Topic:Two Tailed Hypothesis Testing

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Contents 1 Topic Highlights 2 Introduction and Motivation 3 Learning Activities 4 Learning Objectives 5 Topic Notes: What is Hypothesis Testing Anyway? 5.1 Example 1 5.2 Example 2 5.3 Significance level 5.4 Step 1: Define the hypothesis 5.5 Step 2: Sketch the situation 5.6 Step 3: Compute the test statistic (z-score) 5.7 Step 4: Determining whether to reject 5.8 Step 5: Give a conclusion 6 Topic Notes: The Hypothesis Testing Steps 7 Topic Notes: Do I really Get It? 7.1 Example 3 8 Practice Problems 8.1 Practice Problem 1 8.2 Practice Problem 2 (ES) 8.3 Practice Problem 3 9 Footnote

Topic Highlights

(What you will learn)

• What it means to test a hypothesis
• Why you would do it
• How to do it for the two tailed case
  

Introduction and Motivation

(Why learn it)

In this topic we're still talking about inferential statistics. In fact, for the rest of the course we will be talking about cases where the parameters and are unknown and have to be estimated by sample statistics and s.

In the topic More on Confidence Intervals, we saw that a confidence interval can be used to represent the precision with which estimates .

In this topic we will take this concept further by looking at ways to use the sample mean to test claims made about the population mean.

More generally, hypothesis testing has many business applications. One that jumps to mind is the testing required on information collected in a marketing research survey. More on this later.

Learning Activities

(How the levels of understanding will be gained)

Learning activities for this topic

Type	Name	Direction
Lecture and discussion	What is hypothesis testing anyway? (below in this page) The hypothesis testing steps (also below) Do I really get it? (also below)	Instructor-directed
In-class worksheet	Worksheet - Two-tailed hypothesis testing (print this and bring it to class)
Reading	Section 8.4 of Bowerman et al. or: Section 8.1 of Kvanli et al.	Self-directed
Personal activities	e.g. Reflection and review of the learning objectives below	Self-directed

Learning Objectives

(Levels of understanding to be gained)

Learning objectives for this topic

Level of Understanding	Objective(s)
Very best	Can I do part c of Example 3?
Highly satisfactory	Can I do part b of Example 3 Can I solve Practice Problem 2? Can I solve Practice Problem 1?
Satisfactory	Can I follow the extensive example we did after Example 2?
Maybe just enough to pass	Do I understand the basic concept of hypothesis testing, as described in Example 1 and Example 2? Do I have the steps in the Hypothesis Testing Process on my equation sheet?

Topic Notes: What is Hypothesis Testing Anyway?

As mentioned above in the introduction, hypothesis testing plays an important role in many business applications, most notably in marketing research.

The most important first step you can take is understanding the basic concept. And it's not all that difficult. Let's take a look using an example, which we will use throughout this section of the notes.

Example 1

An advertisement put out recently by a manufacturer of Compact Fluorescent Light (CFL) bulbs indicates that the average life time of their new product is 10000 hours.

Imagine that you run the marketing division of competing company and your boss comes in demanding you find out whether the claim of the first company is correct. After all, your bulbs don't last as long. How would you approach evaluating the claim that = 10000 hours?

Solution
Along the lines of our discussion in the topic An Overview of Confidence Intervals, you should take a sample of the life times of the company's light bulbs, e.g.: The concept is that the sample mean  will provide some information about the true mean:

Example 2

Okay, now suppose:

• You collect a sample, say n=100 bulbs
  
• You test the life times of each (wait approximately 10000 hours!)
  
• You compute the sample mean to find that = 9930
• You also compute the sample standard deviation to find that s = 400 hours

What then? How does this help you test their claim that the true mean = 10000 hours?

This is where the concept of hypothesis testing comes in. The idea is that we're going to define a hypothesis and test it against some condition. Let's start by looking at what that condition is.

Significance level

The condition for a hypothesis test is often defined by something called a significance level. For now you should know the following:

• It is the probability of rejecting our hypothesis when the hypothesis is in fact true, e.g. probability that a device indicates you have a disease when in fact you do not
• It is represented by the symbol
• It is typically in the range: = 0.01 to 0.1

On the standard normal curve,  looks like the following, which explains why it is called two-tailed hypothesis testing:

The green areas are known as the rejection region.

The manager is usually responsible for specifying a significance level based on the issue at hand. We will look more at this in a future topic.

Step 1: Define the hypothesis

The primary, or null, hypothesis represented as H₀. It is a statement concerning the population parameter that the researcher wishes to discredit.

In the case of population means, the null hypothesis takes the form:

H₀: = ₀

where ₀ is the value of the mean to be tested.

In the case described in Example 1, the null hypothesis is:

H₀: = 10000

We also usually define an alternative hypothesis, H_a, which is the complementary statement to H₀. In Example 1, this is:

H_a:   10000

When solving problems of this type, you need to write both the null and alternative hypothesis in your solution.

Be clear however, that the task of hypothesis testing is to either reject H₀ or fail to reject H₀.

Step 2: Sketch the situation

The next step is to sketch the situation, including the rejection region and the z-scores. Let's do this assuming we've been told to use a significance level = 0.100.

First, draw and label the rejection regions:

Then add the z-scores, which you should know because you can figure out that the area between 0 and z is 0.45 (and because of all the z-score values we summarized for your equation sheet):

Computationally, the goal of the test is to find the z-score corresponding to , for which we use the symbol z*, and looking to see whether it falls in one of the rejection regions:

Let's do this now to find out.

