Calculus,” and the conclusion is “I will graduate.”

Example \(\PageIndex{4}\)

Use the statement: All prime numbers are odd.

Rewrite in if-then form, determine the hypothesis and conclusion, and determine whether this is a true statement.

This statement can be rewritten as If a number is prime, then it is odd. The hypothesis is "a number is prime" and the conclusion is "it is odd". This is not a true statement (remember that not all conditional statements will be true!) since 2 is a prime number but it is not odd.

Example \(\PageIndex{5}\)

Determine the hypothesis and conclusion: Sarah will go to the store if Riley does the laundry.

The statement can be rewritten as "If Riley does the laundry then Sarah will go to the store." The hypothesis is "Riley does the laundry" and the conclusion is "Sarah will go to the store."

Determine the hypothesis and the conclusion for each statement.

• If 5 divides evenly into \(x\), then \(x\) ends in 0 or 5.
• If a triangle has three congruent sides, it is an equilateral triangle.
• Three points are coplanar if they all lie in the same plane.
• If \(x=3\), then \(x^2=9\).
• If you take yoga, then you are relaxed.
• All baseball players wear hats.
• I'll learn how to drive when I am 16 years old.
• If you do your homework, then you can watch TV.
• Alternate interior angles are congruent if lines are parallel.
• All kids like ice cream.

Additional Resources

Video: If-Then Statements Principles - Basic

Activities: If-Then Statements Discussion Questions

Study Aids: Conditional Statements Study Guide

Practice: If Then Statements

Real World: If Then Statements

A statement that could be true, which might then be tested.

Example: Sam has a hypothesis that "large dogs are better at catching tennis balls than small dogs". We can test that hypothesis by having hundreds of different sized dogs try to catch tennis balls.

Sometimes the hypothesis won't be tested, it is simply a good explanation (which could be wrong). Conjecture is a better word for this.

Example: you notice the temperature drops just as the sun rises. Your hypothesis is that the sun warms the air high above you, which rises up and then cooler air comes from the sides.

Note: when someone says "I have a theory" they should say "I have a hypothesis", because in mathematics a theory is actually well proven.

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Understanding Hypotheses

'What happens if ... ?' to ' This will happen if'

The experimentation of children continually moves on to the exploration of new ideas and the refinement of their world view of previously understood situations. This description of the playtime patterns of young children very nicely models the concept of 'making and testing hypotheses'. It follows this pattern:

• Make some observations. Collect some data based on the observations.
• Draw a conclusion (called a 'hypothesis') which will explain the pattern of the observations.
• Test out your hypothesis by making some more targeted observations.

So, we have

• A hypothesis is a statement or idea which gives an explanation to a series of observations.

Sometimes, following observation, a hypothesis will clearly need to be refined or rejected. This happens if a single contradictory observation occurs. For example, suppose that a child is trying to understand the concept of a dog. He reads about several dogs in children's books and sees that they are always friendly and fun. He makes the natural hypothesis in his mind that dogs are friendly and fun . He then meets his first real dog: his neighbour's puppy who is great fun to play with. This reinforces his hypothesis. His cousin's dog is also very friendly and great fun. He meets some of his friends' dogs on various walks to playgroup. They are also friendly and fun. He is now confident that his hypothesis is sound. Suddenly, one day, he sees a dog, tries to stroke it and is bitten. This experience contradicts his hypothesis. He will need to amend the hypothesis. We see that

• Gathering more evidence/data can strengthen a hypothesis if it is in agreement with the hypothesis.
• If the data contradicts the hypothesis then the hypothesis must be rejected or amended to take into account the contradictory situation.

• A contradictory observation can cause us to know for certain that a hypothesis is incorrect.
• Accumulation of supporting experimental evidence will strengthen a hypothesis but will never let us know for certain that the hypothesis is true.

In short, it is possible to show that a hypothesis is false, but impossible to prove that it is true!

Whilst we can never prove a scientific hypothesis to be true, there will be a certain stage at which we decide that there is sufficient supporting experimental data for us to accept the hypothesis. The point at which we make the choice to accept a hypothesis depends on many factors. In practice, the key issues are

• What are the implications of mistakenly accepting a hypothesis which is false?
• What are the cost / time implications of gathering more data?
• What are the implications of not accepting in a timely fashion a true hypothesis?

For example, suppose that a drug company is testing a new cancer drug. They hypothesise that the drug is safe with no side effects. If they are mistaken in this belief and release the drug then the results could have a disastrous effect on public health. However, running extended clinical trials might be very costly and time consuming. Furthermore, a delay in accepting the hypothesis and releasing the drug might also have a negative effect on the health of many people.

In short, whilst we can never achieve absolute certainty with the testing of hypotheses, in order to make progress in science or industry decisions need to be made. There is a fine balance to be made between action and inaction.

Hypotheses and mathematics So where does mathematics enter into this picture? In many ways, both obvious and subtle:

• A good hypothesis needs to be clear, precisely stated and testable in some way. Creation of these clear hypotheses requires clear general mathematical thinking.
• The data from experiments must be carefully analysed in relation to the original hypothesis. This requires the data to be structured, operated upon, prepared and displayed in appropriate ways. The levels of this process can range from simple to exceedingly complex.

Very often, the situation under analysis will appear to be complicated and unclear. Part of the mathematics of the task will be to impose a clear structure on the problem. The clarity of thought required will actively be developed through more abstract mathematical study. Those without sufficient general mathematical skill will be unable to perform an appropriate logical analysis.

Using deductive reasoning in hypothesis testing

There is often confusion between the ideas surrounding proof, which is mathematics, and making and testing an experimental hypothesis, which is science. The difference is rather simple:

• Mathematics is based on deductive reasoning : a proof is a logical deduction from a set of clear inputs.
• Science is based on inductive reasoning : hypotheses are strengthened or rejected based on an accumulation of experimental evidence.

Of course, to be good at science, you need to be good at deductive reasoning, although experts at deductive reasoning need not be mathematicians. Detectives, such as Sherlock Holmes and Hercule Poirot, are such experts: they collect evidence from a crime scene and then draw logical conclusions from the evidence to support the hypothesis that, for example, Person M. committed the crime. They use this evidence to create sufficiently compelling deductions to support their hypotheses beyond reasonable doubt . The key word here is 'reasonable'. There is always the possibility of creating an exceedingly outlandish scenario to explain away any hypothesis of a detective or prosecution lawyer, but judges and juries in courts eventually make the decision that the probability of such eventualities are 'small' and the chance of the hypothesis being correct 'high'.

• If a set of data is normally distributed with mean 0 and standard deviation 0.5 then there is a 97.7% certainty that a measurement will not exceed 1.0.
• If the mean of a sample of data is 12, how confident can we be that the true mean of the population lies between 11 and 13?

It is at this point that making and testing hypotheses becomes a true branch of mathematics. This mathematics is difficult, but fascinating and highly relevant in the information-rich world of today.

To read more about the technical side of hypothesis testing, take a look at What is a Hypothesis Test?

You might also enjoy reading the articles on statistics on the Understanding Uncertainty website

This resource is part of the collection Statistics - Maths of Real Life

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If-then statement

• Logical correct I
• Logical correct II

When we previously discussed inductive reasoning we based our reasoning on examples and on data from earlier events. If we instead use facts, rules and definitions then it's called deductive reasoning.

We will explain this by using an example.

If you get good grades then you will get into a good college.

The part after the "if": you get good grades - is called a hypotheses and the part after the "then" - you will get into a good college - is called a conclusion.

Hypotheses followed by a conclusion is called an If-then statement or a conditional statement.

This is noted as

$$p \to q$$

This is read - if p then q.

A conditional statement is false if hypothesis is true and the conclusion is false. The example above would be false if it said "if you get good grades then you will not get into a good college".

If we re-arrange a conditional statement or change parts of it then we have what is called a related conditional.

Our conditional statement is: if a population consists of 50% men then 50% of the population must be women.

If we exchange the position of the hypothesis and the conclusion we get a converse statemen t: if a population consists of 50% women then 50% of the population must be men.

$$q\rightarrow p$$

If both statements are true or if both statements are false then the converse is true. A conditional and its converse do not mean the same thing

If we negate both the hypothesis and the conclusion we get a inverse statemen t: if a population do not consist of 50% men then the population do not consist of 50% women.

$$\sim p\rightarrow \: \sim q$$

The inverse is not true juest because the conditional is true. The inverse always has the same truth value as the converse.

We could also negate a converse statement, this is called a contrapositive statemen t:  if a population do not consist of 50% women then the population do not consist of 50% men.

$$\sim q\rightarrow \: \sim p$$

The contrapositive does always have the same truth value as the conditional. If the conditional is true then the contrapositive is true.

A pattern of reaoning is a true assumption if it always lead to a true conclusion. The most common patterns of reasoning are detachment and syllogism.

If we turn of the water in the shower, then the water will stop pouring.

If we call the first part p and the second part q then we know that p results in q. This means that if p is true then q will also be true. This is called the law of detachment and is noted:

$$\left [ (p \to q)\wedge p \right ] \to q$$

The law of syllogism tells us that if p → q and q → r then p → r is also true.

This is noted:

$$\left [ (p \to q)\wedge (q \to r ) \right ] \to (p \to r)$$

If the following statements are true:

If we turn of the water (p), then the water will stop pouring (q). If the water stops pouring (q) then we don't get wet any more (r).

Then the law of syllogism tells us that if we turn of the water (p) then we don't get wet (r) must be true.

Video lesson

Write a converse, inverse and contrapositive to the conditional

"If you eat a whole pint of ice cream, then you won't be hungry"

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Calcworkshop

Conditional Statement If Then's Defined in Geometry - 15+ Examples!

// Last Updated: January 21, 2020 - Watch Video //

In today’s geometry lesson , you’re going to learn all about conditional statements!

Jenn, Founder Calcworkshop ® , 15+ Years Experience (Licensed & Certified Teacher)

We’re going to walk through several examples to ensure you know what you’re doing.

In addition, this lesson will prepare you for deductive reasoning and two column proofs later on.

Here we go!

What are Conditional Statements?

To better understand deductive reasoning, we must first learn about conditional statements.

A conditional statement has two parts: hypothesis ( if ) and conclusion ( then ).

In fact, conditional statements are nothing more than “If-Then” statements!

Sometimes a picture helps form our hypothesis or conclusion. Therefore, we sometimes use Venn Diagrams to visually represent our findings and aid us in creating conditional statements.

