4: Inferential Theory

Step 1 – How likely is it that a particular data value is success?

Suppose a researcher wanted to determine the proportion of the public that believe the owner is responsible for the accident. The researcher has a hypothesis that the proportion is over 60%. In this case, the hypotheses will be:

• \(H_0: p = 0.60\)
• \(H_1: p > 0.60\)

When collecting data, the order in which the units or people are selected determines the order in which the data is collected. In this case, assigning responsibility to the owner will be considered a success and assigning responsibility to the manufacturer is considered a failure. If the first person believes the owner is responsible, the second person believes the manufacturer is responsible, the third person selects the manufacturer, the fourth, fifth and sixth people all select the owner as the responsible party, then we can convert this information to successes and failure by listing the order in which the data were obtained as SFFSSS.

The strategy that is employed to determine which of two competing hypotheses is better supported is always to assume that the null hypothesis is true. This is a critical point, because if we can assume a condition is true, then we can determine the probability of getting our particular sample result, or more extreme results. This is a p-value.

However, before we can determine the probability of obtaining a sequence such as SFFSSS, we must first find the probability of obtaining a success. For this, we need to explore the concept of probability.

If we assume that the null hypothesis is true, then the proportion of people who believe the owner is responsible is 0.60. What does that mean? It means that if there are exactly 100 people, then exactly 60 of them hold the owner responsible and 40 of them do not.

If our goal is to find the probability of SFFSSS, then we must first find the probability of getting a success (owner). If there are 100 people in the population and 60 of these select the owner, then the probabiltiy of selecting a person who chooses the owner is \(P(owner) = \dfrac{60}{100} = 0.60\). Notice that this probability exactly equals the proportion defined in the null hypothesis. This is not a coincidence and it will happen every time because the proportion in the null hypothesis is used to generate a theoretical population, which was used to find the probability. The first important step in the process of testing a hypothesis is to realize that the probability of any data being a success is equal to the proportion defined in the null hypothesis, assuming that sampling is done with replacement, or that the population size is extremely large so that removing a unit from the population does not change the probability a significant amount.

Step 2 - How likely is it that a particular data value is a failure?

Now that we know how to find the probability of success, we must find the probability of failure. For this, we will again digress to the rules of probability.

Recall that the original hypothesis for the responsibility in an accident is that: H0: p = 0.60. We have established that the probability of success is 0.60. The probability of failure is 0.40 since it is the complement of the probability of success and 1 – 0.60 = 0.40.

Step 3 - How likely is it that a sample consists of a specific sequence of successes and failures?

We now know that the probability of success is identical to the proportion defined by the null hypothesis and the probability of failure is the complement. But these probabilities apply to only one selection. What happens when more than one is selected? To find that probability, we must learn the last of the probability rules:

\[P(A\ and\ B) = P(A)P(B)\]

We are now ready to answer the question of what is the probability that we would get the exact sequence of people if the first person believes the owner is responsible, the second person believes the manufacturer is responsible, the third person selects the manufacturer, the fourth, fifth and sixth people all select the owner as the responsible party. Because there are six people selected, then this is an “and” problem and can be written as P(S and F and F and S and S and S) or more concisely, leaving out the word “and” but retaining it by implication, we write P(SFFSSS). Remember that P(S) = 0.6 and P(F) = 0.4

P(SFFSSS) = P(S)P(F)P(F)P(S)P(S)P(S) = (0.6)(0.4)(0.4)(0.6)(0.6)(0.6) = 0.0207.

To summarize, if the null hypothesis is true, then 60% of the people believe the owner is responsible for accidents. Under these conditions, if a sample of six people is taken, with replacement, then the probability of getting this exact sequence of successes and failures is 0.0207.

Step 4 - How likely is it that a sample would contain a specific number of successes?

Knowing the probability of an exact sequence of successes and failures is not particularly important by itself. It is a stepping-stone to a question of greater importance – what is the probability that four out of six randomly selected people (with replacement) will believe the owner is responsible? This is an important transition in thinking that is being made. It is the transition from thinking about a specific sequence to thinking about the number of successes in a sample.

