CLA 1 Paper - Data Analysis & Business Intelligence

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One-Sample Tests of Hypothesis

Chapter 10

10-1

In this chapter, we develop a procedure to test the validity of a statement about a population parameter. We begin by defining a hypothesis and hypothesis testing. Next, the steps in hypothesis testing are outlined. The we conduct tests of hypothesis testing for means. Finally, we describe possible errors due to sampling in hypothesis testing.

Learning Objectives

LO10-1 Explain the process of testing a hypothesis

LO10-2 Apply the six-step procedure for testing a hypothesis

LO10-3 Distinguish between a one-tailed and a two- tailed test of hypothesis

LO10-4 Conduct a test of a hypothesis about a population mean

LO10-5 Compute and interpret a p-value

LO10-6 Use a t-statistic to test a hypothesis

LO10-7 Compute the probability of a Type II error

10-2

Hypothesis Testing

Hypothesis testing begins with a hypothesis statement about a population parameter

Examples

The mean speed of automobiles passing milepost 150 on the West Virginia Turnpike is 68 mph

The mean cost to remodel a kitchen is $20,000

HYPOTHESIS A statement about a population parameter subject to verification

10-3

Here are examples of statements we might want to test.

Hypothesis Testing (2 of 3)

The objective of hypothesis testing is to verify the validity of a statement about a population parameter

HYPOTHESIS TESTING A procedure based on sample evidence and probability theory to determine whether the hypothesis is a reasonable statement.

10-4

There are six steps in the hypothesis testing procedure. We will discuss each of the steps in detail.

Step 1 of the Six-Step Process

State the null hypothesis (H0) and the alternate hypothesis (H1)

The null hypothesis always includes the equal sign

For example; =, ≥, or ≤ will be used in H0

The alternate hypothesis never includes the equal sign

For example; ≠, <, or > is used in H1

NULL HYPOTHESIS A statement about the value of a population parameter developed for the purpose of testing numerical evidence.

ALTERNATE HYPOTHESIS A statement that is accepted if the sample data provide sufficient evidence that the null hypothesis is false.

10-5

The null hypothesis is a statement that is not rejected unless our sample data provide convincing evidence that it is false. Failing to reject the null hypothesis does not prove that H0 is true; it means we have failed to disprove H0. H1 is what you will conclude if you reject the null hypothesis. We turn to the alternate hypothesis only if the data suggest the null hypothesis is untrue.

Step 2 of the Process

LEVEL OF SIGNIFICANCE The probability of rejecting the null hypothesis when it is true.

10-6

Next, you select the level of significance,

Sometimes called the level of risk

Can be any value between 0 and 1

Traditionally,

.05 is used for consumer research projects

.01 for quality assurance

.10 for political polling

There is not one level of significance that is applied to all tests; the researcher must decide on the level of significance before formulating the decision rule or collecting the sample. It is risk you take of rejecting the null hypothesis when it is really true. It is designated with the Greek letter alpha, .

Possible Error in Hypothesis Testing

Since the researcher cannot study every item or individual in the population, error is possible

Type I error is designated with the Greek letter alpha,

Type II error is designated with the Greek letter beta,

TYPE I ERROR Rejecting the null hypothesis, H0, when it is true.

TYPE II ERROR Not rejecting the null hypothesis when it is false.

10-7

There are two types of error that can occur; a Type I error and a Type II error. We choose and we can calculate . While it is not likely that a Type II error occurs, it is possible due to the process of random sampling from a population. Here is a summary of the decisions a researcher could make and the possible consequences.

Step 3 of the Process

Then, select the test statistic

In hypothesis testing for the mean, , when is known, the test statistic z is computed with the following formula

We can determine whether the distance between and is statistically significant by finding the number of standard deviations is from

TEST STATISTIC A value, determined from sample information, used to determine whether to reject the null hypothesis.

10-8

There are many test statistics. In this chapter we use z and t as the test statistics, based on whether or not we know the population standard deviation. Use formula (10-1) when you know the population standard deviation.

Step 4 of the Process

Formulate the decision rule

The decision rule is a statement of specific conditions under which the null hypothesis is rejected and the conditions under which it is not rejected

The region or area of rejection defines the location of all the values that are either so large or so small that their probability of occurrence under a true null hypothesis is remote

CRITICAL VALUE The dividing point between the region where the null hypothesis is rejected and the region where it is not rejected.

10-9

See the next slide for an explanation of the critical value.