But to verify statements are correct, we take a deeper look at our if-then statements. This is why we form the converse , inverse , and contrapositive of our conditional statements.

What is the Converse of a Statement?

Well, the converse is when we switch or interchange our hypothesis and conclusion.

Conditional Statement : “If today is Wednesday, then yesterday was Tuesday.”

Hypothesis : “If today is Wednesday” so our conclusion must follow “Then yesterday was Tuesday.”

So the converse is found by rearranging the hypothesis and conclusion, as Math Planet accurately states.

Converse : “If yesterday was Tuesday, then today is Wednesday.”

What is the Inverse of a Statement?

Now the inverse of an If-Then statement is found by negating (making negative) both the hypothesis and conclusion of the conditional statement.

So using our current conditional statement, “If today is Wednesday, then yesterday was Tuesday”.

Inverse : “If today is not Wednesday, then yesterday was not Tuesday.”

What is a Contrapositive?

And the contrapositive is formed by interchanging the hypothesis and conclusion and then negating both.

Contrapositive : “If yesterday was not Tuesday, then today is not Wednesday”

What is a Biconditional Statement?

A statement written in “if and only if” form combines a reversible statement and its true converse. In other words the conditional statement and converse are both true.

Continuing with our initial condition, “If today is Wednesday, then yesterday was Tuesday.”

Biconditional : “Today is Wednesday if and only if yesterday was Tuesday.”

Examples of Conditional Statements

In the video below we will look at several harder examples of how to form a proper statement, converse, inverse, and contrapositive. And here’s a big hint…

Whenever you see “con” that means you switch! It’s like being a con-artist!

Moreover, we will detail the process for coming up with reasons for our conclusions using known postulates. We will review the ten postulates that we have learned so far, and add a few more problems dealing with perpendicular lines, planes, and perpendicular bisectors.

After this lesson, we will be ready to tackle deductive reasoning head-on, and feel confident as we march onward toward learning two-column proofs!

Conditional Statements – Lesson & Examples (Video)

• Introduction to conditional statements
• 00:00:25 – What are conditional statements, converses, and biconditional statements? (Examples #1-2)
• 00:05:21 – Understanding venn diagrams (Examples #3-4)
• 00:11:07 – Supply the missing venn diagram and conditional statement for each question (Examples #5-8)
• Exclusive Content for Member’s Only
• 00:17:48 – Write the statement and converse then determine if they are reversible (Examples #9-12)
• 00:29:17 – Understanding the inverse, contrapositive, and symbol notation
• 00:35:33 – Write the statement, converse, inverse, contrapositive, and biconditional statements for each question (Examples #13-14)
• 00:45:40 – Using geometry postulates to verify statements (Example #15)
• 00:53:23 – What are perpendicular lines, perpendicular planes and the perpendicular bisector?
• 00:56:26 – Using the figure, determine if the statement is true or false (Example #16)
• Practice Problems with Step-by-Step Solutions
• Chapter Tests with Video Solutions

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Definition Of Hypothesis

Hypothesis is the part of a conditional statement just after the word if.

Examples of Hypothesis

In the conditional, "If all fours sides of a quadrilateral measure the same, then the quadrilateral is a square" the hypothesis is "all fours sides of a quadrilateral measure the same".

Video Examples: Hypothesis  

Solved Example on Hypothesis

Ques:  in the example above, is the hypothesis "all fours sides of a quadrilateral measure the same" always, never, or sometimes true.

A. always B. never C. sometimes Correct Answer: C

Step 1: The hypothesis is sometimes true. Because, its true only for a square and a rhombus, not for the other quadrilaterals rectangle, parallelogram, or trapezoid.

Related Worksheet

• Identifying-and-Describing-Right-Triangles-Gr-4
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• Interpreting-Multiplication-as-Scaling-or-Resizing-Gr-5
• Parallel-Lines-and-Transversals-Gr-8

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Professor: Erika L.C. King Email: [email protected] Office: Lansing 304 Phone: (315)781-3355

The majority of statements in mathematics can be written in the form: "If A, then B." For example: "If a function is differentiable, then it is continuous". In this example, the "A" part is "a function is differentiable" and the "B" part is "a function is continuous." The "A" part of the statement is called the "hypothesis", and the "B" part of the statement is called the "conclusion". Thus the hypothesis is what we must assume in order to be positive that the conclusion will hold.

Whenever you are asked to state a theorem, be sure to include the hypothesis. In order to know when you may apply the theorem, you need to know what constraints you have. So in the example above, if we know that a function is differentiable, we may assume that it is continuous. However, if we do not know that a function is differentiable, continuity may not hold. Some theorems have MANY hypotheses, some of which are written in sentences before the ultimate "if, then" statement. For example, there might be a sentence that says: "Assume n is even." which is then followed by an if,then statement. Include all hypotheses and assumptions when asked to state theorems and definitions!

Still have questions? Please ask!

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3.6: Mathematical Induction - An Introduction

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• Harris Kwong
• State University of New York at Fredonia via OpenSUNY

Mathematical induction can be used to prove that an identity is valid for all integers \(n\geq1\). Here is a typical example of such an identity: \[1+2+3+\cdots+n = \frac{n(n+1)}{2}.\] More generally, we can use mathematical induction to prove that a propositional function \(P(n)\) is true for all integers \(n\geq a\).

Principal of Mathematical Induction (PMI)

Given a propositional function \(P(n)\) defined for integers \(n\), and a fixed integer \(a.\)

Then, if these two conditions are true

• \(P(a)\) is true.
• if \(P(k)\) is true for some integer \(k\geq a\), then \(P(k+1)\) is also true.

then the  \(P(n)\) is true for all integers \(n\geq a\).

Outline for Mathematical Induction

To show that a propositional function \(P(n)\) is true for all integers \(n\geq a\), follow these steps:

• Base Step : Verify that \(P(a)\) is true.
• Assume \(P(n)\) is true for an arbitrary integer, \(k\) with  \(k\geq a\).  This is the inductive hypothesis .
• With this assumption (the inductive hypothesis) , show \(P(k+1)\) is true.
• Conclude, by the Principle of Mathematical Induction (PMI) that \(P(n)\) is true for all integers \(n\geq a\).

The base step is also called the basis step or the  anchor step or the initial step . 

The base step and the inductive step, together, prove that \[P(a) \Rightarrow P(a+1) \Rightarrow P(a+2) \Rightarrow \cdots\,.\] Therefore, \(P(n)\) is true for all integers \(n\geq a\). Compare induction to falling dominoes. When the first domino falls, it knocks down the next domino. The second domino in turn knocks down the third domino. Eventually, all the dominoes will be knocked down. But it will not happen unless these conditions are met:

• The first domino must fall to start the motion. If it does not fall, no chain reaction will occur. This is the base step.
• The distance between adjacent dominoes must be set up correctly. Otherwise, a certain domino may fall down without knocking over the next. Then the chain reaction will stop, and will never be completed. Maintaining the right inter-domino distance ensures that \(P(k)\Rightarrow P(k+1)\) for each integer \(k\geq a\).

To prove the implication \[P(k) \Rightarrow P(k+1)\] in the inductive step, we need to carry out two steps: assuming that \(P(k)\) is true, then using it to prove \(P(k+1)\) is also true. So we can refine an induction proof into a 3-step procedure:

• Verify that \(P(a)\) is true.
• Assume that \(P(k)\) is true for some integer \(k\geq a\).
• Show that \(P(k+1)\) is also true.

The second step, the assumption that \(P(k)\) is true, is referred to as the inductive hypothesis.  This is how a mathematical induction proof may look:

The idea behind mathematical induction is rather simple. However, it must be delivered with precision.

• Be sure to say “Assume \(P(n)\) holds for some integer \(k\geq a\).” Do not say “Assume it holds for all integers \(k\geq a\).” If we already know the result holds for all \(k\geq a\), then there is no need to prove anything at all.
• Be sure to specify the requirement \(k\geq a\). This ensures that the chain reaction of the falling dominoes starts with the first one.
• Do not say “let \(n=k\)” or “let \(n=k+1\).” The point is, you are not assigning the value of \(k\) and \(k+1\) to \(n\). Rather, you are assuming that the statement is true when \(n\) equals \(k\), and using it to show that the statement also holds when \(n\) equals \(k+1\).

Some proofs by induction 

\(1+2+3+\cdots+n\).

Use mathematical induction to show proposition \(P(n)\) :   \[1+2+3+\cdots+n = \frac{n(n+1)}{2}\] for all integers \(n\geq1\).

Base Step:  consider n = 1

On the Left-Hand Side (LHS) we get 1.  On the Right-Hand Side ( RHS) we get \(\frac{1(1+1)}{2}=\frac{2}{2}=1.\)  Thus \(P(n)\) is true for \(n =1.\)

Inductive step: Assume  \(P(n)\) is true for \(n =k, k \geq 1.\)  In other words, \(P(k)\) is true so our  inductive hypothesis is  \[1+2+3+\cdots+k = \frac{k(k+1)}{2}.\]   

Consider the left-hand side of \(P(k+1)\).   \[1+2+3+\cdots+(k+1) = 1+2+\cdots+k+(k+1),\]

we can regroup this as

\[1+2+3+\cdots+(k+1) = [1+2+\cdots+k]+(k+1),\]

so that \(1+2+\cdots+k\) can be replaced by \(\frac{k(k+1)}{2}\), by the inductive hypothesis.

Using the inductive hypothesis, we find

\[\begin{aligned} 1+2+3+\cdots+(k+1) &=& 1+2+3+\cdots+k+(k+1) \\ &=& \frac{k(k+1)}{2}+(k+1) \\ &=& (k+1)\left(\frac{k}{2}+1\right) \\ &=& (k+1)\cdot\frac{k+2}{2}\\ &=& \frac{(k+1)(k+2)}{2}. \end{aligned}\]

Therefore, the identity also holds when \(n=k+1\).

Thus, by the Principle of Mathematical Induction (PMI),  \[1+2+3+\cdots+n = \frac{n(n+1)}{2}\] for all integers \(n\geq1\).

We can use the summation notation (also called the sigma notation ) to abbreviate a sum. For example, the sum in the last example can be written as

\[\sum_{i=1}^n i.\]

The letter \(i\) is the index of summation . By putting \(i=1\) under \(\sum\) and \(n\) above, we declare that the sum starts with \(i=1\), and ranges through \(i=2\), \(i=3\), and so on, until \(i=n\). The quantity that follows \(\sum\) describes the pattern of the terms that we are adding in the summation. Accordingly,

\[\sum_{i=1}^{10} i^2 = 1^2+2^2+3^2+\cdots+10^2.\]

In general, the sum of the first \(n\) terms in a sequence \(\{a_1,a_2,a_3,\ldots\,\}\) is denoted \(\sum_{i=1}^n a_i\). Observe that

\[\sum_{i=1}^{k+1} a_i = \left(\sum_{i=1}^k a_i\right) + a_{k+1},\]

which provides the link between \(P(k+1)\) and \(P(k)\) in an induction proof.

