2: Interval Notation

2: Interval Notation

Confidence Intervals

A confidence interval gives an estimated range of values which is likely to include an unknown population parameter, the estimated range being calculated from a given set of sample data. ( Definition taken from Valerie J. Easton and John H. McColl's Statistics Glossary v1.1 )

The level C of a confidence interval gives the probability that the interval produced by the method employed includes the true value of the parameter .

Suppose a student measuring the boiling temperature of a certain liquid observes the readings (in degrees Celsius) 102.5, 101.7, 103.1, 100.9, 100.5, and 102.2 on 6 different samples of the liquid. He calculates the sample mean to be 101.82. If he knows that the standard deviation for this procedure is 1.2 degrees, what is the confidence interval for the population mean at a 95% confidence level?

In other words, the student wishes to estimate the true mean boiling temperature of the liquid using the results of his measurements. If the measurements follow a normal distribution, then the sample mean will have the distribution N( , ) . Since the sample size is 6, the standard deviation of the sample mean is equal to 1.2/sqrt(6) = 0.49. The selection of a confidence level for an interval determines the probability that the confidence interval produced will contain the true parameter value. Common choices for the confidence level C are 0.90, 0.95, and 0.99. These levels correspond to percentages of the area of the normal density curve. For example, a 95% confidence interval covers 95% of the normal curve -- the probability of observing a value outside of this area is less than 0.05. Because the normal curve is symmetric, half of the area is in the left tail of the curve, and the other half of the area is in the right tail of the curve. As shown in the diagram to the right, for a confidence interval with level C , the area in each tail of the curve is equal to (1- C )/2. For a 95% confidence interval, the area in each tail is equal to 0.05/2 = 0.025.

The value z * representing the point on the standard normal density curve such that the probability of observing a value greater than z * is equal to p is known as the upper p critical value of the standard normal distribution. For example, if p = 0.025, the value z * such that P(Z > z * ) = 0.025, or P(Z < z * ) = 0.975, is equal to 1.96. For a confidence interval with level C , the value p is equal to (1- C )/2. A 95% confidence interval for the standard normal distribution, then, is the interval (-1.96, 1.96), since 95% of the area under the curve falls within this interval.

Confidence Intervals for Unknown Mean and Known Standard Deviation

Note: This interval is only exact when the population distribution is normal. For large samples from other population distributions, the interval is approximately correct by the Central Limit Theorem. In the example above, the student calculated the sample mean of the boiling temperatures to be 101.82, with standard deviation 0.49. The critical value for a 95% confidence interval is 1.96, where (1-0.95)/2 = 0.025. A 95% confidence interval for the unknown mean is ((101.82 - (1.96*0.49)), (101.82 + (1.96*0.49))) = (101.82 - 0.96, 101.82 + 0.96) = (100.86, 102.78).

As the level of confidence decreases, the size of the corresponding interval will decrease. Suppose the student was interested in a 90% confidence interval for the boiling temperature. In this case, C = 0.90, and (1- C )/2 = 0.05. The critical value z * for this level is equal to 1.645, so the 90% confidence interval is ((101.82 - (1.645*0.49)), (101.82 + (1.645*0.49))) = (101.82 - 0.81, 101.82 + 0.81) = (101.01, 102.63) An increase in sample size will decrease the length of the confidence interval without reducing the level of confidence. This is because the standard deviation decreases as n increases. The margin of error m of a confidence interval is defined to be the value added or subtracted from the sample mean which determines the length of the interval: m = z * .

Suppose in the example above, the student wishes to have a margin of error equal to 0.5 with 95% confidence. Substituting the appropriate values into the expression for m and solving for n gives the calculation n = (1.96*1.2/0.5)² = (2.35/0.5)² = 4.7² = 22.09. To achieve a 95% confidence interval for the mean boiling point with total length less than 1 degree, the student will have to take 23 measurements.