Critical Value

The sampling distribution of the statistic z follows the normal distribution

Here, an of .05 is used in a one-tailed test

The value 1.645 separates the regions where the null hypothesis is rejected and where it is not rejected

The value 1.645 is the critical value

10-10

A one-tailed test (right-tailed) is used in this chart; this and how we obtain the critical value will be covered in more detail later. The area where the null hypothesis is not rejected is to the left of 1.645 and the area of rejection is to the right of 1.645.

Steps 5 & 6 of the Six-Step Process

Step 5 Make a decision

First, select a sample and compute the value of the test statistic

Compare the value of the test statistic to the critical value

Then, make the decision regarding the null hypothesis

Step 6 Interpret the results

What can we say or report based on the results of the statistical test?

10-11

If the results of the test result in a decision not to reject the null hypothesis, we can say that based on the sample data, the difference between the sample mean and the hypothesized population mean was not large enough to reject the null hypothesis. On the other hand, if the results of the test result in a decision to reject the null hypothesis, you have disproved the null hypothesis with a stated probability of a Type I error, . So, there is a small probability that the decision to reject the null hypothesis was an error due to random sampling.

Hypothesis Testing (3 of 3)

10-12

One-Tailed and Two-Tailed Tests

H0: ≥ 60,000 miles

H1: < 60,000 miles

with an

Left-tailed test

H0: = $65,000 per year

H1: ≠ $65,000 per year

with an

Two-tailed test

Note that the total area in the normal distribution is 1.0000.

10-13

One way to determine the location of the rejection region is to look at the direction the inequality sign in the alternate hypothesis is pointing. A test is one-tailed when the alternate hypothesis states a direction, either “less than” (left-tailed) or “greater than” (right-tailed). In the case of “not equal to,” the mean could be either “greater than” or “less than”, so it is a two-tailed test. In a two-tailed test, divided equally into the two tails of the sampling distribution, here .05 ÷ 2 = .025 in each tail.

Two-Tailed Test Example, Known

Jamestown Steel Company manufactures and assembles desks and other office equipment at several plants in New York State. At the Fredonia plant, the weekly production of the Model A325 desk follows a normal distribution with a mean of 200 and a standard deviation of 16. New production methods have been introduced and the vice president of manufacturing would like to investigate whether there has been a change in weekly production of the Model A325. Is the mean number of desks produced different from 200 at the .01 significance level?

Step 1: State the null hypothesis and alternate hypothesis.

H0: = 200 desks

H1: ≠ 200 desks

Step 2: Select the level of significance. Here = .01

Step 3: Select the test statistic. In this example, we’ll use z

10-14

This is a two-tailed test because the alternate hypothesis does not state a direction. The level of significance, .01, is given. When testing a hypothesis about a population mean, if the population follows the normal distribution and the population standard deviation is known, the test statistic is z and is determined from formula 10-1.

Two-Tailed Test Example, Known (2 of 3)

Step 4: Formulate the decision rule by first determining the critical values of z.

Decision Rule: If the computed value of z is not between −2.576 and 2.576, reject the null hypothesis. If z falls between −2.576 and 2.576, do not reject the null hypothesis.

10-15

Since this is a two-tailed test, the level of significance, .01, is divided in two. So .01 ÷ 2 = .005 and .005 is placed in each tail and becomes the region of rejection. To find the critical value, using the Student’s t distribution table in Appendix B.5, move to the top margin called Level of Significance for Two-Tailed Tests, Then find the column for 0.01 and move to the last row labeled infinity, . The z value in this cell is 2.576 so the critical values are ± 2.576.

Two-Tailed Test Example, Known (3 of 3)

Step 5: Take sample, compute the test statistic, make decision.

The mean number of desks produced last year (50 weeks because the plant was shut down 2 weeks for vacation) is 203.5. The standard deviation of the population is 16 desks per week. Compute z with formula 10-1.

= = 1.547

Decision: Because 1.547 does not fall in the rejection region, we decide not to reject H0.

Step 6: Interpret the result.

We did not reject the null hypothesis, so we have failed to show that the population mean has changed from 200 per week.

10-16

The sample information fails to indicate that the new production methods resulted in a change in the 200-desks-per-week production rate. However, we did not prove the assembly rate is still 200 per week; we failed to disprove it, which is not the same thing as proving it to be true.