\(\sum_{i=1}^n i^2\)

Example \(\PageIndex{2}\)

Use mathematical induction to show that, for all integers \(n\geq1\), \[\sum_{i=1}^n i^2 = 1^2+2^2+3^2+\cdots+n^2 = \frac{n(n+1)(2n+1)}{6}.\]

Base Step: When \(n=1\), the left-hand side reduces to \(1^2=1\), and the right-hand side becomes \(\frac{1\cdot2\cdot3}{6}=1\); hence, the identity holds when \(n=1\). Inductive Step: Assume it holds when \(n=k\) for some integer \(k\geq1\); that is, assume for some integer \(k\geq1\) that \[\sum_{i=1}^k i^2 = \frac{k(k+1)(2k+1)}{6}\] . Consider \(n=k+1\).   \[\sum_{i=1}^{k+1} i^2 =1^2+2^2+3^2+\cdots+k^2+(k+1)^2. \] From the inductive hypothesis, we find \[\sum_{i=1}^{k+1} i^2 = \sum_{i=1}^k i^2 + (k+1)^2\] \[=\frac{k(k+1)(2k+1)}{6}+(k+1)^2\]  \[=\frac{k(k+1)(2k+1)+6(k+1)^2}{6}\] \[\frac{(k+1)[k(2k+1)+6(k+1)]}{6}\] \[\frac{(k+1)(2k^2+7k+6)}{6}\] \[\frac{(k+1)(k+2)(2k+3)}{6}\] \[\frac{(k+1)(k+2)(2(k+1)+1)}{6}.\] Therefore, the identity also holds when \(n=k+1\).  Thus, by PMI for all integers \(n\geq1\), \[\sum_{i=1}^n i^2 = 1^2+2^2+3^2+\cdots+n^2 = \frac{n(n+1)(2n+1)}{6}.\]

hands-on exercise \(\PageIndex{1}\label{he:induct1-01}\)

It is time for you to write your own induction proof. Prove that \[1\cdot2 + 2\cdot3 + 3\cdot4 + \cdots + n(n+1) = \frac{n(n+1)(n+2)}{3}\] for all integers \(n\geq1\).

hands-on exercise \(\PageIndex{2}\label{he:induct1-02}\)

Use induction to prove that, for all positive integers \(n\), \[1\cdot2\cdot3 + 2\cdot3\cdot4 + \cdots + n(n+1)(n+2) = \frac{n(n+1)(n+2)(n+3)}{4}.\]

hands-on exercise \(\PageIndex{3}\label{he:sumfourn}\)

Use induction to prove that, for all positive integers \(n\), \[1+4^1+4^2+\cdots+4^n = \frac{4^{n+1}-1}{3}.\]

All three steps in an induction proof must be completed; otherwise, the proof may not be correct.

Example \(\PageIndex{3}\label{eg:induct1-03}\)

Can we just use examples?

Never attempt to prove \(P(k)\Rightarrow P(k+1)\) by examples alone . Consider \[P(n): \qquad n^2+n+11 \mbox{ is prime}.\] In the inductive step, we want to prove that \[P(k) \Rightarrow P(k+1) \qquad\mbox{ for ANY } k\geq1.\] The following table verifies that it is true for \(1\leq k\leq 9\): \[\begin{array}{|*{10}{c|}} \hline n & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ \hline n^2+n+11 & 13 & 17 & 23 & 31 & 41 & 53 & 67 & 83 & 101 \\ \hline \end{array}\] Nonetheless, when \(n=10\), \(n^2+n+11=121\) is composite. So \(P(9) \Rightarrow P(10)\) is false. The inductive step breaks down when \(k=9\).

Example \(\PageIndex{4}\label{eg:induct1-04}\)

The base step is equally important . Consider proving \[P(n): \qquad 3n+2 = 3q \mbox{ for some integer $q$}\] for all \(n\in\mathbb{N}\). Assume \(P(k)\) is true for some integer \(k\geq1\); that is, assume \(3k+2=3q\) for some integer \(q\). Then \[3(k+1)+2 = 3k+3+2 = 3+3q = 3(1+q).\] Therefore, \(3(k+1)+2\) can be written in the same form. This proves that \(P(k+1)\) is also true. Does it follow that \(P(n)\) is true for all integers \(n\geq1\)? We know that \(3n+2\) cannot be written as a multiple of 3. What is the problem?

The problem is: we need \(P(k)\) to be true for at least one value of \(k\) so as to start the sequence of implications \[P(1) \Rightarrow P(2), \qquad P(2) \Rightarrow P(3), \qquad P(3) \Rightarrow P(4), \qquad\ldots\] The induction fails because we have not established the basis step. In fact, \(P(1)\) is false. Since the first domino does not fall, we cannot even start the chain reaction.

Thus far, we have learned how to use mathematical induction to prove identities. In general, we can use mathematical induction to prove a statement about \(n\). This statement can take the form of an identity, an inequality, or simply a verbal statement about \(n\). We shall learn more about mathematical induction in the next few sections.

Summary and Review

• Mathematical induction can be used to prove that a statement about \(n\) is true for all integers \(n\geq a\).
• We have to complete three steps.
• In the base step, verify the statement for \(n=a\).
• In the inductive hypothesis, assume that the statement holds when \(n=k\) for some integer \(k\geq a\).
• In the inductive step, use the information gathered from the inductive hypothesis to prove that the statement also holds when \(n=k+1\).
• Be sure to complete all three steps.
• Pay attention to the wording. At the beginning, follow the template closely. When you feel comfortable with the whole process, you can start venturing out on your own.

Exercises 

Exercise \(\PageIndex{1}\label{ex:induct1-01}\)

Use induction to prove that \[1^3+2^3+3^3+\cdots+n^3 = \frac{n^2(n+1)^2}{4}\] for all integers \(n\geq1\).

Exercise \(\PageIndex{2}\)

Use induction to prove that the following identity holds for all integers \(n\geq1\): \[1+3+5+\cdots+(2n-1) = n^2.\]

Base Case: consider \(n=1\).  \(2(1)-1=1\) and \(1^2=1\) so the LHS & RHS are both 1. This works for  \(n=1\).

Inductive Step: Assume this works for some integer, \(k \geq 1.\) In other words,  \(1+3+5+\cdots+(2k-1) = k^2.\)  ( Inductive Hypothesis )

Consider the case of  \(n=k+1.\)   \(1+3+5+\cdots +(2k-1)+(2(k+1)-1)\)

 \[=k^2+(2(k+1)-1) \text{   by inductive hypothesis}\] \[=k^2+2k+2-1=k^2+2k+1=(k+1)^2 \text{   by algebra} \]

\(1+3+5+\cdots+(2(k+1)-1)=(k+1)^2\); assuming our proposition works for \(k\) it will also work for \(k+1.\)

By PMI, \(1+3+5+\cdots+(2n-1) = n^2\)  for all integers,  \(n\geq1\).

Exercise \(\PageIndex{3}\label{ex:induct1-03}\)

Use induction to show that \[1+\frac{1}{3}+\frac{1}{3^2}+\cdots+\frac{1}{3^n} = \frac{3}{2}\left(1-\frac{1}{3^{n+1}}\right)\] for all positive integers \(n\).

Exercise \(\PageIndex{4}\label{ex:induct1-04}\)

Use induction to establish the following identity for any integer \(n\geq1\): \[1-3+9-\cdots+(-3)^n = \frac{1-(-3)^{n+1}}{4}.\]

Exercise \(\PageIndex{5}\label{ex:induct1-05}\)

Use induction to show that, for any integer \(n\geq1\): \[\sum_{i=1}^n i\cdot i! = (n+1)!-1.\]

Exercise \(\PageIndex{6}\label{ex:induct1-06}\)

Use induction to prove the following identity for integers \(n\geq1\): \[\sum_{i=1}^n \frac{1}{(2i-1)(2i+1)} = \frac{n}{2n+1}.\]

Exercise \(\PageIndex{7}\)

Prove \(2^{2n}-1\) is divisible by 3, for all integers \(n\geq0.\)

Base Case: consider \(n=0\).  \(2^{2(0)}-1=1-1=0.\)  \(0\) is divisible by 3 because 0 = 0(3).

Inductive Step: Assume this works for some integer, \(k \geq 0.\) In other words, \(2^{2k}-1\) is divisible by 3. ( Inductive Hypothesis )

Since \(2^{2k}-1\) is divisible by 3, there exists some integer, m such that \(2^{2k}-1=3m,\)  by definition of divides.

Consider the case of  \(n=k+1.\)  By algebra: \[2^{2(k+1)}-1=2^{2k+2}-1=2^{2k}\cdot 2^2-1=2^{2k}\cdot 4 -1=2^{2k}\cdot (3+1)-1=3 \cdot 2^{2k}+2^{2k}-1\] \[=3 \cdot 2^{2k}+3m \text{   by inductive hypothesis}\]

\[=3(2^{2k}+m) \text{   by algebra}\]

\(2^{2(k+1)}-1=3(2^{2k}+m)\) and  \((2^{2k}+m)\in \mathbb{Z}\) since the integers are closed under addition and multiplication.  

So, \(2^{2(k+1)}-1\) is divisible by 3 by the definition of divisible.

Thus assuming our proposition works for \(k\) it will also work for \(k+1.\)

By PMI,  \(2^{2n}-1\) is divisible by 3, for all integers \(n\geq0.\)

Exercise \(\PageIndex{8}\label{ex:induct1-08}\)

Evaluate \(\sum_{i=1}^n \frac{1}{i(i+1)}\) for a few values of \(n\). What do you think the result should be? Use induction to prove your conjecture.

Exercise \(\PageIndex{9}\label{ex:induct1-09}\)

Use induction to prove that \[\sum_{i=1}^n (2i-1)^3 = n^2(2n^2-1)\] whenever \(n\) is a positive integer.