When data are collected, researchers don’t care about the order of the data, only how many successes there were in the sample. We need to find a way to transition from the probability of getting particular sequence of successes and failures such as SFFSSS to finding the probability of getting four successes from a sample of size 6. This transition will make use of the commutative property of multiplication, the P(A or B) rule and combinatorics (counting methods).

At the end of Step 3 we found that P(SFFSSS) = 0.0207.

• What do you think will be the probability of P(SSSFSF)?
• What do you think will be the probability of P(SSSSFF)?

The answer to both these questions is 0.0207 because all of these sequences contain 4 successes and two failures and since the probability is found by multiplying the probabilities of success and failure in sequence and since multiplication is commutative (order doesn’t matter) then

(0.6)(0.4)(0.4)(0.6)(0.6)(0.6)= (0.6)(0.6)(0.6)(0.4)(0.6)(0.4)= (0.6)(0.6)(0.6)(0.6)(0.4)(0.4) = 0.020736.

If the question now changes from what is the probability of a sequence to what is the probability of 4 successes in a sample of size 6, then we have to consider all the different ways in which four successes can be arranged. We could get 4 successes if the sequence of our selection is SFFSSS or SSSFSF or SSSSFF or numerous other possibilities. Because we are sampling one time and because there are many possible outcomes we could have, this is an “or” problem that uses an expanded version of the formula P(A or B) = P(A) + P(B). This can be written as:

P(4 out of 6) = P(SFFSSS or SSSFSF or SSSSFF or ...) = P(SFFSSS) + P(SSSFSF) + P(SSSSFF) + ...

In other words, we can add the probability of each of these orders. However, since the probability of each of these orders is the same (0.0207) then this process would be much quicker if we simply multiplied 0.0207 by the number of orders that are possible. The question we must answer then is how many ways are there to arrange four successes and 2 failures? To answer this, we must explore the field of combinatorics, which provide various counting methods.

Recall that 6 people were selected and 4 thought the owner should be responsible. We saw that the probability of any sequence of 4 successes and 2 failures, such as SFFSSS or SSSFSF or SSSSFF is 0.0207 if the null hypothesis is p = 0.60. If we knew the number of combinations of these 4 successes and 2 failures, we could multiply that number times the probability of any specific sequence to get the probability of 4 successes in a sample of size 6.

Using the formula for nCr, we get: \(_6C_4 = \dfrac{6!}{(6 - 4)!4!} = 15\) combinations.

Therefore, the probability of 4 successes in a sample of size 6 is 15*0.020736 = 0.31104. This means that if the null hypothesis of p=0.60 is true and six people are asked, there is a probability of 0.311 that four of those people will believe the owner is responsible.

We are now ready to make the transition to distributions. The following table summarizes our journey to this point.

Step 5 – How can the exact p-value be found using the binomial distribution?

Recall that in chapter 2, we determined which hypothesis was supported by finding the p-value. If the p-value was small, less than the level of significance, we concluded that the data supported the alternative hypothesis. If the p-value was larger than the level of significance, we concluded that the data supported the null hypothesis. The p-value is the probability of getting the data, or more extreme data, assuming the null hypothesis is true.

In Step 4 we found the probability of getting the data (for example, four successes out of 6) but we haven’t found the probability of getting more extreme data yet. To do so, we must now create distributions. A distribution shows the probability of all the possible outcomes from an experiment. For categorical data, we make a discrete distribution.

Before looking at the distribution that is relevant to the problem of responsibility for accidents, a general discussion of distributions will be provided.

In chapter 4 you learned to make histograms. Histograms show the distribution of the data. In chapter 4 you also learned about means and standard deviations. Distributions have means and standard deviations too.

To demonstrate the concepts, we will start with a simple example. Suppose that someone has two routes used for running. One route is 2 miles long and the other route is 5 miles long. Below is the running schedule for last week.