Confidence Intervals for Unknown Mean and Unknown Standard Deviation

For a population with unknown mean and unknown standard deviation, a confidence interval for the population mean, based on a simple random sample (SRS) of size n , is + t * , where t * is the upper (1- C )/2 critical value for the t distribution with n-1 degrees of freedom, t(n-1) . Example

The dataset "Normal Body Temperature, Gender, and Heart Rate" contains 130 observations of body temperature, along with the gender of each individual and his or her heart rate. Using the MINITAB "DESCRIBE" command provides the following information: To find a 95% confidence interval for the mean based on the sample mean 98.249 and sample standard deviation 0.733, first find the 0.025 critical value t * for 129 degrees of freedom. This value is approximately 1.962, the critical value for 100 degrees of freedom (found in Table E in Moore and McCabe). The estimated standard deviation for the sample mean is 0.733/sqrt(130) = 0.064, the value provided in the SE MEAN column of the MINITAB descriptive statistics. A 95% confidence interval, then, is approximately ((98.249 - 1.962*0.064), (98.249 + 1.962*0.064)) = (98.249 - 0.126, 98.249+ 0.126) = (98.123, 98.375).

For a more precise (and more simply achieved) result, the MINITAB "TINTERVAL" command, written as follows, gives an exact 95% confidence interval for 129 degrees of freedom: According to these results, the usual assumed normal body temperature of 98.6 degrees Fahrenheit is not within a 95% confidence interval for the mean.

2: Interval Notation

Many of our applications in this chapter will revolve around minimum and maximum values of a function. While we can all visualize the minimum and maximum values of a function we want to be a little more specific in our work here. In particular, we want to differentiate between two types of minimum or maximum values. The following definition gives the types of minimums and/or maximums values that we’ll be looking at.

Definition

1. We say that (fleft( x ight)) has an absolute (or global) maximum at (x = c) if(fleft( x ight) le fleft( c ight)) for every (x) in the domain we are working on.

Note that when we say an “open interval around(x = c)” we mean that we can find some interval (left( ight)), not including the endpoints, such that (a < c < b). Or, in other words, (c) will be contained somewhere inside the interval and will not be either of the endpoints.

Also, we will collectively call the minimum and maximum points of a function the extrema of the function. So, relative extrema will refer to the relative minimums and maximums while absolute extrema refer to the absolute minimums and maximums.

Now, let’s talk a little bit about the subtle difference between the absolute and relative in the definition above.

We will have an absolute maximum (or minimum) at (x = c) provided (fleft( c ight)) is the largest (or smallest) value that the function will ever take on the domain that we are working on. Also, when we say the “domain we are working on” this simply means the range of (x)’s that we have chosen to work with for a given problem. There may be other values of (x) that we can actually plug into the function but have excluded them for some reason.

A relative maximum or minimum is slightly different. All that’s required for a point to be a relative maximum or minimum is for that point to be a maximum or minimum in some interval of (x)’s around (x = c). There may be larger or smaller values of the function at some other place, but relative to (x = c), or local to (x = c), (fleft( c ight)) is larger or smaller than all the other function values that are near it.

Note as well that in order for a point to be a relative extrema we must be able to look at function values on both sides of (x = c) to see if it really is a maximum or minimum at that point. This means that relative extrema do not occur at the end points of a domain. They can only occur interior to the domain.

There is actually some debate on the preceding point. Some folks do feel that relative extrema can occur on the end points of a domain. However, in this class we will be using the definition that says that they can’t occur at the end points of a domain. This will be discussed in a little more detail at the end of the section once we have a relevant fact taken care of.

It’s usually easier to get a feel for the definitions by taking a quick look at a graph.

For the function shown in this graph we have relative maximums at (x = b) and (x = d). Both of these points are relative maximums since they are interior to the domain shown and are the largest point on the graph in some interval around the point. We also have a relative minimum at (x = c) since this point is interior to the domain and is the lowest point on the graph in an interval around it. The far-right end point, (x = e), will not be a relative minimum since it is an end point.