Exercise \(\PageIndex{10}\label{ex:induct1-10}\)

Use induction to show that, for any integer \(n\geq1\): \[1^2-2^2+3^2-\cdots+(-1)^{n-1}n^2 = (-1)^{n-1}\,\frac{n(n+1)}{2}.\]

Exercise \(\PageIndex{11}\label{ex:induct1-11}\)

Use mathematical induction to show that \[\sum_{i=1}^n \frac{i+4}{i(i+1)(i+2)} = \frac{n(3n+7)}{2(n+1)(n+2)}\] for all integers \(n\geq1\).

Exercise \(\PageIndex{12}\)

Use mathematical induction to show that \[3+\sum_{i=1}^n (3+5i) = \frac{(n+1)(5n+6)}{2}\] for all integers \(n\geq1\).

No answer here at this time.

Right Triangles

Rules, Formula and more

Pythagorean Theorem

The sum of the squares of the lengths of the legs equals the square of the length of the hypotenuse .

Usually, this theorem is expressed as $$ A^2 + B^2 = C^2 $$ .

Right Triangle Properties

A right triangle has one $$ 90^{\circ} $$ angle ($$ \angle $$ B in the picture on the left) and a variety of often-studied formulas such as:

• The Pythagorean Theorem
• Trigonometry Ratios (SOHCAHTOA)
• Pythagorean Theorem vs Sohcahtoa (which to use)

SOHCAHTOA only applies to right triangles ( more here ) .

A Right Triangle's Hypotenuse

The hypotenuse is the largest side in a right triangle and is always opposite the right angle.

In the triangle above, the hypotenuse is the side AB which is opposite the right angle, $$ \angle C $$.

Online tool calculates the hypotenuse (or a leg) using the Pythagorean theorem.

Practice Problems

Below are several practice problems involving the Pythagorean theorem, you can also get more detailed lesson on how to use the Pythagorean theorem here .

Find the length of side t in the triangle on the left.

Substitute the two known sides into the Pythagorean theorem's formula : A² + B² = C²

What is the value of x in the picture on the left?

Set up the Pythagorean Theorem : 14 2 + 48 2 = x 2 2,500 = X 2

$$ x = \sqrt{2500} = 50 $$

$$ x^2 = 21^2 + 72^2 \\ x^2= 5625 \\ x = \sqrt{5625} \\ x =75 $$

Find the length of side X in the triangle on on the left?

Substitue the two known sides into the pythagorean theorem's formula : $$ A^2 + B^2 = C^2 \\ 8^2 + 6^2 = x^2 \\ x = \sqrt{100}=10 $$

What is x in the triangle on the left?

x 2 + 4 2 = 5 2 x 2 + 16 = 25 x 2 = 25 - 16 = 9 x = 3

Ultimate Math Solver (Free) Free Algebra Solver ... type anything in there!

Popular pages @ mathwarehouse.com.

15 Hypothesis Examples

A hypothesis is defined as a testable prediction , and is used primarily in scientific experiments as a potential or predicted outcome that scientists attempt to prove or disprove (Atkinson et al., 2021; Tan, 2022).

In my types of hypothesis article, I outlined 13 different hypotheses, including the directional hypothesis (which makes a prediction about an effect of a treatment will be positive or negative) and the associative hypothesis (which makes a prediction about the association between two variables).

This article will dive into some interesting examples of hypotheses and examine potential ways you might test each one.

Hypothesis Examples

1. “inadequate sleep decreases memory retention”.

Field: Psychology

Type: Causal Hypothesis A causal hypothesis explores the effect of one variable on another. This example posits that a lack of adequate sleep causes decreased memory retention. In other words, if you are not getting enough sleep, your ability to remember and recall information may suffer.

How to Test:

To test this hypothesis, you might devise an experiment whereby your participants are divided into two groups: one receives an average of 8 hours of sleep per night for a week, while the other gets less than the recommended sleep amount.

During this time, all participants would daily study and recall new, specific information. You’d then measure memory retention of this information for both groups using standard memory tests and compare the results.

Should the group with less sleep have statistically significant poorer memory scores, the hypothesis would be supported.

Ensuring the integrity of the experiment requires taking into account factors such as individual health differences, stress levels, and daily nutrition.

Relevant Study: Sleep loss, learning capacity and academic performance (Curcio, Ferrara & De Gennaro, 2006)

2. “Increase in Temperature Leads to Increase in Kinetic Energy”

Field: Physics

Type: Deductive Hypothesis The deductive hypothesis applies the logic of deductive reasoning – it moves from a general premise to a more specific conclusion. This specific hypothesis assumes that as temperature increases, the kinetic energy of particles also increases – that is, when you heat something up, its particles move around more rapidly.

This hypothesis could be examined by heating a gas in a controlled environment and capturing the movement of its particles as a function of temperature.

You’d gradually increase the temperature and measure the kinetic energy of the gas particles with each increment. If the kinetic energy consistently rises with the temperature, your hypothesis gets supporting evidence.

Variables such as pressure and volume of the gas would need to be held constant to ensure validity of results.

3. “Children Raised in Bilingual Homes Develop Better Cognitive Skills”

Field: Psychology/Linguistics

Type: Comparative Hypothesis The comparative hypothesis posits a difference between two or more groups based on certain variables. In this context, you might propose that children raised in bilingual homes have superior cognitive skills compared to those raised in monolingual homes.

Testing this hypothesis could involve identifying two groups of children: those raised in bilingual homes, and those raised in monolingual homes.

Cognitive skills in both groups would be evaluated using a standard cognitive ability test at different stages of development. The examination would be repeated over a significant time period for consistency.

If the group raised in bilingual homes persistently scores higher than the other, the hypothesis would thereby be supported.

The challenge for the researcher would be controlling for other variables that could impact cognitive development, such as socio-economic status, education level of parents, and parenting styles.

Relevant Study: The cognitive benefits of being bilingual (Marian & Shook, 2012)

4. “High-Fiber Diet Leads to Lower Incidences of Cardiovascular Diseases”

Field: Medicine/Nutrition

Type: Alternative Hypothesis The alternative hypothesis suggests an alternative to a null hypothesis. In this context, the implied null hypothesis could be that diet has no effect on cardiovascular health, which the alternative hypothesis contradicts by suggesting that a high-fiber diet leads to fewer instances of cardiovascular diseases.

To test this hypothesis, a longitudinal study could be conducted on two groups of participants; one adheres to a high-fiber diet, while the other follows a diet low in fiber.

After a fixed period, the cardiovascular health of participants in both groups could be analyzed and compared. If the group following a high-fiber diet has a lower number of recorded cases of cardiovascular diseases, it would provide evidence supporting the hypothesis.

Control measures should be implemented to exclude the influence of other lifestyle and genetic factors that contribute to cardiovascular health.

Relevant Study: Dietary fiber, inflammation, and cardiovascular disease (King, 2005)

5. “Gravity Influences the Directional Growth of Plants”

Field: Agronomy / Botany

Type: Explanatory Hypothesis An explanatory hypothesis attempts to explain a phenomenon. In this case, the hypothesis proposes that gravity affects how plants direct their growth – both above-ground (toward sunlight) and below-ground (towards water and other resources).

The testing could be conducted by growing plants in a rotating cylinder to create artificial gravity.

Observations on the direction of growth, over a specified period, can provide insights into the influencing factors. If plants consistently direct their growth in a manner that indicates the influence of gravitational pull, the hypothesis is substantiated.

It is crucial to ensure that other growth-influencing factors, such as light and water, are uniformly distributed so that only gravity influences the directional growth.

6. “The Implementation of Gamified Learning Improves Students’ Motivation”

Field: Education

Type: Relational Hypothesis The relational hypothesis describes the relation between two variables. Here, the hypothesis is that the implementation of gamified learning has a positive effect on the motivation of students.

To validate this proposition, two sets of classes could be compared: one that implements a learning approach with game-based elements, and another that follows a traditional learning approach.

The students’ motivation levels could be gauged by monitoring their engagement, performance, and feedback over a considerable timeframe.

If the students engaged in the gamified learning context present higher levels of motivation and achievement, the hypothesis would be supported.

Control measures ought to be put into place to account for individual differences, including prior knowledge and attitudes towards learning.

Relevant Study: Does educational gamification improve students’ motivation? (Chapman & Rich, 2018)

7. “Mathematics Anxiety Negatively Affects Performance”

Field: Educational Psychology

Type: Research Hypothesis The research hypothesis involves making a prediction that will be tested. In this case, the hypothesis proposes that a student’s anxiety about math can negatively influence their performance in math-related tasks.

To assess this hypothesis, researchers must first measure the mathematics anxiety levels of a sample of students using a validated instrument, such as the Mathematics Anxiety Rating Scale.

Then, the students’ performance in mathematics would be evaluated through standard testing. If there’s a negative correlation between the levels of math anxiety and math performance (meaning as anxiety increases, performance decreases), the hypothesis would be supported.

It would be crucial to control for relevant factors such as overall academic performance and previous mathematical achievement.

8. “Disruption of Natural Sleep Cycle Impairs Worker Productivity”

Field: Organizational Psychology

Type: Operational Hypothesis The operational hypothesis involves defining the variables in measurable terms. In this example, the hypothesis posits that disrupting the natural sleep cycle, for instance through shift work or irregular working hours, can lessen productivity among workers.

To test this hypothesis, you could collect data from workers who maintain regular working hours and those with irregular schedules.

Measuring productivity could involve examining the worker’s ability to complete tasks, the quality of their work, and their efficiency.

If workers with interrupted sleep cycles demonstrate lower productivity compared to those with regular sleep patterns, it would lend support to the hypothesis.

Consideration should be given to potential confounding variables such as job type, worker age, and overall health.

9. “Regular Physical Activity Reduces the Risk of Depression”

Field: Health Psychology

Type: Predictive Hypothesis A predictive hypothesis involves making a prediction about the outcome of a study based on the observed relationship between variables. In this case, it is hypothesized that individuals who engage in regular physical activity are less likely to suffer from depression.

Longitudinal studies would suit to test this hypothesis, tracking participants’ levels of physical activity and their mental health status over time.

The level of physical activity could be self-reported or monitored, while mental health status could be assessed using standard diagnostic tools or surveys.

If data analysis shows that participants maintaining regular physical activity have a lower incidence of depression, this would endorse the hypothesis.

However, care should be taken to control other lifestyle and behavioral factors that could intervene with the results.

Relevant Study: Regular physical exercise and its association with depression (Kim, 2022)

10. “Regular Meditation Enhances Emotional Stability”

Type: Empirical Hypothesis In the empirical hypothesis, predictions are based on amassed empirical evidence . This particular hypothesis theorizes that frequent meditation leads to improved emotional stability, resonating with numerous studies linking meditation to a variety of psychological benefits.