A distribution of the amount run each day is shown below.

The mean can be found by adding all the daily runs and dividing by 7. The mean is 3.286 miles per day. Because the same distances are repeated on different days, a weighted mean can also be used. In this case, the weight is the number of times a particular distance was run. A weighted mean takes advantage of multiplication instead of addition. Thus, instead of calculating: \(\dfrac{2 + 2 + 2 + 2 + 5 + 5 + 5}{7} = 3.286\). we can multiply each number by the number of times it occurs then divide by the number of occurrences: \(\dfrac{4 \cdot 2 + 3 \cdot 5}{4 + 3} = 3.286\). The formula for a weighted mean is:

\[\dfrac{\sum w \cdot x}{\sum w}\]

The same graph is presented below, but this time there are percentages above the bars.

Instead of using counts as the weight, the percentages (actually the proportions) can be used as the weight. Thus 57.143% can be written as 0.57143. Likewise, 42.857% can be written 0.42857.

Substituting into the formula gives: \(\dfrac{0.57143 \cdot 2 + 0.42857 \cdot 5}{0.57143 + 0.42857} = 3.286\). Notice that the denominator adds to 1. Therefore, if the weight is the proportion of times that a value occurs, then the mean of a distribution that uses percentages can be found using the formula:

\[\mu = \sum P(x) x\]

This mean, which is also known as the expected value, is the sum of the probability of a value times the value. There is no need for dividing, as is customary when finding means, because we would always just divide by 1.

Recall from chapter 4 that the standard deviation is the square root of the variance. The variance is \(\sigma^2 = \sum[(x - \mu)^2 \cdot P(x)]\). The standard deviation is \(\sigma = \sqrt{\sum[(x - \mu)^2 \cdot P(x)]}\).

\(\sigma = \sqrt{\sum[(x - \mu)^2 \cdot P(x)]}\).

\(\sigma = \sqrt{(2 - 3.286)^2 \cdot 0.57143 + (5 - 3.286)^2 \cdot 0.42857} = 1.485\)

The self-driving car problem will show us one way in which we encounter discrete distributions. In fact, it will result in the creation of a special kind of discrete distribution called the Binomial distribution, which will be defined after exploring the concepts that lead to its creation.

When testing a hypothesis about proportions of successes, there are two random variables that are of interest to us. The first random variable is specific to the data that we will collect. For example, in research about who is responsible for accidents caused by autonomous cars, the random variable would be “responsible party”. There would be two possible values for this random variable, owner or car manufacturer. We have been considering the owner to be a success and the manufacturer to be a failure. The data the researchers collect is about this random variable. However, creating distributions and finding probabilities and p-values requires a shift of our focus to a different random variable. This second random variable is about the number of successes in a sample of size n. In other words, if six people are asked, how many of them think the owner is responsible? It is possible that none of them think the owner is responsible, or one thinks the owner is responsible, or two, or three, or four, or five, or all six think the owner is responsible. Therefore, in a sample of size 6, the random variable for the number of successes can have the values of 0,1,2,3,4,5,6. We have already found that the probability of getting four successes is 0.3110. We will now find the probability of getting 0,1,2,3,5,6 successes, assuming the hypotheses are still \(H_0: p = 0.60\), \(H_1: p > 0.60\). This will allow us to create the discrete binomial distribution of all possible outcomes.

Find the probability of 0 successes

The only way to have 0 successes is to have all failures, thus we are seeking P(FFFFFF).

P(FFFFFF) = P(F)P(F)P(F)P(F)P(F)P(F) = (0.4)(0.4)(0.4)(0.4)(0.4)(0.4) = 0.004096.

Since there is only one combination for 0 successes, then the probability of 0 successes is 0.0041.

Find the probability of 1 success

We know that all combinations have the same probability, so we may as well create the simplest combination for 1 success. This would be SFFFFF.