The function will have an absolute maximum at (x = d) and an absolute minimum at (x = a). These two points are the largest and smallest that the function will ever be. We can also notice that the absolute extrema for a function will occur at either the endpoints of the domain or at relative extrema. We will use this idea in later sections so it’s more important than it might seem at the present time.

Let’s take a quick look at some examples to make sure that we have the definitions of absolute extrema and relative extrema straight.

Since this function is easy enough to graph let’s do that. However, we only want the graph on the interval (left[ < - 1,2> ight]). Here is the graph,

Note that we used dots at the end of the graph to remind us that the graph ends at these points.

We can now identify the extrema from the graph. It looks like we’ve got a relative and absolute minimum of zero at (x = 0) and an absolute maximum of four at (x = 2). Note that (x = - 1) is not a relative maximum since it is at the end point of the interval.

This function doesn’t have any relative maximums.

As we saw in the previous example functions do not have to have relative extrema. It is completely possible for a function to not have a relative maximum and/or a relative minimum.

Here is the graph for this function.

In this case we still have a relative and absolute minimum of zero at (x = 0). We also still have an absolute maximum of four. However, unlike the first example this will occur at two points, (x = - 2) and (x = 2).

Again, the function doesn’t have any relative maximums.

As this example has shown there can only be a single absolute maximum or absolute minimum value, but they can occur at more than one place in the domain.

In this case we’ve given no domain and so the assumption is that we will take the largest possible domain. For this function that means all the real numbers. Here is the graph.

In this case the graph doesn’t stop increasing at either end and so there are no maximums of any kind for this function. No matter which point we pick on the graph there will be points both larger and smaller than it on either side so we can’t have any maximums (of any kind, relative or absolute) in a graph.

We still have a relative and absolute minimum value of zero at (x = 0).

So, some graphs can have minimums but not maximums. Likewise, a graph could have maximums but not minimums.

Here is the graph for this function.

This function has an absolute maximum of eight at (x = 2) and an absolute minimum of negative eight at (x = - 2). This function has no relative extrema.

So, a function doesn’t have to have relative extrema as this example has shown.

Again, we aren’t restricting the domain this time so here’s the graph.

In this case the function has no relative extrema and no absolute extrema.

As we’ve seen in the previous example functions don’t have to have any kind of extrema, relative or absolute.

We’ve not restricted the domain for this function. Here is the graph.

Cosine has extrema (relative and absolute) that occur at many points. Cosine has both relative and absolute maximums of 1 at

[x = ldots - 4pi ,, - 2pi ,,,0,,,2pi ,,,4pi , ldots ]

Cosine also has both relative and absolute minimums of -1 at

[x = ldots - 3pi ,, - pi ,,,pi ,,,3pi , ldots ]

As this example has shown a graph can in fact have extrema occurring at a large number (infinite in this case) of points.

We’ve now worked quite a few examples and we can use these examples to see a nice fact about absolute extrema. First let’s notice that all the functions above were continuous functions. Next notice that every time we restricted the domain to a closed interval (i.e. the interval contains its end points) we got absolute maximums and absolute minimums. Finally, in only one of the three examples in which we did not restrict the domain did we get both an absolute maximum and an absolute minimum.

These observations lead us the following theorem.

Extreme Value Theorem

Suppose that (fleft( x ight)) is continuous on the interval (left[ ight]) then there are two numbers (a le c,d le b) so that (fleft( c ight)) is an absolute maximum for the function and (fleft( d ight)) is an absolute minimum for the function.

So, if we have a continuous function on an interval (left[ ight]) then we are guaranteed to have both an absolute maximum and an absolute minimum for the function somewhere in the interval. The theorem doesn’t tell us where they will occur or if they will occur more than once, but at least it tells us that they do exist somewhere. Sometimes, all that we need to know is that they do exist.

This theorem doesn’t say anything about absolute extrema if we aren’t working on an interval. We saw examples of functions above that had both absolute extrema, one absolute extrema, and no absolute extrema when we didn’t restrict ourselves down to an interval.