Earlier studies reported some correlations, but to test this hypothesis directly, you’d organize an experiment where one group meditates regularly over a set period while a control group doesn’t.

Both groups’ emotional stability levels would be measured at the start and end of the experiment using a validated emotional stability assessment.

If regular meditators display noticeable improvements in emotional stability compared to the control group, the hypothesis gains credit.

You’d have to ensure a similar emotional baseline for all participants at the start to avoid skewed results.

11. “Children Exposed to Reading at an Early Age Show Superior Academic Progress”

Type: Directional Hypothesis The directional hypothesis predicts the direction of an expected relationship between variables. Here, the hypothesis anticipates that early exposure to reading positively affects a child’s academic advancement.

A longitudinal study tracking children’s reading habits from an early age and their consequent academic performance could validate this hypothesis.

Parents could report their children’s exposure to reading at home, while standardized school exam results would provide a measure of academic achievement.

If the children exposed to early reading consistently perform better acadically, it gives weight to the hypothesis.

However, it would be important to control for variables that might impact academic performance, such as socioeconomic background, parental education level, and school quality.

12. “Adopting Energy-efficient Technologies Reduces Carbon Footprint of Industries”

Field: Environmental Science

Type: Descriptive Hypothesis A descriptive hypothesis predicts the existence of an association or pattern related to variables. In this scenario, the hypothesis suggests that industries adopting energy-efficient technologies will resultantly show a reduced carbon footprint.

Global industries making use of energy-efficient technologies could track their carbon emissions over time. At the same time, others not implementing such technologies continue their regular tracking.

After a defined time, the carbon emission data of both groups could be compared. If industries that adopted energy-efficient technologies demonstrate a notable reduction in their carbon footprints, the hypothesis would hold strong.

In the experiment, you would exclude variations brought by factors such as industry type, size, and location.

13. “Reduced Screen Time Improves Sleep Quality”

Type: Simple Hypothesis The simple hypothesis is a prediction about the relationship between two variables, excluding any other variables from consideration. This example posits that by reducing time spent on devices like smartphones and computers, an individual should experience improved sleep quality.

A sample group would need to reduce their daily screen time for a pre-determined period. Sleep quality before and after the reduction could be measured using self-report sleep diaries and objective measures like actigraphy, monitoring movement and wakefulness during sleep.

If the data shows that sleep quality improved post the screen time reduction, the hypothesis would be validated.

Other aspects affecting sleep quality, like caffeine intake, should be controlled during the experiment.

Relevant Study: Screen time use impacts low‐income preschool children’s sleep quality, tiredness, and ability to fall asleep (Waller et al., 2021)

14. Engaging in Brain-Training Games Improves Cognitive Functioning in Elderly

Field: Gerontology

Type: Inductive Hypothesis Inductive hypotheses are based on observations leading to broader generalizations and theories. In this context, the hypothesis deduces from observed instances that engaging in brain-training games can help improve cognitive functioning in the elderly.

A longitudinal study could be conducted where an experimental group of elderly people partakes in regular brain-training games.

Their cognitive functioning could be assessed at the start of the study and at regular intervals using standard neuropsychological tests.

If the group engaging in brain-training games shows better cognitive functioning scores over time compared to a control group not playing these games, the hypothesis would be supported.

15. Farming Practices Influence Soil Erosion Rates

Type: Null Hypothesis A null hypothesis is a negative statement assuming no relationship or difference between variables. The hypothesis in this context asserts there’s no effect of different farming practices on the rates of soil erosion.

Comparing soil erosion rates in areas with different farming practices over a considerable timeframe could help test this hypothesis.

If, statistically, the farming practices do not lead to differences in soil erosion rates, the null hypothesis is accepted.

However, if marked variation appears, the null hypothesis is rejected, meaning farming practices do influence soil erosion rates. It would be crucial to control for external factors like weather, soil type, and natural vegetation.

The variety of hypotheses mentioned above underscores the diversity of research constructs inherent in different fields, each with its unique purpose and way of testing.

While researchers may develop hypotheses primarily as tools to define and narrow the focus of the study, these hypotheses also serve as valuable guiding forces for the data collection and analysis procedures, making the research process more efficient and direction-focused.

Hypotheses serve as a compass for any form of academic research. The diverse examples provided, from Psychology to Educational Studies, Environmental Science to Gerontology, clearly demonstrate how certain hypotheses suit specific fields more aptly than others.

It is important to underline that although these varied hypotheses differ in their structure and methods of testing, each endorses the fundamental value of empiricism in research. Evidence-based decision making remains at the heart of scholarly inquiry, regardless of the research field, thus aligning all hypotheses to the core purpose of scientific investigation.

Testing hypotheses is an essential part of the scientific method . By doing so, researchers can either confirm their predictions, giving further validity to an existing theory, or they might uncover new insights that could potentially shift the field’s understanding of a particular phenomenon. In either case, hypotheses serve as the stepping stones for scientific exploration and discovery.

Atkinson, P., Delamont, S., Cernat, A., Sakshaug, J. W., & Williams, R. A. (2021).  SAGE research methods foundations . SAGE Publications Ltd.

Curcio, G., Ferrara, M., & De Gennaro, L. (2006). Sleep loss, learning capacity and academic performance.  Sleep medicine reviews ,  10 (5), 323-337.

Kim, J. H. (2022). Regular physical exercise and its association with depression: A population-based study short title: Exercise and depression.  Psychiatry Research ,  309 , 114406.

King, D. E. (2005). Dietary fiber, inflammation, and cardiovascular disease.  Molecular nutrition & food research ,  49 (6), 594-600.

Marian, V., & Shook, A. (2012, September). The cognitive benefits of being bilingual. In Cerebrum: the Dana forum on brain science (Vol. 2012). Dana Foundation.

Tan, W. C. K. (2022). Research Methods: A Practical Guide For Students And Researchers (Second Edition) . World Scientific Publishing Company.

Waller, N. A., Zhang, N., Cocci, A. H., D’Agostino, C., Wesolek‐Greenson, S., Wheelock, K., … & Resnicow, K. (2021). Screen time use impacts low‐income preschool children’s sleep quality, tiredness, and ability to fall asleep. Child: care, health and development, 47 (5), 618-626.

Chris Drew (PhD)

Dr. Chris Drew is the founder of the Helpful Professor. He holds a PhD in education and has published over 20 articles in scholarly journals. He is the former editor of the Journal of Learning Development in Higher Education. [Image Descriptor: Photo of Chris]

• Chris Drew (PhD) https://helpfulprofessor.com/author/chris-drew-phd/ 5 Top Tips for Succeeding at University
• Chris Drew (PhD) https://helpfulprofessor.com/author/chris-drew-phd/ 50 Durable Goods Examples
• Chris Drew (PhD) https://helpfulprofessor.com/author/chris-drew-phd/ 100 Consumer Goods Examples
• Chris Drew (PhD) https://helpfulprofessor.com/author/chris-drew-phd/ 30 Globalization Pros and Cons

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A hypothesis is a proposition that is consistent with known data, but has been neither verified nor shown to be false.

In statistics, a hypothesis (sometimes called a statistical hypothesis) refers to a statement on which hypothesis testing will be based. Particularly important statistical hypotheses include the null hypothesis and alternative hypothesis .

In symbolic logic , a hypothesis is the first part of an implication (with the second part being known as the predicate ).

In general mathematical usage, "hypothesis" is roughly synonymous with " conjecture ."

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Unit 12: Significance tests (hypothesis testing)

About this unit.

Significance tests give us a formal process for using sample data to evaluate the likelihood of some claim about a population value. Learn how to conduct significance tests and calculate p-values to see how likely a sample result is to occur by random chance. You'll also see how we use p-values to make conclusions about hypotheses.

The idea of significance tests

• Simple hypothesis testing (Opens a modal)
• Idea behind hypothesis testing (Opens a modal)
• Examples of null and alternative hypotheses (Opens a modal)
• P-values and significance tests (Opens a modal)
• Comparing P-values to different significance levels (Opens a modal)
• Estimating a P-value from a simulation (Opens a modal)
• Using P-values to make conclusions (Opens a modal)
• Simple hypothesis testing Get 3 of 4 questions to level up!
• Writing null and alternative hypotheses Get 3 of 4 questions to level up!
• Estimating P-values from simulations Get 3 of 4 questions to level up!

Error probabilities and power

• Introduction to Type I and Type II errors (Opens a modal)
• Type 1 errors (Opens a modal)
• Examples identifying Type I and Type II errors (Opens a modal)
• Introduction to power in significance tests (Opens a modal)
• Examples thinking about power in significance tests (Opens a modal)
• Consequences of errors and significance (Opens a modal)
• Type I vs Type II error Get 3 of 4 questions to level up!
• Error probabilities and power Get 3 of 4 questions to level up!

Tests about a population proportion

• Constructing hypotheses for a significance test about a proportion (Opens a modal)
• Conditions for a z test about a proportion (Opens a modal)
• Reference: Conditions for inference on a proportion (Opens a modal)
• Calculating a z statistic in a test about a proportion (Opens a modal)
• Calculating a P-value given a z statistic (Opens a modal)
• Making conclusions in a test about a proportion (Opens a modal)
• Writing hypotheses for a test about a proportion Get 3 of 4 questions to level up!
• Conditions for a z test about a proportion Get 3 of 4 questions to level up!
• Calculating the test statistic in a z test for a proportion Get 3 of 4 questions to level up!
• Calculating the P-value in a z test for a proportion Get 3 of 4 questions to level up!
• Making conclusions in a z test for a proportion Get 3 of 4 questions to level up!

Tests about a population mean

• Writing hypotheses for a significance test about a mean (Opens a modal)
• Conditions for a t test about a mean (Opens a modal)
• Reference: Conditions for inference on a mean (Opens a modal)
• When to use z or t statistics in significance tests (Opens a modal)
• Example calculating t statistic for a test about a mean (Opens a modal)
• Using TI calculator for P-value from t statistic (Opens a modal)
• Using a table to estimate P-value from t statistic (Opens a modal)
• Comparing P-value from t statistic to significance level (Opens a modal)
• Free response example: Significance test for a mean (Opens a modal)
• Writing hypotheses for a test about a mean Get 3 of 4 questions to level up!
• Conditions for a t test about a mean Get 3 of 4 questions to level up!
• Calculating the test statistic in a t test for a mean Get 3 of 4 questions to level up!
• Calculating the P-value in a t test for a mean Get 3 of 4 questions to level up!
• Making conclusions in a t test for a mean Get 3 of 4 questions to level up!