P(SFFFFF) = P(S)P(F)P(F)P(F)P(F)P(F) = (0.6)(0.4)(0.4)(0.4)(0.4)(0.4) = \((0.6)^{1}(0.4)^{5}\)= 0.006144.

How many combinations are there for 1 success? This can be found using \(_6C_1\).

\(_6C_1 = \dfrac{6!}{(6 - 1)!1!} = 6\) combinations. Does this answer seem reasonable? Consider there are only 6 places in which the success could happen.

The probability of 1 success is then 6*0.006144 = 0.036864 or if we round to four decimal places, 0.0369.

Find the probability of 2 successes.

Instead of doing this problem in steps as was done for the prior examples, it will be demonstrated by combining steps.

\(_6C_2 P\)(SSFFFF)

\(\dfrac{6!}{(6 - 2)! 2!} (0.6)(0.6)(0.4)(0.4)(0.4)(0.4) = \dfrac{6!}{(6 - 2)! 2!} (0.6)^2(0.4)^4 =\)

15(0.009216) = 0.13824 or with rounding to four decimal places 0.1382.

Find the probability of 3 successes using the combined steps. (Now its your turn).

Find the probability of 5 successes.

Find the probability of 6 successes.

When all the probabilities have been found, we can create a table that shows the values the random variable can take and their probabilities. We will define the random variables for the number of successes as X with the possible values defined as x.

Do your values for 3,5, and 6 successes agree with the values found in the table?

A graph of this distribution can lead to a better understanding of it. This graph is called a probability mass function, which is shown using a stick graph. It is a way to graph discrete distributions, since there cannot be any values in between the numbers on the x-axis. The heights of the bar correspond to the probability of getting the number of successes.

There are three things you should notice about this distribution. First, it is a complete distribution. That is, in a sample of size six, it is only possible to have 0,1,2,3,4,5, or 6 successes and all of those have been included in the graph. The second thing to notice is that all the probabilities have values between 0 and 1. This should be expected because probabilities must always be between 0 and 1. The final thing to notice, which may not be obvious at first, is that if you add all the probabilities, the sum will be 1. The sum of all complete probability distributions should equal 1, \(\sum P(x) =1\). If you add all the probabilities and they don’t equal one, but are very close, it could be because of rounding, not because you did anything wrong.

Now that you have the ability to create a complete binomial distribution, you are ready to test a hypothesis. This will be demonstrated with the autonomous car example that has been used throughout this chapter.

Suppose a researcher wanted to determine the proportion of people who believe the owner is responsible. The researcher may have had a hypothesis that the proportion of people who believe the owner is responsible for accidents is over 60%. In this case, the hypotheses will be: H0: p = 0.60 and H1: p > 0.60. The level of significance will be 0.10 because only a small sample size will be used. In this case, the sample size will be 6.

With this sample size we have already seen what the binomial distribution will be like. We also know that the direction of the extreme is to the right because the alternative hypothesis uses a greater than symbol.

The researcher randomly selects 6 people. Of these, four say the owner is responsible. Which hypothesis is supported by this data?

The p-value is the probability the researcher would get four or more people claiming the owner is responsible. From the table we created earlier, we see the probability of getting 4 people who think

the owner is responsible is 0.3110, the probability of getting 5 is 0.1866 and the probability of getting 6 is 0.0467. If we add these together, we find the probability of getting 4 or more is 0.5443. Since this probability is larger than our level of significance, we conclude the data supports the null hypothesis and is therefore not significant. The conclusion that would be written is: At the 0.10 level of significance, the proportion of people who think the owner is responsible is not significantly greater than 0.60 (\(x = 4\), \(p = 0.5443\), \(n = 6\)). Remember that in statistical conclusions, the p is the p-value, not the sample proportion.

In this example, the data were not significant and so the researcher could not claim the proportion of people who think the owner is responsible is greater than 0.60. A sample size of 6 is very small for categorical data and therefore it is difficult to arrive at any significant results. If the data are changed so that instead of getting 4 out of 6 people, the researcher gets 400 out of 600, does the conclusion change? Use 1 – binomcdf(600,0.6,399) to find the p-value for this situation.