Step 3: Compute the test statistic (z-score)

Finding the test statistic z* just means finding the z-score for . You should be familiar enough now to recognize that the following applies for this:

Because we don't know the population standard devation , we'll approximate the above equation with the following:

Plugging in the numbers from Example 2 we get:

z* = (9930 - 10000) / (400/sqrt(100)) = -1.75

Step 4: Determining whether to reject

So, you can add the following to your sketch, showing that because -1.75 is less than -1.645, z* does fall in the lower rejection region as follows:

This means that the null hypothesis should be rejected. You could write something like:

• I reject H₀ because -1.75 is less than -1.645

Step 5: Give a conclusion

Now just give a succint conclusion, e.g.:

• Based on a sample of 100 and a significance level of 0.10, the average life time of the CFL bulbs is not 10000.

Notes:

• If you reject H₀, you can say that there is a difference (as we did above)
• But if you don't reject H₀, you should NEVER say that you proved there is no difference - you didn't prove that! - you can only say that we cannot reject their claim
  

That's it!

In the case presented in Example 1, you could now tell your boss that the claim made by the competing company is not supported by your data at the 0.10 significance level.

Topic Notes: The Hypothesis Testing Steps

Now that you've been through it once, let's summarize the basic steps:

1. Define the hypotheses, i.e.:

H₀: = some number

H_a: that number

2. Sketch the situation, e g.:

3. Compute the test statistic, i.e. using:

4. Determine whether to reject

5. Give a conclusion

These steps should be on your equation sheet.

You should do each of these every time you are asked a two-tailed test of the mean.

Topic Notes: Do I really Get It?

A good test of whether you really understand is to compare the hypothesis testing situation to the confidence interval situation.

Let's do that with another example.

Example 3

a) Compute the confidence interval for the level of confidence corresponding to the significance level =0.10 in Example 2. Use the same values of n = 100, = 9930, s = 400.

b) Does the confidence interval support your hypothesis testing conclusion obtained above?

c) Sketch both situations together

Solution to a)

From the topic More on Confidence Intervals, you should recognize that: CI = [ - t (), + t () ] You can then find df = n-1 = 99 When you look at Table A.5 you will see that there is no entry for df = 99, so you will have to make use of the fact that the normal curve is very close to the t curve for n > 30. This allows you to use: CI = [ - z (), + z () ] A significance level of 0.10 corresponds to a confidence level of 0.90 or 90%. Convince yourself of this using the following sketch: z is therefore = -1.645. Filling in the values, you get: CI = [ 9930 - 1.645*400/sqrt(100), 9930 + 1.645*400/sqrt(100) ] = [ 9864.21, 9995.79 ]

Solution to b)
We found that: CI = [ 9864.21, 9995.79 ] This is range in which we are 90% confident the true mean will fall. The company is claiming that 0 = 10000, which does not call in this range. So, yes, the confidence interval supports the conclusion I came to with hypothesis testing.

Solution to c)
The following sketch shows both situations. Notice that the complementary nature of the two: In the first case, we are testing if 0 is outside the confidence interval (which it is here) In the second case we are testing whether is in one of the rejection regions (which it is here)

Practice Problems

Practice Problem 1

Do exercise 8.7 on page 317 of Kvanli et al. The solution is in the back of the book.

For those who don't have this textbook, this problem is very similar to the examples seen above.

Practice Problem 2 (ES)

You work for a manufacturer of Compact Fluorescent Light (CFL) bulbs that has been selling standard CFL bulbs for many years. The average life time of its standard bulbs is 10000 hours.

The company has recently come out with a new type of bulb that it believes last longer on average. You take a sample of 65 of the new bulbs and find that the average life time is 10100 hours and the standard deviation is 450 hours.

a) Using a 5% level of significance, determine whether there is a significant difference in the average life time of the new bulbs

b) Find the confidence interval for the average lifetime for a 95% confidence level

c) Do the hypothesis test and confidence interval lead you to the same conclusion?

Solution to a)

Step 1: Define the hypotheses: H0: = 10000 Ha: 10000 Step 2: Sketch the situation: Step 3: Compute the test statistic: = (10100 - 10000) / (450/sqrt(65)) = 1.7916 Step 4: Determine whether to reject: Since 1.7916 is less than 1.96, we don't reject H0 Step 5: Give a conclusion Based on the sample of 65 of the new bulbs and a significance level of 5%, the average life time of the new bulbs is not significantly different from 10000

Solution to b)
Do this to find: CI = [ 9990.6034,  10209.3966 ]

Solution to c)
For the confidence interval, we would notice that 0 = 10000 is in the confidence interval, i.e. between 9990.6034 and 10209.3966, so we would not reject. So yes, both methods lead us to the same conclusion.

Practice Problem 3

A donut factory has just completed contract negotiations with the union representing its workers. The company claims that the workers' average salary is now $400 per week. In a survey of 28 workers, the average wage was found to be $378.86 and the standard deviation was $49.20. The wages are known to be normally distributed.

Is the company's claim incorrect? Use a level of significance of 2%.

Solution

Given: = 378.86 s = 49.20 n = 28[1] Step 1: Hypotheses: H0: = 0 = 400 Ha: 400 Step 2: Sketch the situation Here, you should show that the area on the outsides is 0.01 and that z=2.33 Step 3: Compute the test statistic: z* = -2.274 Step 4: Determine whether to reject: Do not reject H0 since -2.33 < -2.274 Step 5: Draw a conclusion: Based on the sample of 28 and a significance level of 2%, the average salary is not significantly different from $400. Evidence was not found that the company's claim is incorrect.

Footnote

1. ↑ In a later topic, we will learn that n=28 corresponds to the small sample case, which is solved slightly differently than this (using the t-distribution). Don't worry about this for now, and solve this using the z-distribution.