More significance testing videos

• Hypothesis testing and p-values (Opens a modal)
• One-tailed and two-tailed tests (Opens a modal)
• Z-statistics vs. T-statistics (Opens a modal)
• Small sample hypothesis test (Opens a modal)
• Large sample proportion hypothesis testing (Opens a modal)

Hypothesis Testing

Hypothesis testing is a tool for making statistical inferences about the population data. It is an analysis tool that tests assumptions and determines how likely something is within a given standard of accuracy. Hypothesis testing provides a way to verify whether the results of an experiment are valid.

A null hypothesis and an alternative hypothesis are set up before performing the hypothesis testing. This helps to arrive at a conclusion regarding the sample obtained from the population. In this article, we will learn more about hypothesis testing, its types, steps to perform the testing, and associated examples.

What is Hypothesis Testing in Statistics?

Hypothesis testing uses sample data from the population to draw useful conclusions regarding the population probability distribution . It tests an assumption made about the data using different types of hypothesis testing methodologies. The hypothesis testing results in either rejecting or not rejecting the null hypothesis.

Hypothesis Testing Definition

Hypothesis testing can be defined as a statistical tool that is used to identify if the results of an experiment are meaningful or not. It involves setting up a null hypothesis and an alternative hypothesis. These two hypotheses will always be mutually exclusive. This means that if the null hypothesis is true then the alternative hypothesis is false and vice versa. An example of hypothesis testing is setting up a test to check if a new medicine works on a disease in a more efficient manner.

Null Hypothesis

The null hypothesis is a concise mathematical statement that is used to indicate that there is no difference between two possibilities. In other words, there is no difference between certain characteristics of data. This hypothesis assumes that the outcomes of an experiment are based on chance alone. It is denoted as \(H_{0}\). Hypothesis testing is used to conclude if the null hypothesis can be rejected or not. Suppose an experiment is conducted to check if girls are shorter than boys at the age of 5. The null hypothesis will say that they are the same height.

Alternative Hypothesis

The alternative hypothesis is an alternative to the null hypothesis. It is used to show that the observations of an experiment are due to some real effect. It indicates that there is a statistical significance between two possible outcomes and can be denoted as \(H_{1}\) or \(H_{a}\). For the above-mentioned example, the alternative hypothesis would be that girls are shorter than boys at the age of 5.

Hypothesis Testing P Value

In hypothesis testing, the p value is used to indicate whether the results obtained after conducting a test are statistically significant or not. It also indicates the probability of making an error in rejecting or not rejecting the null hypothesis.This value is always a number between 0 and 1. The p value is compared to an alpha level, \(\alpha\) or significance level. The alpha level can be defined as the acceptable risk of incorrectly rejecting the null hypothesis. The alpha level is usually chosen between 1% to 5%.

Hypothesis Testing Critical region

All sets of values that lead to rejecting the null hypothesis lie in the critical region. Furthermore, the value that separates the critical region from the non-critical region is known as the critical value.

Hypothesis Testing Formula

Depending upon the type of data available and the size, different types of hypothesis testing are used to determine whether the null hypothesis can be rejected or not. The hypothesis testing formula for some important test statistics are given below:

• z = \(\frac{\overline{x}-\mu}{\frac{\sigma}{\sqrt{n}}}\). \(\overline{x}\) is the sample mean, \(\mu\) is the population mean, \(\sigma\) is the population standard deviation and n is the size of the sample.
• t = \(\frac{\overline{x}-\mu}{\frac{s}{\sqrt{n}}}\). s is the sample standard deviation.
• \(\chi ^{2} = \sum \frac{(O_{i}-E_{i})^{2}}{E_{i}}\). \(O_{i}\) is the observed value and \(E_{i}\) is the expected value.

We will learn more about these test statistics in the upcoming section.

Types of Hypothesis Testing

Selecting the correct test for performing hypothesis testing can be confusing. These tests are used to determine a test statistic on the basis of which the null hypothesis can either be rejected or not rejected. Some of the important tests used for hypothesis testing are given below.

Hypothesis Testing Z Test

A z test is a way of hypothesis testing that is used for a large sample size (n ≥ 30). It is used to determine whether there is a difference between the population mean and the sample mean when the population standard deviation is known. It can also be used to compare the mean of two samples. It is used to compute the z test statistic. The formulas are given as follows:

• One sample: z = \(\frac{\overline{x}-\mu}{\frac{\sigma}{\sqrt{n}}}\).
• Two samples: z = \(\frac{(\overline{x_{1}}-\overline{x_{2}})-(\mu_{1}-\mu_{2})}{\sqrt{\frac{\sigma_{1}^{2}}{n_{1}}+\frac{\sigma_{2}^{2}}{n_{2}}}}\).

Hypothesis Testing t Test

The t test is another method of hypothesis testing that is used for a small sample size (n < 30). It is also used to compare the sample mean and population mean. However, the population standard deviation is not known. Instead, the sample standard deviation is known. The mean of two samples can also be compared using the t test.

• One sample: t = \(\frac{\overline{x}-\mu}{\frac{s}{\sqrt{n}}}\).
• Two samples: t = \(\frac{(\overline{x_{1}}-\overline{x_{2}})-(\mu_{1}-\mu_{2})}{\sqrt{\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}}}}\).

Hypothesis Testing Chi Square

The Chi square test is a hypothesis testing method that is used to check whether the variables in a population are independent or not. It is used when the test statistic is chi-squared distributed.

One Tailed Hypothesis Testing

One tailed hypothesis testing is done when the rejection region is only in one direction. It can also be known as directional hypothesis testing because the effects can be tested in one direction only. This type of testing is further classified into the right tailed test and left tailed test.

Right Tailed Hypothesis Testing

The right tail test is also known as the upper tail test. This test is used to check whether the population parameter is greater than some value. The null and alternative hypotheses for this test are given as follows:

\(H_{0}\): The population parameter is ≤ some value

\(H_{1}\): The population parameter is > some value.

If the test statistic has a greater value than the critical value then the null hypothesis is rejected

Left Tailed Hypothesis Testing

The left tail test is also known as the lower tail test. It is used to check whether the population parameter is less than some value. The hypotheses for this hypothesis testing can be written as follows:

\(H_{0}\): The population parameter is ≥ some value

\(H_{1}\): The population parameter is < some value.

The null hypothesis is rejected if the test statistic has a value lesser than the critical value.

Two Tailed Hypothesis Testing

In this hypothesis testing method, the critical region lies on both sides of the sampling distribution. It is also known as a non - directional hypothesis testing method. The two-tailed test is used when it needs to be determined if the population parameter is assumed to be different than some value. The hypotheses can be set up as follows:

\(H_{0}\): the population parameter = some value

\(H_{1}\): the population parameter ≠ some value

The null hypothesis is rejected if the test statistic has a value that is not equal to the critical value.

Hypothesis Testing Steps

Hypothesis testing can be easily performed in five simple steps. The most important step is to correctly set up the hypotheses and identify the right method for hypothesis testing. The basic steps to perform hypothesis testing are as follows:

• Step 1: Set up the null hypothesis by correctly identifying whether it is the left-tailed, right-tailed, or two-tailed hypothesis testing.
• Step 2: Set up the alternative hypothesis.
• Step 3: Choose the correct significance level, \(\alpha\), and find the critical value.
• Step 4: Calculate the correct test statistic (z, t or \(\chi\)) and p-value.
• Step 5: Compare the test statistic with the critical value or compare the p-value with \(\alpha\) to arrive at a conclusion. In other words, decide if the null hypothesis is to be rejected or not.

Hypothesis Testing Example

The best way to solve a problem on hypothesis testing is by applying the 5 steps mentioned in the previous section. Suppose a researcher claims that the mean average weight of men is greater than 100kgs with a standard deviation of 15kgs. 30 men are chosen with an average weight of 112.5 Kgs. Using hypothesis testing, check if there is enough evidence to support the researcher's claim. The confidence interval is given as 95%.

Step 1: This is an example of a right-tailed test. Set up the null hypothesis as \(H_{0}\): \(\mu\) = 100.

Step 2: The alternative hypothesis is given by \(H_{1}\): \(\mu\) > 100.

Step 3: As this is a one-tailed test, \(\alpha\) = 100% - 95% = 5%. This can be used to determine the critical value.

1 - \(\alpha\) = 1 - 0.05 = 0.95

0.95 gives the required area under the curve. Now using a normal distribution table, the area 0.95 is at z = 1.645. A similar process can be followed for a t-test. The only additional requirement is to calculate the degrees of freedom given by n - 1.

Step 4: Calculate the z test statistic. This is because the sample size is 30. Furthermore, the sample and population means are known along with the standard deviation.

z = \(\frac{\overline{x}-\mu}{\frac{\sigma}{\sqrt{n}}}\).

\(\mu\) = 100, \(\overline{x}\) = 112.5, n = 30, \(\sigma\) = 15

z = \(\frac{112.5-100}{\frac{15}{\sqrt{30}}}\) = 4.56

Step 5: Conclusion. As 4.56 > 1.645 thus, the null hypothesis can be rejected.

Hypothesis Testing and Confidence Intervals

Confidence intervals form an important part of hypothesis testing. This is because the alpha level can be determined from a given confidence interval. Suppose a confidence interval is given as 95%. Subtract the confidence interval from 100%. This gives 100 - 95 = 5% or 0.05. This is the alpha value of a one-tailed hypothesis testing. To obtain the alpha value for a two-tailed hypothesis testing, divide this value by 2. This gives 0.05 / 2 = 0.025.

Related Articles:

• Probability and Statistics
• Data Handling

Important Notes on Hypothesis Testing

• Hypothesis testing is a technique that is used to verify whether the results of an experiment are statistically significant.
• It involves the setting up of a null hypothesis and an alternate hypothesis.
• There are three types of tests that can be conducted under hypothesis testing - z test, t test, and chi square test.
• Hypothesis testing can be classified as right tail, left tail, and two tail tests.