1 – binomcdf(600,0.6,399) = ____________________

Write the concluding sentence:

Step 6 - How can the approximate p-value be found using the normal approximation to the binomial distribution?

When a hypothesis is tested using the binomial distribution, an exact p-value is found. It is exact because the binomial distribution is created from every combination of successes and failures that is possible for a sample of size n. There are other methods for determining the p-value that will give an approximate p-value. In fact, the typical method that is used to test hypotheses about proportions will give an approximate p-value. You may wonder why a method that gives an approximate p-value is used instead of the method that gives an exact p-value. This will be explained after the next two methods have been demonstrated. Before these can be demonstrated, we need to learn about a different distribution called the normal distribution.

Return to Step 6: Apply this concept of probability to the hypothesis about the responsibility for a self-driving car in an accident.

Remember that the hypotheses for the autonomous car problem are: \(H_0: p = 0.60\), \(H_1: p > 0.60\). In the original problem, the researcher found that 4 out of 6 people thought the owner was responsible. Which hypothesis does this data support if the level of significance is 0.10?

This hypothesis test will be done using a method called the Normal Approximation to the Binomial Distribution.

The first step is to find the mean and standard deviation of the binomial distribution (which was done earlier but is now repeated):

\(\mu = np = 6(0.6) = 3.6\)

\(\sigma = \sqrt{npq} = \sqrt{6(0.6)(0.4)} = 1.2\)

Draw and label a normal curve with a mean of 3.6 and a standard deviation of 1.2.

Find the z score if the data is 4.

\(z = \dfrac{x - \mu}{\sigma}\) \(z = \dfrac{4 - 3.6}{1.2} = 0.33\)

From the table, the area to the left is \(A_L = 0.6255\). Since the direction of the extreme is to the right, subtract the area to the left from 1 to get \(A_R = 0.3745\). This is the p-value.

Since this value is greater than the level of significance, if the calculator generated p-value is used, the conclusion will be written as: At the 0.10 level of significance, the proportion of people who think the owner is responsible is not significantly more than 0.60 (z = 0.33, p = 0.3694, n = 6).

Let us now take a moment to compare the p-value from the Normal Approximation to the Binomial Distribution (0.3694) to the exact p-value found using the Binomial Distribution (0.5443). While these p- values are not very close to each other, the conclusion that is drawn is the same. The reason they are not very close is because a sample size of 6 is very small and the normal approximation is not very good with a small sample size.

Test the hypothesis again if the researcher finds that 400 out of 600 of the people believe the owner is responsible for accidents.

\(\mu = np = 600(0.6) = 360\) This indicates that if lots of samples of 600 people were sampled the average number of people who think the owner is responsible would be 360.

\(\sigma = \sqrt{npq} = \sqrt{600(0.6)(0.4)} = 12\)

Draw a label a curve with a mean of 360 and a standard deviation of 12.

Find the z score if the data is 400.

\(z = \dfrac{x - \mu}{\sigma}\) \(z = \dfrac{400 - 360}{12} = 3.33\)

Using the table, the area to the left is \(A_L = 0.9996\). Since the direction of the extreme is to the right, subtract the area to the left from 1 to get \(A_R = 0.0004\). More precisely, it is 0.000430.

This time when the results of the Normal Approximation to the Binomial Distribution (0.000430) are compared to the results of the binomial distribution (0.000443), they are very close. This is because the sample size is larger.

In general, if \(np \ge 5\) and \(nq \ge 5\), then the normal approximation makes a good, but not perfect estimate for the binomial distribution. When a sample of size 6 was used, \(np = 3.6\) which is less than 5. Also, \(nq = 6(0.4) = 2.4\), which is less than 5, too. Therefore, using the normal approximation for samples that small is not a good strategy.