The requirement that a function be continuous is also required in order for us to use the theorem. Consider the case of

This function is not continuous at (x = 0) as we move in towards zero the function is approaching infinity. So, the function does not have an absolute maximum. Note that it does have an absolute minimum however. In fact the absolute minimum occurs twice at both (x = - 1) and (x = 1).

If we changed the interval a little to say,

the function would now have both absolute extrema. We may only run into problems if the interval contains the point of discontinuity. If it doesn’t then the theorem will hold.

We should also point out that just because a function is not continuous at a point that doesn’t mean that it won’t have both absolute extrema in an interval that contains that point. Below is the graph of a function that is not continuous at a point in the given interval and yet has both absolute extrema.

This graph is not continuous at (x = c), yet it does have both an absolute maximum ((x = b)) and an absolute minimum ((x = c)). Also note that, in this case one of the absolute extrema occurred at the point of discontinuity, but it doesn’t need to. The absolute minimum could just have easily been at the other end point or at some other point interior to the region. The point here is that this graph is not continuous and yet does have both absolute extrema

The point of all this is that we need to be careful to only use the Extreme Value Theorem when the conditions of the theorem are met and not misinterpret the results if the conditions aren’t met.

In order to use the Extreme Value Theorem we must have an interval that includes its endpoints, often called a closed interval, and the function must be continuous on that interval. If we don’t have a closed interval and/or the function isn’t continuous on the interval then the function may or may not have absolute extrema.

We need to discuss one final topic in this section before moving on to the first major application of the derivative that we’re going to be looking at in this chapter.

Fermat’s Theorem

If (fleft( x ight)) has a relative extrema at (x = c) and (f'left( c ight)) exists then (x = c) is a critical point of (fleft( x ight)). In fact, it will be a critical point such that (f'left( c ight) = 0).

To see the proof of this theorem see the Proofs From Derivative Applications section of the Extras chapter.

Also note that we can say that (f'left( c ight) = 0) because we are also assuming that (f'left( c ight)) exists.

This theorem tells us that there is a nice relationship between relative extrema and critical points. In fact, it will allow us to get a list of all possible relative extrema. Since a relative extrema must be a critical point the list of all critical points will give us a list of all possible relative extrema.

Consider the case of (fleft( x ight) = ). We saw that this function had a relative minimum at (x = 0) in several earlier examples. So according to Fermat’s theorem (x = 0) should be a critical point. The derivative of the function is,

Sure enough (x = 0) is a critical point.

Be careful not to misuse this theorem. It doesn’t say that a critical point will be a relative extrema. To see this, consider the following case.

[fleft( x ight) = hspace<0.25in>hspace<0.25in>f'left( x ight) = 3]

Clearly (x = 0) is a critical point. However, we saw in an earlier example this function has no relative extrema of any kind. So, critical points do not have to be relative extrema.

Also note that this theorem says nothing about absolute extrema. An absolute extrema may or may not be a critical point.

Before we leave this section we need to discuss a couple of issues.

First, Fermat’s Theorem only works for critical points in which (f'left( c ight) = 0). This does not, however, mean that relative extrema won’t occur at critical points where the derivative does not exist. To see this consider (fleft( x ight) = left| x ight|). This function clearly has a relative minimum at (x = 0) and yet in a previous section we showed in an example that (f'left( 0 ight)) does not exist.

What this all means is that if we want to locate relative extrema all we really need to do is look at the critical points as those are the places where relative extrema may exist.

Finally, recall that at that start of the section we stated that relative extrema will not exist at endpoints of the interval we are looking at. The reason for this is that if we allowed relative extrema to occur there it may well (and in fact most of the time) violate Fermat’s Theorem. There is no reason to expect end points of intervals to be critical points of any kind. Therefore, we do not allow relative extrema to exist at the endpoints of intervals.

How do you exclude numbers in interval notation?

For example, take f(x)=x+2x&minus3. We can see that its domain is all real numbers except 3. In interval notation that is written (&minus&infin,3)&cup(3,&infin). It is not as easy to see what the the range must be.