Examples on Hypothesis Testing

• Example 1: The average weight of a dumbbell in a gym is 90lbs. However, a physical trainer believes that the average weight might be higher. A random sample of 5 dumbbells with an average weight of 110lbs and a standard deviation of 18lbs. Using hypothesis testing check if the physical trainer's claim can be supported for a 95% confidence level. Solution: As the sample size is lesser than 30, the t-test is used. \(H_{0}\): \(\mu\) = 90, \(H_{1}\): \(\mu\) > 90 \(\overline{x}\) = 110, \(\mu\) = 90, n = 5, s = 18. \(\alpha\) = 0.05 Using the t-distribution table, the critical value is 2.132 t = \(\frac{\overline{x}-\mu}{\frac{s}{\sqrt{n}}}\) t = 2.484 As 2.484 > 2.132, the null hypothesis is rejected. Answer: The average weight of the dumbbells may be greater than 90lbs
• Example 2: The average score on a test is 80 with a standard deviation of 10. With a new teaching curriculum introduced it is believed that this score will change. On random testing, the score of 38 students, the mean was found to be 88. With a 0.05 significance level, is there any evidence to support this claim? Solution: This is an example of two-tail hypothesis testing. The z test will be used. \(H_{0}\): \(\mu\) = 80, \(H_{1}\): \(\mu\) ≠ 80 \(\overline{x}\) = 88, \(\mu\) = 80, n = 36, \(\sigma\) = 10. \(\alpha\) = 0.05 / 2 = 0.025 The critical value using the normal distribution table is 1.96 z = \(\frac{\overline{x}-\mu}{\frac{\sigma}{\sqrt{n}}}\) z = \(\frac{88-80}{\frac{10}{\sqrt{36}}}\) = 4.8 As 4.8 > 1.96, the null hypothesis is rejected. Answer: There is a difference in the scores after the new curriculum was introduced.
• Example 3: The average score of a class is 90. However, a teacher believes that the average score might be lower. The scores of 6 students were randomly measured. The mean was 82 with a standard deviation of 18. With a 0.05 significance level use hypothesis testing to check if this claim is true. Solution: The t test will be used. \(H_{0}\): \(\mu\) = 90, \(H_{1}\): \(\mu\) < 90 \(\overline{x}\) = 110, \(\mu\) = 90, n = 6, s = 18 The critical value from the t table is -2.015 t = \(\frac{\overline{x}-\mu}{\frac{s}{\sqrt{n}}}\) t = \(\frac{82-90}{\frac{18}{\sqrt{6}}}\) t = -1.088 As -1.088 > -2.015, we fail to reject the null hypothesis. Answer: There is not enough evidence to support the claim.

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FAQs on Hypothesis Testing

What is hypothesis testing.

Hypothesis testing in statistics is a tool that is used to make inferences about the population data. It is also used to check if the results of an experiment are valid.

What is the z Test in Hypothesis Testing?

The z test in hypothesis testing is used to find the z test statistic for normally distributed data . The z test is used when the standard deviation of the population is known and the sample size is greater than or equal to 30.

What is the t Test in Hypothesis Testing?

The t test in hypothesis testing is used when the data follows a student t distribution . It is used when the sample size is less than 30 and standard deviation of the population is not known.

What is the formula for z test in Hypothesis Testing?

The formula for a one sample z test in hypothesis testing is z = \(\frac{\overline{x}-\mu}{\frac{\sigma}{\sqrt{n}}}\) and for two samples is z = \(\frac{(\overline{x_{1}}-\overline{x_{2}})-(\mu_{1}-\mu_{2})}{\sqrt{\frac{\sigma_{1}^{2}}{n_{1}}+\frac{\sigma_{2}^{2}}{n_{2}}}}\).

What is the p Value in Hypothesis Testing?

The p value helps to determine if the test results are statistically significant or not. In hypothesis testing, the null hypothesis can either be rejected or not rejected based on the comparison between the p value and the alpha level.

What is One Tail Hypothesis Testing?

When the rejection region is only on one side of the distribution curve then it is known as one tail hypothesis testing. The right tail test and the left tail test are two types of directional hypothesis testing.

What is the Alpha Level in Two Tail Hypothesis Testing?

To get the alpha level in a two tail hypothesis testing divide \(\alpha\) by 2. This is done as there are two rejection regions in the curve.

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8.4: Hypothesis Test Examples for Proportions

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• In a hypothesis test problem, you may see words such as "the level of significance is 1%." The "1%" is the preconceived or preset \(\alpha\).
• The statistician setting up the hypothesis test selects the value of α to use before collecting the sample data.
• If no level of significance is given, a common standard to use is \(\alpha = 0.05\).
• When you calculate the \(p\)-value and draw the picture, the \(p\)-value is the area in the left tail, the right tail, or split evenly between the two tails. For this reason, we call the hypothesis test left, right, or two tailed.
• The alternative hypothesis, \(H_{a}\), tells you if the test is left, right, or two-tailed. It is the key to conducting the appropriate test.
• \(H_{a}\) never has a symbol that contains an equal sign.
• Thinking about the meaning of the \(p\)-value: A data analyst (and anyone else) should have more confidence that he made the correct decision to reject the null hypothesis with a smaller \(p\)-value (for example, 0.001 as opposed to 0.04) even if using the 0.05 level for alpha. Similarly, for a large p -value such as 0.4, as opposed to a \(p\)-value of 0.056 (\(\alpha = 0.05\) is less than either number), a data analyst should have more confidence that she made the correct decision in not rejecting the null hypothesis. This makes the data analyst use judgment rather than mindlessly applying rules.

Full Hypothesis Test Examples

Example \(\PageIndex{7}\)

Joon believes that 50% of first-time brides in the United States are younger than their grooms. She performs a hypothesis test to determine if the percentage is the same or different from 50% . Joon samples 100 first-time brides and 53 reply that they are younger than their grooms. For the hypothesis test, she uses a 1% level of significance.

Set up the hypothesis test:

The 1% level of significance means that α = 0.01. This is a test of a single population proportion .

\(H_{0}: p = 0.50\)  \(H_{a}: p \neq 0.50\)

The words "is the same or different from" tell you this is a two-tailed test.

Calculate the distribution needed:

Random variable: \(P′ =\) the percent of of first-time brides who are younger than their grooms.

Distribution for the test: The problem contains no mention of a mean. The information is given in terms of percentages. Use the distribution for P′ , the estimated proportion.

\[P' - N\left(p, \sqrt{\frac{p-q}{n}}\right)\nonumber \]

\[P' - N\left(0.5, \sqrt{\frac{0.5-0.5}{100}}\right)\nonumber \]

where \(p = 0.50, q = 1−p = 0.50\), and \(n = 100\)

Calculate the p -value using the normal distribution for proportions:

\[p\text{-value} = P(p′ < 0.47 or p′ > 0.53) = 0.5485\nonumber \]

where \[x = 53, p' = \frac{x}{n} = \frac{53}{100} = 0.53\nonumber \].

Interpretation of the \(p\text{-value})\: If the null hypothesis is true, there is 0.5485 probability (54.85%) that the sample (estimated) proportion \(p'\) is 0.53 or more OR 0.47 or less (see the graph in Figure).

\(\mu = p = 0.50\) comes from \(H_{0}\), the null hypothesis.

\(p′ = 0.53\). Since the curve is symmetrical and the test is two-tailed, the \(p′\) for the left tail is equal to \(0.50 – 0.03 = 0.47\) where \(\mu = p = 0.50\). (0.03 is the difference between 0.53 and 0.50.)

Compare \(\alpha\) and the \(p\text{-value}\):

Since \(\alpha = 0.01\) and \(p\text{-value} = 0.5485\). \(\alpha < p\text{-value}\).

Make a decision: Since \(\alpha < p\text{-value}\), you cannot reject \(H_{0}\).

Conclusion: At the 1% level of significance, the sample data do not show sufficient evidence that the percentage of first-time brides who are younger than their grooms is different from 50%.

The \(p\text{-value}\) can easily be calculated.

Press STAT and arrow over to TESTS . Press 5:1-PropZTest . Enter .5 for \(p_{0}\), 53 for \(x\) and 100 for \(n\). Arrow down to Prop and arrow to not equals \(p_{0}\). Press ENTER . Arrow down to Calculate and press ENTER . The calculator calculates the \(p\text{-value}\) (\(p = 0.5485\)) and the test statistic (\(z\)-score). Prop not equals .5 is the alternate hypothesis. Do this set of instructions again except arrow to Draw (instead of Calculate ). Press ENTER . A shaded graph appears with \(\(z\) = 0.6\) (test statistic) and \(p = 0.5485\) (\(p\text{-value}\)). Make sure when you use Draw that no other equations are highlighted in \(Y =\) and the plots are turned off.

The Type I and Type II errors are as follows:

The Type I error is to conclude that the proportion of first-time brides who are younger than their grooms is different from 50% when, in fact, the proportion is actually 50%. (Reject the null hypothesis when the null hypothesis is true).

The Type II error is there is not enough evidence to conclude that the proportion of first time brides who are younger than their grooms differs from 50% when, in fact, the proportion does differ from 50%. (Do not reject the null hypothesis when the null hypothesis is false.)

Exercise \(\PageIndex{7}\)

A teacher believes that 85% of students in the class will want to go on a field trip to the local zoo. She performs a hypothesis test to determine if the percentage is the same or different from 85%. The teacher samples 50 students and 39 reply that they would want to go to the zoo. For the hypothesis test, use a 1% level of significance.

First, determine what type of test this is, set up the hypothesis test, find the \(p\text{-value}\), sketch the graph, and state your conclusion.

Since the problem is about percentages, this is a test of single population proportions.

• \(H_{0} : p = 0.85\)
• \(H_{a}: p \neq 0.85\)
• \(p = 0.7554\)

Because \(p > \alpha\), we fail to reject the null hypothesis. There is not sufficient evidence to suggest that the proportion of students that want to go to the zoo is not 85%.

Example \(\PageIndex{8}\)

Suppose a consumer group suspects that the proportion of households that have three cell phones is 30%. A cell phone company has reason to believe that the proportion is not 30%. Before they start a big advertising campaign, they conduct a hypothesis test. Their marketing people survey 150 households with the result that 43 of the households have three cell phones.

Set up the Hypothesis Test:

\(H_{0}: p = 0.30, H_{a}: p \neq 0.30\)

Determine the distribution needed:

The random variable is \(P′ =\) proportion of households that have three cell phones.

The distribution for the hypothesis test is \(P' - N\left(0.30, \sqrt{\frac{(0.30 \cdot 0.70)}{150}}\right)\)

Exercise 9.6.8.2

a. The value that helps determine the \(p\text{-value}\) is \(p′\). Calculate \(p′\).

a. \(p' = \frac{x}{n}\) where \(x\) is the number of successes and \(n\) is the total number in the sample.