One-Tailed Test

Before:

A two-tailed test

H0: = 200 desks

H1: ≠ 200 desks

Now:

A one-tailed test

H0: ≤ 200 desks

H1: > 200 desks

10-17

Suppose instead of wanting to know if there had been a change in the mean number of desks assembled, the vice president wanted to know if there had been an increase in the number of units assembled. Can we conclude, because of the improved production methods, that the mean number of desks assembled in the last 50 weeks was more than 200? Use .

The critical values for a one-tailed test are different from a two-tailed test at the same significance level. In the two-tailed test, we split the significance level in half and put half in the lower tail and half in the upper tail. In a one-tailed test, we put all the rejection region in one tail. Using Appendix B.5 again, move to the top heading called “Level of Significance, select the column with = .01, and move to the last row, which is labeled z value is 2.326.

The p-Value in Hypothesis Testing

Compare the p-value with the level of significance,

If the p-value is smaller than the significance level, reject H0

If the p-value is larger than , H0 is not rejected

A p-value not only results in a decision about H0, but gives additional insight about the strength of that decision

p-VALUE The probability of observing a sample value as extreme as, or more extreme than the value observed, given that the null hypothesis is true.

10-18

A p-value is the probability that the value of the test statistic is as extreme as the value computed, when the null hypothesis is true. A very small p-value, such as .001, indicates that there is little likelihood that H0 is true. On the other hand, a p-value of .2033 means that H0 is not rejected, and there is little likelihood that it is false.

Finding a p-Value

In the previous example about desk production, the computed z was 1.547 and H0 was not rejected

Round the computed z-value to two decimal places, 1.55

Using the z-table, find the probability of finding a z-value of 1.55 or more by .5000 − .4394 = .0606

Since this is a two-tailed test 2(.0606) = .1212

In this chart, we can easily compare the p-value with the level of significance

10-19

A p-value is the probability that the value of the test statistic is as extreme as the value computed, when the null hypothesis is true. The probability of finding a sample mean greater than 203.5 when the population mean is 200 is .0606. The two-tailed p-value is 2(.0606) = .1212. Thus, the p-value of .1212 is greater than the significance level of .01 so H0 is not rejected.

Hypothesis Testing, Unknown

When testing a hypothesis about a population mean

The major characteristics of the t distribution are

It is a continuous distribution

It is bell-shaped and symmetrical

There is a family of t distributions, depending on the number of degrees of freedom

It is flatter, or more spread out, than the standard normal distribution

10-20

If the population standard deviation is not known, s is substituted for . The test statistic is the t distribution, and its value is determined with formula 10-2.

Hypothesis Testing, Unknown Example

The Myrtle Beach International Airport provides a cell phone parking lot where people can wait for a message to pick up arriving passengers. To decide if the cell phone lot has enough parking places, the manager of airport parking needs to know if the mean time in the lot is more than 15 minutes. A sample of 12 recent customers showed they were in the lot the following lengths of time, in minutes (see below).

At the .05 significance level, is it reasonable to conclude that the mean time in the lot is more than 15 minutes?

Step 1: State the null hypothesis and the alternate hypothesis

H0: μ ≤ 15

H1: μ > 15

10-21

We use the six-step hypothesis testing procedure. This is a one-tailed test, a left-tailed test.

Hypothesis Testing, Unknown Example (2 of 3)

Step 2: Select the level of significance; we will use .05

Step 3: Select the test statistic; we will use t

Step 4: Formulate the decision rule; reject H0 if t is less than 1.796

10-22

Using Appendix B.5, this is a one-tailed test, so find the row at the top labeled, Level of Significance for One-Tailed Test,

Because −1.82 lies in the region to the right of the critical value of −2.485, the null hypothesis is not rejected at the .01 significance level. We have not disproved the null hypothesis.

Hypothesis Testing, Unknown Example (3 of 3)

10-23

Step 5: Take sample, make decision

t = = = 2.818

The test statistic of 8.818 is greater than our critical value of 1.796. Therefore, our decision is: Do not reject H0

Step 6: Interpret the result; The test results do not allow the claims manager to conclude the cost-cutting measures have been effective.

Type I and Type II Errors

Type I error occurs when a true null hypothesis is rejected

The probability of making a Type I error is equal to the level of significance,

A Type II error occurs when a false null hypothesis is not rejected

The probability of making a Type II error is designated,

The likelihood of a Type II error must be calculated comparing the hypothesized distribution to an alternate distribution based on sample results and can be calculated with this formula

10-24

There are two types of errors that can occur in a test of hypothesis, Type I (we choose) and Type II (we calculate). Formula 10-3 is used to calculate the probability of making a Type II error and finds the number of standard errors (the z value) between a critical value (designated x bar sub c) and the sample mean that we assume to be the true population mean (designated 1).