One may also ask, what is an example of interval notation? Interval Notation. A notation for representing an interval as a pair of numbers. The numbers are the endpoints of the interval. Parentheses and/or brackets are used to show whether the endpoints are excluded or included. For example, [3, 8) is the interval of real numbers between 3 and 8, including 3 and excluding 8.

Correspondingly, how do you write all real numbers except 0 in interval notation?

A set including all real numbers except a single number. For example, we can express the set, 0>, using interval notation as, (&minus&infin, 0) &cup (0, &infin).

Examples

This section comprises some examples of problems involving inequalities that may arise

Solve the following inequalities 1) 3 x + 5 ≤ 6 x + 14

When solving inequalities, the final answer is sometimes required to be in interval notation. For this problem that is [ − 3 , ∞ )

2) Here we can solve each inequality individually, and x has to satisfy both inequalities. Thus, we have to solve − 3 < 2 x + 5 ਊnd  2 x + 5 ≤ 10 >2x+5leq 10>

For the first, we get -8 < 2x and -4 < x. For the last one we have 2 x ≤ 5 ਊnd  x ≤ 5 2 >xleq <2>>>

3) For this problem we need the numerator and denominator to both be positive or both be negative. So we want to solve when 3x - 5 > 0. Notice that we do not include 3x - 5 = 0 since we cannot divide by 0. Solving this inequality we find x > 5 3 <3>>> . In interval notation we have ( 5 3 , ∞ ) <3>>,infty )>

How to Write the Interval Notation of Solutions to Inequalities:

Step 1. Graph the solution set on the number line. Use an open dot () at the boundary point/s excluded in the solution. Use a closed dot (●) at the boundary point/s included in the solution.

Step 2. Write the interval notation beginning with the lower-bound then the upper-bound. Use square brackets ([ ]) to indicate the inclusion of the boundaries in the solution, or parentheses ( ) to indicate their exclusion.

NOTE: Infinities (-∞, +∞) are always enclosed with a parenthesis. Also, solution sets can have one square bracket and one parenthesis on either side, depending on the inclusion or exclusion of the boundaries.

For example, look at the image below.

x > 4
Solution:
Step 1.
Graph the solution. Use an open dot at 4 and shade all real numbers greater than 4. Put positive infinity above to indicate that the solution set is unbounded to the right of the number line (or all positive real numbers).
Step 2. Write the interval notation using a parenthesis because the lower-bound (4) is not included, then put positive infinity as the upper-bound, which is automatically enclosed with a parenthesis.
Interval Notation: (4, +∞)

x ≤ 4
Solution:
Step 1. Graph the solution. Use a closed dot at 4 and shade all real numbers below 4, including 4. Put negative infinity above to indicate that the solution set is unbounded to the left of the number line (or all negative real numbers).
Step 2. Write the interval notation. Use a parenthesis with the lower-bound (-∞) and a square bracket with the upper-bound (4).
Interval Notation: (-∞, 4]

Brief Introduction to Algebra Online

Interval Notation

Have your youngsters asked you for help on their algebra research, and you haven’t done algebra because Mrs. Tabais 6th period algebra course in senior high school or you have a significant algebra examination showing up and also you cannot master it. Well, do not stress out excessive, as you can now discover algebra worksheets, algebra calculators as well as popular algebra solvers on the internet, which will certainly assist you through the difficult learning procedure.

An algebra worksheet is an excellent method to develop your math skills, and also a way for a future mathematics test, or get some beneficial algebra tips. Algebra worksheets typically consist of hundreds of troubles and equations that you can utilize to check on your own. Generally, the site supplying the algebra worksheets will certainly grade your answers for you, or offer a response secret.

For algebra software application tools that will aid fix algebra formulas, algebra calculators might be the answer you are looking for.

Survey of Algebra and Probability: MAT 101

When we solve an equation we find a single value for our variable. With inequalities we will give a range of values for our variable. To do this we will not use an equals sign, but one of the following symbols:

WeBWorK: Entering Inequality Symbols.

