Data Analysis & Business Intelligence

Question:

Briefly define the following concepts and provide numerical examples:

The Null hypothesis
Type I error
Type II error
Test statistic
Critical values
p-value

Note:

1. Define the words in the own words. Do not quote from the textbook.

2. Need to write at least 2 paragraphs

3. Need to include the information from the textbook as the reference.

4. Need to include at least 1 peer reviewed article as the reference.

5. Please find the textbook and related power point in the attachment

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

Possible Error in Hypothesis Testing

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

Step 3 of the Process

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

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

Hypothesis Testing (3 of 3)

10-12

One-Tailed and Two-Tailed Tests

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

10-13

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?

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.

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

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

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

The p-Value in Hypothesis Testing

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.

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

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.

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

10-23

Step 5: Take sample, make decision

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

10-24

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

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

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

Data Analysis & Business Intelligence

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