Type II Error Example

Western Wire Products purchases steel bars to make cotter pins. Past experience indicates that the mean tensile strength of all incoming shipments is 10,000 psi and that the standard deviation is 400 psi. To monitor the quality of the cotter pins, samples of 100 pins are randomly selected and tested for their strength. In our hypothesis testing procedure the hypotheses are:

H0: μ = 10,000

H1: μ ≠ 10,000

Using a 0.05 significance level, accept the shipment if the sample mean strength falls between the critical values 9.922 psi and 10.078 psi. If the sample mean does not fall between the critical values, we conclude the shipment does not meet the quality standard.

10-25

The graph indicates that the probability of a Type II error is .2912. We can also use Formula 10-3 to calculate the probability of making a Type II error and will get the same result. The number of standard errors between the mean of the incoming lot (9,900) and critical value (9,922) is computed with formula 10-3 (see previous slide); where x bar sub c represents the critical value (9,922) and 1 represents the assumed population mean (9,900). Recall that = 400 and n = 100, resulting in a z = .55. The area under the curve with a z value of .55 is .2088 so the area beyond this is .5000-.2088 = .2912, so .2912 is the probability of making a Type II error.

Type II Error Example (2 of 2)

10-26

The sample mean, 9.900 psi, is not within the specified range. To calculate the probability of a Type II error, assume the sample mean is the true mean (see graph B). Determine the probability of the sample mean falling between 9.900 and 9.922. Then subtract this probability from .5000 to arrive at the probability of making a Type II error, .2912

The number of standard errors between the mean of the incoming lot (9,900) and critical value (9,922) is computed with formula 10-3 (see previous slide); where x bar sub c represents the critical value (9,922) and 1 represents the assumed population mean (9,900). Recall that = 400 and n = 100, resulting in a z = .55. The area under the curve with a z value of .55 is .2088 so the area beyond this is .5000-.2088 = .2912, so .2912 is the probability of making a Type II error.

Graph A. Given that the population mean is 10,000 psi, designated μ0, with a standard deviation of 400, the distribution shows the regions where the hypothesis is rejected and where it is not rejected, that is, whether the shipment meets the quality standard for tensile strength.

Graph B to determine the probability of a Type II error, β. From the quality standards we know that 9,922 psi is used to reject the null hypothesis. Any sample mean greater than 9,922 and less than 10,078 is accepted. If the distribution is really centered on 9,900 psi, it is possible to find sample means more than 9,922, and we would fail to reject the null hypothesis, μ = 10,000. This is the area in Graph B labeled “Probability of β” where the null hypothesis would not be rejected. The graph indicates that the probability of a Type II error is .2912.

Chapter 10 Practice Problems

10-27

Question 7

10-28

A recent national survey found that high school students watched an average (mean) of 6.8 movies per month with a population standard deviation of 1.8. The distribution of number of movies watched per month follows the normal distribution. A random sample of 36 college students revealed that the mean number of movies watched last month was 6.2. At the .05 significance level, can we conclude that college students watch fewer movies a month than high school students?

LO10-2,3,4

Question 13

10-29

The mean income per person in the United States is $60,000, and the distribution of incomes follows a normal distribution. A random sample of 10 residents of Wilmington, Delaware, had a mean of $70,000 with a standard deviation of $10,000. At the .05 level of significance, is that enough evidence to conclude that residents of Wilmington, Delaware, have more income than the national average?

LO10-2,3,6

Question 19

10-30

A Washington, D.C., “think tank” announces the typical teenager sent 67 text messages per day in 2017. To update that estimate, you phone a sample of 12 teenagers and ask them how many text messages they sent the previous day. Their responses were:

At the .05 level, can you conclude that the mean number is greater than 67? Compute the p-value and describe what it tells you.

LO10-2,3,5,6

Question 23

10-31

The management of KSmall Industries is considering a new method of assembling a computer. The current assembling method requires a mean time of 60 minutes with a standard deviation of 2.7 minutes. Using the new method, the mean assembly time for a random sample of 24 computers was 58 minutes.

Using the .10 level of significance, can we conclude that the assembly time using the new method is faster?

What is the probability of a Type II error?

LO10-27