Step 7 – Find the approximate p-value using the Sampling Distribution of Sample Proportions

Up to this point the discussion has been about the number of people. When sampling that produces categorical data is done, these numbers or counts can also be represented as proportions by dividing the number of successes by the sample size. Thus, instead of the researcher saying that 4 out of 6 people believe the owner is responsible, the researcher could say that 66.7% of the people believe the owner is responsible. This leads to the concept of looking at proportions rather than counts which means that instead of the distribution being made up of the number of successes, represented by x, it is made up of the sample proportion of successes represented by \(\hat{p}\).

Remember that the hypotheses for the people who think the owner is responsible are: \(H_0: p = 0.60\), \(H_1: p > 0.60\). In the original problem, the researcher found that 4 out of 6 people think the owner is responsible. Which hypothesis does this data support if the level of significance is 0.10?

Since \(\mu_{\hat{p}} = p\) then the mean is 0.60 (from the null hypothesis).

Since \(\sigma_{\hat{p}} = \sqrt{\dfrac{p(1-p)}{n}} = \sqrt{\dfrac{0.6(0.4)}{6}} = 0.2\) then the standard deviation is 0.2.

Draw a label a normal curve with a mean of 0.6 and a

standard deviation of 0.2.
If the data is 4, then the sample proportion,

\(\hat{p} = \dfrac{x}{n} = \dfrac{4}{6} = 0.6667\)

Find the z score if the data is 4.

\(z = \dfrac{\hat{p} - p}{\sqrt{\dfrac{p(1-p)}{n}}}\) \(z = \dfrac{0.6667 - 0.6}{0.2} = 0.33\)

The area to the left is \(A_L = 0.6304\). Since the direction of the extreme is to the right, subtract the area to the left from 1 to get \(A_R = 0.3696\).

Compare this result to the result found when using the Normal Approximation to the Binomial Distribution. Notice that both results are exactly the same. This should happen every time, provided there isn’t any rounding of numbers. The reason this has happened is because the number of successes can be represented as counts or proportions. The distributions are the same, although the x-axis is labeled differently. Divide the z scores for the normal approximation by the sample size 6 and you will get the z scores for the sampling distribution.

Test the hypothesis again if the researcher finds that 400 out of 600 of the people believe the owner is responsible.

Since \(\mu_{\hat{p}} = p\) then the mean is 0.60 (from the null hypothesis).

Since \(\sigma_{\hat{p}} = \sqrt{\dfrac{p(1-p)}{n}} = \sqrt{\dfrac{0.6(0.4)}{600}} = 0.02\) then the standard deviation is 0.02.

If the data is 400, then the sample proportion, \(\hat{p} = \dfrac{x}{n} = \dfrac{400}{600} = 0.66667\)

\(z = \dfrac{\hat{p} - p}{\sqrt{\dfrac{p(1-p)}{n}}}\) \(z = \dfrac{0.66667 - 0.6}{0.02} = 3.33\)

The area to the left is \(A_L = 0.9996\). Since the direction of the extreme is to the right, subtract the area to the left from 1 to get \(A_R = 0.0004\). More precisely, it is 0.000430.

Conclusion for testing hypotheses about categorical data.

By this time, many students are wondering why there are three methods and why the binomial distribution method isn’t the only one that is used since it produces an exact p-value. One justification of using the last method is comparing the results of surveys or other data. Imagine if one news organization reported their results of a survey as 670 out of 1020 were in favor while another organization reported they found 630 out of 980 were in favor. A comparison between these would be difficult without converting them to proportions, therefore, the third method, which uses proportions, is the method of choice. When the sample size is sufficiently large, there is not much difference between the methods. For smaller samples, it may be more appropriate to use the binomial distribution.

Making inferences using quantitative Data

The strategy for making inferences with quantitative data uses sampling distributions in the same way that they were used for making inferences about proportions. In that case, the normal distribution was used to model the distribution of sample proportions, pˆ . With quantitative data, we find the mean, therefore the normal distribution will be used to model the distribution of sample means, x .