\(x = 43, n = 150\)

\(p′ = 43150\)

Exercise 9.6.8.3

b. What is a success for this problem?

b. A success is having three cell phones in a household.

Exercise 9.6.8.4

c. What is the level of significance?

c. The level of significance is the preset \(\alpha\). Since \(\alpha\) is not given, assume that \(\alpha = 0.05\).

Exercise 9.6.8.5

d. Draw the graph for this problem. Draw the horizontal axis. Label and shade appropriately.

Calculate the \(p\text{-value}\).

d. \(p\text{-value} = 0.7216\)

Exercise 9.6.8.6

e. Make a decision. _____________(Reject/Do not reject) \(H_{0}\) because____________.

e. Assuming that \(\alpha = 0.05, \alpha < p\text{-value}\). The decision is do not reject \(H_{0}\) because there is not sufficient evidence to conclude that the proportion of households that have three cell phones is not 30%.

Exercise \(\PageIndex{8}\)

Marketers believe that 92% of adults in the United States own a cell phone. A cell phone manufacturer believes that number is actually lower. 200 American adults are surveyed, of which, 174 report having cell phones. Use a 5% level of significance. State the null and alternative hypothesis, find the p -value, state your conclusion, and identify the Type I and Type II errors.

• \(H_{0}: p = 0.92\)
• \(H_{a}: p < 0.92\)
• \(p\text{-value} = 0.0046\)

Because \(p < 0.05\), we reject the null hypothesis. There is sufficient evidence to conclude that fewer than 92% of American adults own cell phones.

• Type I Error: To conclude that fewer than 92% of American adults own cell phones when, in fact, 92% of American adults do own cell phones (reject the null hypothesis when the null hypothesis is true).
• Type II Error: To conclude that 92% of American adults own cell phones when, in fact, fewer than 92% of American adults own cell phones (do not reject the null hypothesis when the null hypothesis is false).

The next example is a poem written by a statistics student named Nicole Hart. The solution to the problem follows the poem. Notice that the hypothesis test is for a single population proportion. This means that the null and alternate hypotheses use the parameter \(p\). The distribution for the test is normal. The estimated proportion \(p′\) is the proportion of fleas killed to the total fleas found on Fido. This is sample information. The problem gives a preconceived \(\alpha = 0.01\), for comparison, and a 95% confidence interval computation. The poem is clever and humorous, so please enjoy it!

Example \(\PageIndex{9}\)

My dog has so many fleas,

They do not come off with ease. As for shampoo, I have tried many types Even one called Bubble Hype, Which only killed 25% of the fleas, Unfortunately I was not pleased.

I've used all kinds of soap, Until I had given up hope Until one day I saw An ad that put me in awe.

A shampoo used for dogs Called GOOD ENOUGH to Clean a Hog Guaranteed to kill more fleas.

I gave Fido a bath And after doing the math His number of fleas Started dropping by 3's! Before his shampoo I counted 42.

At the end of his bath, I redid the math And the new shampoo had killed 17 fleas. So now I was pleased.

Now it is time for you to have some fun With the level of significance being .01, You must help me figure out

Use the new shampoo or go without?

\(H_{0}: p \leq 0.25\)   \(H_{a}: p > 0.25\)

In words, CLEARLY state what your random variable \(\bar{X}\) or \(P′\) represents.

\(P′ =\) The proportion of fleas that are killed by the new shampoo

State the distribution to use for the test.

\[N\left(0.25, \sqrt{\frac{(0.25){1-0.25}}{42}}\right)\nonumber \]

Test Statistic: \(z = 2.3163\)

Calculate the \(p\text{-value}\) using the normal distribution for proportions:

\[p\text{-value} = 0.0103\nonumber \]

In one to two complete sentences, explain what the p -value means for this problem.

If the null hypothesis is true (the proportion is 0.25), then there is a 0.0103 probability that the sample (estimated) proportion is 0.4048 \(\left(\frac{17}{42}\right)\) or more.

Use the previous information to sketch a picture of this situation. CLEARLY, label and scale the horizontal axis and shade the region(s) corresponding to the \(p\text{-value}\).

Indicate the correct decision (“reject” or “do not reject” the null hypothesis), the reason for it, and write an appropriate conclusion, using complete sentences.

Conclusion: At the 1% level of significance, the sample data do not show sufficient evidence that the percentage of fleas that are killed by the new shampoo is more than 25%.

Construct a 95% confidence interval for the true mean or proportion. Include a sketch of the graph of the situation. Label the point estimate and the lower and upper bounds of the confidence interval.

Confidence Interval: (0.26,0.55) We are 95% confident that the true population proportion p of fleas that are killed by the new shampoo is between 26% and 55%.

This test result is not very definitive since the \(p\text{-value}\) is very close to alpha. In reality, one would probably do more tests by giving the dog another bath after the fleas have had a chance to return.

Example \(\PageIndex{11}\)

In a study of 420,019 cell phone users, 172 of the subjects developed brain cancer. Test the claim that cell phone users developed brain cancer at a greater rate than that for non-cell phone users (the rate of brain cancer for non-cell phone users is 0.0340%). Since this is a critical issue, use a 0.005 significance level. Explain why the significance level should be so low in terms of a Type I error.

We will follow the four-step process.

• \(H_{0}: p \leq 0.00034\)
• \(H_{a}: p > 0.00034\)

If we commit a Type I error, we are essentially accepting a false claim. Since the claim describes cancer-causing environments, we want to minimize the chances of incorrectly identifying causes of cancer.

• We will be testing a sample proportion with \(x = 172\) and \(n = 420,019\). The sample is sufficiently large because we have \(np = 420,019(0.00034) = 142.8\), \(nq = 420,019(0.99966) = 419,876.2\), two independent outcomes, and a fixed probability of success \(p = 0.00034\). Thus we will be able to generalize our results to the population.

Figure 9.6.11.

Figure 9.6.12.

• Since the \(p\text{-value} = 0.0073\) is greater than our alpha value \(= 0.005\), we cannot reject the null. Therefore, we conclude that there is not enough evidence to support the claim of higher brain cancer rates for the cell phone users.

Example \(\PageIndex{12}\)

According to the US Census there are approximately 268,608,618 residents aged 12 and older. Statistics from the Rape, Abuse, and Incest National Network indicate that, on average, 207,754 rapes occur each year (male and female) for persons aged 12 and older. This translates into a percentage of sexual assaults of 0.078%. In Daviess County, KY, there were reported 11 rapes for a population of 37,937. Conduct an appropriate hypothesis test to determine if there is a statistically significant difference between the local sexual assault percentage and the national sexual assault percentage. Use a significance level of 0.01.

We will follow the four-step plan.

• We need to test whether the proportion of sexual assaults in Daviess County, KY is significantly different from the national average.
• \(H_{0}: p = 0.00078\)
• \(H_{a}: p \neq 0.00078\)

Figure 9.6.13.

Figure 9.6.14.

• Since the \(p\text{-value}\), \(p = 0.00063\), is less than the alpha level of 0.01, the sample data indicates that we should reject the null hypothesis. In conclusion, the sample data support the claim that the proportion of sexual assaults in Daviess County, Kentucky is different from the national average proportion.

The hypothesis test itself has an established process. This can be summarized as follows:

• Determine \(H_{0}\) and \(H_{a}\). Remember, they are contradictory.
• Determine the random variable.
• Determine the distribution for the test.
• Draw a graph, calculate the test statistic, and use the test statistic to calculate the \(p\text{-value}\). (A z -score and a t -score are examples of test statistics.)
• Compare the preconceived α with the p -value, make a decision (reject or do not reject H 0 ), and write a clear conclusion using English sentences.

Notice that in performing the hypothesis test, you use \(\alpha\) and not \(\beta\). \(\beta\) is needed to help determine the sample size of the data that is used in calculating the \(p\text{-value}\). Remember that the quantity \(1 – \beta\) is called the Power of the Test . A high power is desirable. If the power is too low, statisticians typically increase the sample size while keeping α the same.If the power is low, the null hypothesis might not be rejected when it should be.

• Data from Amit Schitai. Director of Instructional Technology and Distance Learning. LBCC.
• Data from Bloomberg Businessweek . Available online at http://www.businessweek.com/news/2011- 09-15/nyc-smoking-rate-falls-to-record-low-of-14-bloomberg-says.html.
• Data from energy.gov. Available online at http://energy.gov (accessed June 27. 2013).
• Data from Gallup®. Available online at www.gallup.com (accessed June 27, 2013).
• Data from Growing by Degrees by Allen and Seaman.
• Data from La Leche League International. Available online at www.lalecheleague.org/Law/BAFeb01.html.
• Data from the American Automobile Association. Available online at www.aaa.com (accessed June 27, 2013).
• Data from the American Library Association. Available online at www.ala.org (accessed June 27, 2013).
• Data from the Bureau of Labor Statistics. Available online at http://www.bls.gov/oes/current/oes291111.htm .
• Data from the Centers for Disease Control and Prevention. Available online at www.cdc.gov (accessed June 27, 2013)
• Data from the U.S. Census Bureau, available online at quickfacts.census.gov/qfd/states/00000.html (accessed June 27, 2013).
• Data from the United States Census Bureau. Available online at www.census.gov/hhes/socdemo/language/.
• Data from Toastmasters International. Available online at http://toastmasters.org/artisan/deta...eID=429&Page=1 .
• Data from Weather Underground. Available online at www.wunderground.com (accessed June 27, 2013).
• Federal Bureau of Investigations. “Uniform Crime Reports and Index of Crime in Daviess in the State of Kentucky enforced by Daviess County from 1985 to 2005.” Available online at http://www.disastercenter.com/kentucky/crime/3868.htm (accessed June 27, 2013).
• “Foothill-De Anza Community College District.” De Anza College, Winter 2006. Available online at research.fhda.edu/factbook/DA...t_da_2006w.pdf.
• Johansen, C., J. Boice, Jr., J. McLaughlin, J. Olsen. “Cellular Telephones and Cancer—a Nationwide Cohort Study in Denmark.” Institute of Cancer Epidemiology and the Danish Cancer Society, 93(3):203-7. Available online at http://www.ncbi.nlm.nih.gov/pubmed/11158188 (accessed June 27, 2013).
• Rape, Abuse & Incest National Network. “How often does sexual assault occur?” RAINN, 2009. Available online at www.rainn.org/get-information...sexual-assault (accessed June 27, 2013).

Contributors and Attributions

Barbara Illowsky and Susan Dean (De Anza College) with many other contributing authors. Content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at http://cnx.org/contents/[email protected] .