Type the two symbols together:

>= for (geq) (greater than or equal to)

<= for (leq) (less than or equal to)

The expression (x lt 4) this means our variable (x) can be any number smaller than (4) such as (-2, 0, 3, 3.9) or even (3.999999999) as long as it is smaller than (4 ext<.>) In other words, (x lt 4) is the set of all numbers less than (4 ext<.>) 4 is NOT less than 4. We write (4 less 4 ext<.>) However (4) IS less than or equal to (4 ext<.>) We write (4leq 4 ext<.>)

The expression (y geq -2,) means that the variable (y) can be any number greater than or equal to (-2,) such as (5, 0,-1,-1.9999,) or even (-2 ext<.>) In other words, (x geq -2) is the set of all numbers greater than or equal to (-2 ext<.>)

It is often useful to draw a picture of the solutions to the inequality on a number line. We will start from the value in the problem and bold the lower part of the number line if the variable is smaller than the number, and bold the upper part of the number line if the variable is larger. The value itself we will mark with an open or closed circle: open for less than or greater than, and a closed circle for less than or equal to or greater than or equal to.

Once the graph is drawn we can quickly convert the graph into what is called . Interval notation gives two numbers, the first is the smallest value (furthest left on the number line), the second is the largest value (furthest right on the number line). We will use square brackets if the inequality includes or equal to (so either (leq) or (geq)). We will use round brackets if the inequality is strictly less than or greater than (so either (lt) or (gt)). If there is no largest value, we can use (infty) (infinity). If there is no smallest value, we can use (-infty) (negative infinity). If we use either positive or negative infinity we will always use a round bracket by the symbol.

Example 2.B.1 . Relating an Inequality, Graph and Interval.

Graph the inequality (xgeq 4) and give the interval notation.

Start at (4) and shade to the right. Use a closed circle for greater than or equal to.

Interval notation: ([4,infty ) checkmark)

WeBWorK: Entering Intervals.

Type [4,inf) for the interval ([4,infty ) ext<.>)

Example 2.B.2 . Relating an Inequality, Graph and Interval.

Graph the inequality (xlt -4) and give the interval notation.

Start at (-4) and shade to the left. Use an open circle for less than.

Interval notation: ((-infty, -4) checkmark)

WeBWorK: Entering an Infinity Symbol.

Type (-inf,-4) for the interval ((-infty, -4) ext<.>)

Example 2.B.3 . Relating an Inequality, Graph and Interval.

Graph the inequality (-3lt x lt 1) and give the interval notation.

Start at (-3) and shade to the right to (1 ext<.>) Use open circles on both ends for less than.

Interval notation: ((-3, -1) checkmark)

Subsection 2.B.2 Linear Inequalities

Solving inequalities is very similar to solving equations with one exception. To understand the exception, consider the tools we use to solve an equation: add/subtract, multiply/divide numbers to both sides of the equation to isolate the variable. We consider the inequality (1 lt 3) and notice what happens to the inequality sign as we add, subtract, multiply and divide by both positive and negative numbers.

Interval notation

Mathematicians frequently want to talk about intervals of real numbers such as “all real numbers between (1) and (2) ”, without mentioning a variable. As an example, “The range of the function (f:xmapsto sin x) is all real numbers between (-1) and (1) ”.

A compact notation often used for these intervals of real numbers is as follows:

((1,2)) means all real numbers between (1) and (2) , excluding the endpoints

([1,2]) means all real numbers between (1) and (2) , including the endpoints

We can also write these intervals using set notation as () and () respectively.

If needed, we can also mix the two types of bracket, so ((1,2]) means the interval () and ([1,2)) means () .

The interval “all real numbers greater than (-5) ” is written as ((-5,infty)) , and “all real numbers less than or equal to (7) ” is written as ((-infty,7]) . This does not mean that (infty) is a number it is just a convenient shorthand.

Although the notation ((1,2)) is exactly the same as the notation for coordinates, the two are rarely confused because the context will make it clear which is meant.