To demonstrate this, a small population will be used. This population consists of the 50 states of the United States plus the District of Columbia and the 5 US territories of American Samoa, Guam, Northern Mariana Islands, Puerto Rico and the Virgin Islands, each of which has one representative in Congress with limited voting authority. A histogram showing the distribution of the number of representatives in a state or territory is provided. On the graph is a normal distribution based on the mean of this population being 7.875 representatives and a standard deviation of 9.487. The distribution is positively skewed and cannot be modeled by the normal curve that is on the graph.

Be aware that in reality, the mean and standard deviation, which are parameters, are not known and so we would normally write hypotheses about them. However, for this demonstration, a small population with a known mean and standard deviation are necessary. With this, it is possible to illustrate what happens when repeated samples of the same size are drawn from this population, with replacement, and the means of each sample are found and becomes part of the distribution of sample means.

A sampling distribution of sample means (a distribution of x ) contains all possible sample means that in theory could be obtained if a random selection process was used, with replacement. The number of possible sample means can be found using the fundamental rule of counting. Draw a line to represent each state/territory that would be selected. On the line write the number of options, so that it would look like this:

If our sample size is 40, then there are \(56^{40}\) possible samples that could be selected which is equal to \(8.46 \times 10^{69}\). That is a lot of possible samples. For this demonstration, only 10,000 samples of size 40 will be taken. The distribution of these sample means when this was done is shown in the histogram below.

The mean of all these sample means is 7.8748 and the standard deviation is 1.502. Notice that the mean of all these sample means is almost exactly the same as the mean of the original population. Also notice that the standard deviation of all these sample means is much smaller than the standard deviation of the population. This is summarized in the table below.

The following graph has both the original data and the sample means on it. Notice how the two normal curves are centered at approximately the same place but the curve for the sample means is narrower. This shows that when samples of sufficient size are taken from any population, the means of those samples will be close to the means of the population.

We are now ready to discuss the Central Limit Theorem. This theorem states that for any set of quantitative data with a mean μ and a standard deviation σ, the mean of all possible sample means will equal the mean of the population. The standard deviation of all the sample means, which is also called the standard error, will equal the standard deviation of the population divided by the square root of n. These are shown as:

\[\mu_{\bar{x}} = \mu\]

and

\[\sigma_{\bar{x}} = \dfrac{\sigma}{\sqrt{n}}.\]

It also says the distribution of sample means will be normal if the sample size is sufficiently large (generally considered to be 30 or more). If the original population is normally distributed, than the distribution of sample means will be normally distributed for any sample size.

Before doing an example, it will be important to see the effect of sample sizes. Compare the following 4 graphs that show the distribution of sample means for samples of size 40, 30, 20, and 10.

Notice how the distributions become more skewed as the sample size decreases. Notice also that the mean of the sample means are still approximately equal to the mean of the population but the standard deviations get larger as the sample size gets smaller. This implies there is more variation in sample means with small sample sizes than with large sample sizes.

When making inferences about quantitative data, the basic \(z\) formula \(z = \dfrac{x − μ}{\sigma}\) can now be rewritten knowing that in a distribution of sample means, the results of the sample that have formerly been represented with \(x\) can now be represented with \(\bar{x}\). The mean, \(\mu\) will still be represented with \(\mu\), since \(\mu_{\bar{x}} = \mu\) and the standard deviation \(\sigma\) can now be represented with \(\dfrac{\sigma}{\sqrt{n}}\) since \(\sigma_{\bar{x}} = \dfrac{\sigma}{\sqrt{n}}\)

Therefore, for the sampling distribution of sample means, the \(z\) formula, \(z = \dfrac{x − μ}{\sigma}\) becomes

\[z = \dfrac{\bar{x} - \mu}{\dfrac{\sigma}{\sqrt{n}}}.\]

It is now time to use the central limit theorem to test a hypothesis.